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Department of Philosophy, WITS
PHIL 3009 SYMBOLIC LOGIC COURSE PACK Lecturer: MURALI RAMACHANDRAN
INTRODUCTION: DEDUCTIVE AND INDUCTIVE ARGUMENTS
An argument is collection of propositions, one of which—the conclusion—is presented as
being supported (backed-up) by the others—the premises.
Deductive validity
A (deductively) valid argument is one where it is impossible for the premises to be true
without the conclusion also being true; i.e. where it would be contradictory to affirm the
premises but deny the conclusion. When an argument is valid, we say the premises entail the
conclusion.
Any argument that is not valid is said to be invalid.
Examples:
Argument A
Premises: (1) Tracy is a vegetarian or a smoker.
(2) If she is a smoker, she lied to her dad.
(3) Tracy has never lied to her dad.
Conclusion: (4) So, Tracy is a vegetarian. Valid
Argument B
Premises: (1) Vince is a nerd if Brian is.
(2) Brian isn‟t a nerd unless he supports United.
(3) Brian doesn‟t support United.
Conclusion: (4) Hence, Vince is a nerd. Invalid
Inductive strength
An inductively strong argument is one whose premises would provide positive support for
the conclusion if they were true—the premises render the conclusion more likely.
Argument C
Premises: (1) Kev is an animal-rights activist and
Beth is a butcher.
Conclusion: (2) So, if either is a vegetarian, Kev is.
Inductively strong
Argument D
Premises: (1) Malcolm is an accountant.
(2) Beth was nearly bored to death by some
accountants at a party once.
Conclusion: (3) So, Beth will (probably) find Malcolm
boring too. Inductively weak
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 2
Differences between validity and inductive strength
An argument is (deductively) valid or inductively strong NOT in virtue of the truth of its
premises or conclusion, but in virtue of how the premises are related to the conclusion, i.e.
what degree of support they would provide for the conclusion. So, e.g. an argument with false
premises may still be valid or inductively strong—consider (1) of argument A.
Whether an argument is valid or not is knowable a priori. But inductive strength depends
on background knowledge. The force of argument C, for example, stems from our knowledge
of butchers and animal-rights activists. This suggests that inductive strength is not generally
known a priori.
Adding further premises to a valid argument cannot make it invalid. In contrast, whether an
argument is inductively strong or weak is defeasible: the „strength‟ of the argument may be
altered by adding further premises. E.g. suppose one added the further premise that Beth
comes from a long line of vegetarian butchers in argument C. The conclusion does not seem as
plausible as before. Hence, inductive strength, unlike validity, admits of degrees.
Logical form
Certain valid arguments are valid purely in virtue of their „shape‟. For example, this is the
shape (form) of Argument A:
Premises: V or S
If S, then L
Not-L
Conclusion: V
Any argument of the same form is guaranteed to be valid. Thus, this is called a valid logical
form.
Formal logic is the study of logical form and it is this we shall be concerned with hereafter in
this course, since it provides a fundamental and relatively easy starting point.
Exercises
1. For each of the following arguments state whether it is deductively valid, inductively
strong, or inductively weak giving your reasons.
(a) Lucy won‟t find Logic difficult if she‟s good at Maths, and she got top marks in her
Maths test. So, she‟s not going to find Logic difficult.
(b) Raju‟s parents are strict vegetarians. Now, he wouldn‟t eat meat unless they did. So, he
too must be a vegetarian.
(c) It is highly likely that Sabrina has musical talent. For, the indications are that musical
talent is largely inherited and Sabrina‟s sisters, brothers, parents and grandparents are
all excellent musicians.
(d) Willard always leaves food on his plate when he is ill. He must have been ill last
Monday: he came in here that day (after being fired), had one mouthful of stew and
promptly left!
2. Give an example of strong inductive argument. Now add a premise to make it weak
inductive argument.
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 3
TRUTH-FUNCTIONALITY
Sentence connectives
A sentence connective operates on sentences to form compound (or complex) sentences.
E.g. „… and …‟ operates on pairs of sentences to form compound sentences of the form
„A and B‟; „If … then …‟ operates on pairs of sentences to form sentences of the form „If A
then B‟.
NOTE on negation: „It is not the case that …‟ is treated as a sentence connective even
though it just operates on single sentences. (It is called a one-place connective).
Truth-functional connectives
A sentence connective is truth-functional if the truth-value of any sentence it gives rise to
is wholly determined by the truth-values of the simpler sentences it combines. For example,
conjunction („and‟, &), disjunction („or‟, ) and negation („not‟, ) are all truth-functional
connectives. They are fully defined in the following truth-table:
A
B
A and B
A & B
A or B
A B
not-A
A
T T T T F T F F T F
F T F T T
F F F F T
Explanation
The „A‟ and „B‟ stand for propositions (or declarative sentences—sentences that can be true
or false). „T‟ and „F‟ stand for the truth-values true and false, respectively.
It is assumed that a proposition must be true or false: it can’t be neither.
Now, consider e.g. the truth table for „‟ („or‟). The first row indicates that when A and B are
both true, „A B‟ is true; the second row indicates that when A is true and B is false, „A B‟ is
true, and so on. The only circumstances in which „A B‟ is false is when both A and B are false.
A non-truth-functional connective
„A because B‟ is an example of a sentence connective that is not truth-functional, i.e. its
truth-value is not determined solely by the truth-values of A and B. It can be true when A and
B are both true, as in e.g. „The fire spread because the doors were not closed‟; but it may also
be false when A and B are both true, as in e.g. „The fire spread because Oswald shot
Kennedy‟.
Complex sentences
A simple (or atomic) sentence is one that does not have any connectives (e.g. , , &) in
it. A complex sentence is one that does.
A connective may operate on complex sentences to generate even more complex
sentences, e.g. „(A B) & C‟ is also a sentence.
The brackets (parentheses), „(‟ and „)‟ are required to avoid ambiguity. To give a simple
example, compare the English sentences “Arnie didn‟t go to the party but Belinda did” and
“Arnie and Belinda didn‟t both go to the party”. The first sentence affirms that Belinda didn‟t
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 4
go to the party, but the second would be true if Belinda went and Arnie didn‟t. The sentences
would be expressed in our logical language as: „ A & B‟ and „ (A & B)‟, respectively.
Working out the truth-value of complex sentences
Great news! We can work out the truth-value of any complex sentence if we know the truth-
values of the simple sentences it contains.
Example: What is the truth-value of „(A B) & C‟ if B and C are true and A is false?
We begin by wrting down the truth-values of the simple sentences
(and numbering the steps below—this lets me know the order in
which you have done things). We have to work out the truth-value
of the sentence in brackets first. We begin by putting the truth-
value of „ B‟ under the „‟, then we can work out the truth-value of „A B‟:
Since „A‟ is false, and „ B‟ is false, the truth-value of „A B‟ is
the truth-value of „false false‟, which according to the truth table
of „‟ is false.
The truth-value of the whole sentence „(A B) & C‟ is the value under the main connective, in
this case, the „&‟. The main connective of a sentence is the one with the fewest brackets around it.
(If there is a tie, it is the one that is not „‟.)
The truth-value under the „&‟ is determined by the truth-values of
„(A B)‟ and „C‟; i.e. we want the value of „false & true‟;
according to the truth table definition of „&‟, this is false. So, the
truth-value of the whole sentence is FALSE.
Another example. Work out the truth value of (A & B) if A and B are both true.
Again, we begin by putting the values of the simple sentences below
them (steps 1 and 2). Next comes the „‟ inside the bracket, then the
„&‟, then finally the „‟ outside the bracket, which is the main
connective.
NOTE: step 5 is determined by looking at step 4, not step 1! This is because the truth value of
the bracketed sentence „(A & B)‟ is the value under the „&‟.
Exercises
If A, B and C are true and D and E are false, what are the truth-values of the following
sentences?
SENTENCE T/F
a. C
b. (B & A)
c. (( D & C) E)
d. (A D) & (A & D)
e. (E ( A & C))
(A B) & C F T T
1 2 3
(A B) & C F F F T T
1 5 4 2 3
(A B) & C
F F F T F T
1 5 4 2 6 3
(A & B) T T F F T
5 1 4 3 2
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 5
Working backwards
In some cases, we can work „backwards‟ to find the truth-value of atomic sentences from
the truth-value of complex sentences.
Example: Suppose „ C (S P)‟ is false. What must the values of C, S and P be?
Step 1 is false. The truth table for „‟ shows that „‟ is false only
when both disjuncts are also false; hence steps 2 and 3. If „ C‟ is
false, C must be true; thus, step 4. Finally, again using the table for
„‟, we see that it is false only when both sides are false; hence
steps 5 and 6. So, the whole sentence is false only if C is true and S and P are false.
Exercises
1. If „P & (Q & R)‟ is true, and R is true, what are the truth-values of P and Q?
2. If „ (A & (B A))‟ is false, what are the values A and B?
Another logical connective: ‘’
The truth table below introduces and defines a new connective, „‟:
A B A & B A B A A B
T T T T F T
T F F T F F
F T F T T T
F F F F T T
An „arrow‟ statement, „A B‟, is FALSE only when A is true and B is false; it is TRUE
in all other circumstances. (We‟ll consider what „A B‟ means, what sorts of sentences of
ordinary language it is used to translate or represent shortly.)
Exercises
1. What are the truth values of the following sentences if P is true, and Q and R are false:
1a. P (Q R)
1b. (( Q & (Q P))
2. What are the truth values of P, Q and R if the following sentence is false:
( P Q) (P R)
Using truth tables to determine logical equivalence, tautologies, etc.
Two sentences are logically equivalent if they cannot differ in truth value: any situation
where one is true is a situation where the other is true also. We can establish whether a pair of
SL-sentences are logically equivalent by constructing truth tables. Two SL-sentences are
logically equivalent if the values under their main connectives in the truth table are exactly
the same.
A logically true sentence (a tautology) has „T‟ under its main connective in every row. A
logically false (or self-contradictory) sentence has „F‟ under its main connective in every row.
C (S P) F T F F F F
2 4 1 5 3 6
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 6
A sentence that is neither logically true nor logically false s said to be logically indeterminate
(or contingent).
We can also establish whether a set of SL-sentences is consistent, i.e. whether the
sentences could all be true simultaneously, by constructing a truth table. The set is consistent if
there is a row in the truth table which has a T under the main connective of every sentence in
the set. (It is inconsistent otherwise.)
Examples:
Which of the following pairs of sentences are logically equivalent?
(1a) E E; E (1b) K L; K & ( K L)
(1a) The first section of the table depicts the possible combination
of truth values of the atomic sentences involved. The truth values
under the main connectives match each other. So, the sentences in
(1a) are logically equivalent.
(1b) Again, the first section of the table depicts the
possible combination of truth values of the atomic
sentences. Notice the order in which they are
written: it is important to have such a system to
ensure that all the possible combinations are
covered. These sentences are NOT logically
equivalent. For example, when K is false and L is
true (see third row), the first sentence is true but the second is false.
More examples: Construct truth tables to determine whether the following sentences are
logically true, logically false, or logically indeterminate:
(2a) L (L L)
(2b) (L P) L
L P L (L L) (L P) L
T T
T F
F T
F F
Complete the above truth table and answer the question!
Exercises
1. Construct a full truth table to establish whether the following sentence is logically true,
logically false, or logically indeterminate:
(A & B) A
E E E E
T
F
T F F
F T T
F
T
K L K L K & ( K L)
T T
T F
F T
F F
T
T
T
F
4
T F T
F F F *
F T T *
F T T
3 1 2
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 7
2. Construct a full truth table to establish whether the following pair of sentences are logically
equivalent or not:
L M; M L
Sentential Logic (SL) It is time to introduce you to a simple logical language, Sentential Logic (also referred to as
propositional logic), which consists of atomic sentences and rules for combining sentences
with truth-functional sentence connectives. In such a language it is easy to evaluate the truth
values of complex sentences and to establish the validity or invalidity of arguments. [Our
ultimate goal is to evaluate ordinary-language arguments by way of representing them in
Sentential Logic.]
Vocabulary Atomic sentences: A, B, C, ..., Z, A1, ..., Z1, A2, ..., Z2, ...
Truth-functional connectives:
NAME SYMBOL INTERPRETATION/COMMENTS
NEGATION , ( A) It is not the case that A
CONJUNCTION &, (A & B) A and B (A and B are called conjuncts)
DISJUNCTION (A B) A or B (A and B are called disjuncts)
IMPLICATION , (A B) If A, then B (This is also called a material
conditional; A is called the antecedent and B
the consequent of the conditional)
EQUIVALENCE , (A B) A if and only if B (This is also called a material
bi-conditional)
Punctuation marks: Left and right brackets (parentheses): ( , )
Formation Rules
What counts as an SL-sentence?
(a) Every atomic sentence is an SL-sentence.
And for any SL-sentences A and B, (b)-(f) are also SL-sentences:
(b) A; (c) (A & B); (d) (A B); (e) (A B); (f) (A B)
Bracket convention:
The outermost brackets of a sentence may be dropped. So, e.g. „A (B & C)‟ in place of
„(A (B & C))‟ is permissible, but „ A B‟ in place of „ (A B)‟ is not permissible.
Example: Explain, using the formation rules above, why „ B ( B & A)‟ is an SL-sentence.
i. By clause (a), A and B are SL-sentences.
ii. By clause (b), B is an SL-sentence.
iii. By clause (c), ( B & A) is an SL-sentence.
iv. By clause (e), ( B ( B & A)) is an SL-sentence.
v. Finally, by the bracket convention, B ( B & A) is an SL-sentence.
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 8
Truth-table definition of the connectives
As before, the connectives of SL are fully defined by truth tables. In addition to the connectives
introduced earlier, we have the bi-conditional, „‟:
A A & B A B A B A B
F T T T T T T T T T T T T T
T F T F F T T F T F F T F F
F F T F T T F T T F F T
F F F F F F F T F F T F
Sentential Logic: Translation Before we can evaluate ordinary-language arguments, we need to know how to translate
complex English sentences into SL. Here is an important distinction to bear in mind before we
consider some of the more common translation rules.
Distinction between what is said (asserted) and what is implied or merely conveyed
Our intuitions about translation can be misleading.We must distinguish between the literal
content of an utterance and what is implied by the utterance. Consider e.g. an utterance of the
sentence “I brushed my teeth and took off my specs”. The speaker implies that she brushed
her teeth before taking off her spectacles—but this is not part of what is literally said. For, it
would not be contradictory if she added, “… but not in that order”. The implication is thus
cancelable (the phrase is H.P. Grice‟s). What is literally affirmed is not cancelable in this
sense; e.g. it would be contradictory if she had instead added “… but I didn‟t brush my teeth”.
Her original utterance, “I brushed my teeth and took off my specs”, is literally true if, and only
if, she did both those things (but not necessarily in that order). Hence, the earlier truth-table
definition of “and” stands.
Likewise, if someone says, “Blair is either very rich or very clever”, she may imply that he
is not both, but this is not part of what is literally asserted: it would not be contradictory for
her to add, “In fact, he is both.”
Another example. Suppose someone says, “If I win the lottery, I will retire”. One might
well take the speaker to be implying that if she doesn’t win the lottery, she won’t retire. But
this implication is cancelable: she could have added, “Come to think of it, I might retire even if
I don‟t!”
Cancelability is a useful simple test for distinguishing what is literally said from what is
merely implied or conveyed.
Translation of ‘but’
Consider the statement, „Annie is poor but she is happy‟. The use of the word „but‟ implies
that the two conjuncts in some tension which each other: that Annie‟s being rich is somewhat
surprising or unexpected in light of her being poor. But, the statement is true so long as both
conjuncts are.
More generally, „A but B‟ has the same truth conditions as „A and B‟, and simply gets
translated into SL as „A & B‟.
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 9
Translation of If-statements
A statement of the form „If A, then B‟ is definitely false if A is true and B is false—e.g. „If
Annie goes to the party, Bill will‟ is definitely false if it turns out that Annie goes and Bill
doesn’t. Note that „A B‟ is also false when A is true and B is false.
One of the uses of „‟ in elementary logic is to translate If-statements, e.g. statements of
the form “If A, then B” and “A if B”.
SIMPLE RULE FOR TRANSLATING IF-STATEMENTS WITH „‟:
The If-part goes in front of the arrow.
o So, „If A then B‟ gets translated as „A B‟,
o whereas „A if B‟ gets translated as „B A‟
It may seem counterintuitive to count „If A, then B‟ as true when A is false. But think of
this translation as charitable way of accommodating the assumption that every sentence must
be either true or false.
Translation of ‘Only If’-statements
Now, ask yourself, when is „Annie will go to the party only if Bill does‟ definitely false?
— Answer: if it turns out that Annie goes and Bill doesn‟t.
More generally, a statement of the form „A only if B‟ is definitely false in the same
circumstances that „If A, B‟ is definitely false: i.e. when A is true and B is false.
So, „A only if B‟ gets translated into SL as „A B‟ as well.
So, as far as SL is concerned, „If Annie goes to the party, Bill will‟ and „Annie will go to
the party only if Bill does‟ say the same thing—at any rate, have the same truth conditions.
This will seem counterintuitive: surely the first implies that Annie‟s going would be responsible
for Bill‟s going, whereas the only-if statements implies that Annie‟s going would depend on
Bill‟s going? These casual implications are lost in the SL-translations.
Example. Let us take (1) as our candidate for translation:
(1) If Adam doesn‟t pass his driving test and Bianca does, he will get upset.
The first step is to assign a (capital) letter to each of the „atomic’ propositions—the simplest,
„non-negative‟ propositions—that the sentence involves. There are three atomic propositions
in this case:
A : Adam passes his driving test.
B : Bianca passes her driving test.
U : Adam will get upset.
Using these letters, we see that what sentence (1) affirms (getting ungrammatical awhile) is:
(1a) If not-A and B, then U.
Now, (1a) contains three connectives: „if ... then‟, „not‟ and „and‟. Which one is the main
connective? I.e., is the sentence fundamentally an „if ... then‟-sentence, a „not‟-sentence or an
„and‟-sentence? Clearly, it is the first. So the main connective of the SL-translation of (1a) will
be „‟; that is, the SL-translation will be of the form: *** U.
Next, what goes in place of ‘***’? The SL-translation of „not-A and B‟, which is: „ A & B‟;
and we need brackets around this, because „&‟ is not the main connective of the whole
sentence. So the SL-translation of (1) is:
(1SL) ( A & B) U
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 10
Translation Exercises 1
Using A, B and U to stand for the same propositions as above, symbolize the following
statements as sentences of SL:
a. Bianca passed her driving test but Adam didn‟t.
b. Neither Adam nor Bianca passed their driving test.
c. Adam will not get upset if he passes.
d. Adam will get upset only if he fails and she passes.
Translation of ‘Unless’-statements
The statement „The patient will die unless the doctor arrives soon‟ basically affirms that the
patient will die if the doctor does not arrive soon. More generally, „A unless B‟ should be read
as „A if not-B‟.
Given our simple rule for translating if-statements (that the if-bit goes in front of the
arrow), the SL-translation is „A if not-B‟ is „B A‟
SIMPLE RULE FOR TRANSLATING UNLESS WITH AN ARROW: JUST REPLACE IT WITH „IF-NOT‟
o So, e.g. „Unless A, B‟ should be translated as „If not-A, B‟, i.e. „A B‟
„A B‟ is actually equivalent to „A B‟—(check it!).
So, AN EVEN SIMPLER RULE: JUST TREAT „UNLESS‟ AS DISJUNCTION, „‟.
o So, e.g. both „A unless B‟ and „Unless A, B‟ can be translated as „A B‟
„But‟, „If‟, „Only if‟, „if and only if‟ and „Unless‟ are some of the common sentence
connectives in English; you are going to have to figure out how to best translate the less
common ones yourself!
Translation Exercises 2
Symbolize the following statements as sentences of SL, using the suggested abbreviations:
KEY
A : Anna went to the party. D : Dora went to the party.
B : Barry went to the movies. E : Anna enjoyed the party.
C : Cheryl got drunk at the party.
(a) Dora didn‟t go to the party.
(b) Anna went to the party but didn‟t enjoy it.
(c) If Anna went to the party, Barry went to the movies.
(d) Either Anna or Dora went to the party.
(e) Anna and Dora didn‟t both go the party.
(f) Cheryl got drunk at the party if Anna went to it.
(g) Anna went to the party only if Barry went to the movies.
(h) Dora went to the party so long as Anna did.
(i) Cheryl didn‟t get drunk at the party unless Anna or Dora went to it.
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 11
Using Truth Tables to Establish SL-Validity We are going to see how to establish whether an argument is valid or not using a complete
truth table. Later, we will consider a simple shortcut method that can save a lot of time.
Consider the following argument:
Sarah is in Paris if Dave is. She is in Cambridge if Dave is not in Paris. As we know,
she definitely is not in Cambridge. So, both Sarah and Dave are in Paris.
(S = Sarah is in Paris; D = Dave is in Paris; C = Sarah is in Cambridge)
The full truth table method
The first step is to translate the argument into Sentential Logic:
D S, D C, C |= S & D
(„|=‟ is the entailment sign: it separates the premises from the conclusion).
A B A & B A B A A B
T T T T F T
T F F T F F
F T F T T T
F F F F T T
The figure below shows a full truth table for the argument (using the truth table definitions
of the connectives above):
C D S D S D C C S & D
T T T T F T T F T
T T F F F T T F F
T F T T T T T F F
T F F T T T T F F
F T T T F T F T T *
F T F F F T F T F
F F T T T F F T F
F F F T T F F T F
Note: The left-hand columns list the sentence letters involved in alphabetical order. The
truth values in the columns below are arranged so that all the possible combinations are
exhausted. The conclusion of the argument appears in the right-most column.
The asterisks indicate the rows in which the ALL the premises are true. (In this case, there
is only one such row, but in other cases there may be many, or even none.)
The table shows that the argument is VALID: because on all the rows in which the
premises are true, the conclusion is also true.
An argument is only INVALID if there is at least one row in its truth table where ALL the
premises are true but the conclusion is false. (Note: if there are some or many rows with true
premises and true conclusion and just one row with true premises and false conclusion, the
argument is still INVALID.)
Note: if there is NO row with all the premises coming out true, then the argument is
VALID.
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 12
Exercises on full truth table method
Translate the following arguments into sentences of Sentential Logic and then construct full
truth tables to establish whether or not they are valid:
(1) Either Ernie has a cold or he enjoys blowing his nose. If the latter, he is weird. Hence,
since Ernie does not have a cold, we can conclude that he is weird.
(C = Ernie has a cold; E = Ernie enjoys blowing his nose; W = Ernie is weird)
[Translation of ‘since’. You can translate ‘Since A, B’ as ‘A B’; but, alternatively—
and this is what I would recommend you do generally: take the since-bit, ‘A’, to be an
additional premise and take the conclusion of the argument to simply be ‘B’.]
(2) Bianca won‟t buy a car unless she passes her test. But she‟ll pass only if her partner
does. So, Bianca will buy a car if her partner passes the test.
(B = Bianca will pass; P = Bianca‟s partner will pass; C = Bianca will buy a car)
(3) Lord Dweeb is not guilty if the butler is lying. But if Dweeb is guilty, he was being
blackmailed. So, either the butler is telling the truth or Dweeb was being blackmailed.
(G = Dweeb is guilty; B = Dweeb was being blackmailed; L = the butler is lying).
The shortcut method The strategy of the shortcut method is try and make the premises true and the conclusion
false by assigning truth values. If you are thereby forced into assiging contradictory truth
values somewhere along the line, you have shown the argument to be valid, because you have
shown it is impossible for the premises to be true and the conclusion false. But if can assign
the truth values without any resulting contradiction, you have shown the argument to be
invalid.
Let us see the method used with the earlier example.
Always begin by assigning the premises true
and the conclusion false and number the
assignments underneath.
Then see if any other assignments are forced
upon you. In this case, step 3 dictates that C is
false, so we can put that down (step 5).
Steps 2 and 5 force you to assign false to
„ D‟ (step 6). This means that D must be true
(steps 7 and 8).
Steps 7 and 1 force you to assign true to S
(steps 9 and 10).
D S, D C, C |= S & D T T T F
1 2 3 4
D S, D C, C |= S & D T T F T F
1 2 5 3 4
D S, D C, C |= S & D T T F T F T F T
7 1 6 2 5 3 4 8
D S, D C, C |= S & D T T T F T F T T F T
7 1 9 6 2 5 3 10 4 8
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 13
But now we have arrived at a contradiction: for steps 10, 4 and 8 cannot all be correct
(true & true should be true, not false). So, we have shown that it is impossible for the
premises to be true and the conclusion false: i.e. the argument is VALID.
Another example: Determine whether or not the following SL-argument is valid:
R S, S E, E |= R
Here are the various stages:
First, assign the premises true and the conclusion false.
(Notice that we assign „E‟ a truth value at step 3. It has no
connective but it is still a premise, so it must be assigned true.)
Next, since step 3 assigns true to E, we can write this down
under the E in the second premise (step 5). Steps 2 and 5 can
only be correct if S is true (step 6).
Now we can put true under S in the first premise (step 7).
We have R as false at step 4, so we can put this down under R
in the first premise (step 8).
We have finished assigning the values. There is no contradiction anywhere: F T should
be true, as the first premise has it; likewise, T T should be true, as the second premise has it;
and the assignments to E and R do not conflict with any of the earlier values. So, we have
shown it is possible for the premises of this argument to be true while the conclusion is false.
That is all that is required for the argument to be INVALID.
When an argument is invalid, you need to specify truth values of the atomic
sentences which result in true premises and false conclusion.
o So in the above example, final answer should be something like this: When R
is true and E and S are false, the premises are true but the conclusion is false,
so the argument in INVALID.
The usefulness of the shortcut method is clear when we have an argument with four or
more atomic sentences. Consider e.g.: G F, B F, B R |= R G
Steps 7 and 12 contradict step 1. So, it
is not possible for the premises to be true
while the conclusion is false. Hence, the
argument is VALID.
(A full truth table for the above argument, with 5 sentence letters, would require 32 rows! The
shortcut method makes the task easy.)
Sometimes, the initial assignment of values does not fully determine the truth values of all
the atomic sentences. Consider e.g.: (P W) (P & W) |= P
This is far as one can go from the initial assignment
(steps 1 and 2). Step 5 follows from step 3 because
both F T and F F are true.
R S, S E, E |= R T T T F
1 2 3 4
R S, S E, E |= R T T T T T F
1 6 2 5 3 4
R S, S E, E |= R F T T T T T T F
8 1 7 6 2 5 3 4
G F, B F, B R ╞═ R G
F T F T T T F T F T F F
7 1 12 10 2 11 9 3 8 5 4 6
(P W) (P & W) ╞═ P
F T T F F
3 5 1 4 2
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 14
But in this case, we have enough information to see that the argument is invalid. For, step
5 guarantees the correctness of step 1—because a disjunction (an „or‟ statement) is true if
either disjunct (i.e. either side) is true. So we need go no further. We have shown that it is
possible for the premises to be true and the conclusion to be false; namely, when „P‟ is false,
regardless of the value of „W‟. Hence, the argument is INVALID.
Exercises on the shortcut method:
1. Use the shortcut method to establish whether the following SL-arguments are valid or
invalid:
(a) J M, J ( J & M) |═ M (M & J)
(b) (M & J), J K |= K M
2. Translate the following arguments into Sentential Logic, and then determine whether
they are valid or invalid using the shortcut method:
2a If Blair resigns and there is an immediate general election, then Cameron will become
Prime Minister if the newspapers support him. Clearly, if Blair resigns, there will be an
immediate general election. And, equally clearly, the newspapers will support Cameron
if Blair resigns. So, unless Blair does not resign, Cameron will become Prime Minister.
(Use the following key: B = Blair will resign; C = Cameron will become PM;
E = There will be a general election; N = The newspapers will support Cameron)
2b Belinda is not guilty if Andrew‟s testimony is true. And if Belinda is not guilty,
Charlene lied. Since Charlene did not lie, Andrew‟s testimony is not true.
(B = Belinda is guilty; A = Andrew‟s testimony is true; C = Charlene lied)
2c. Brains‟ plan will work if and only if Lady Penelope‟s car does not break down and
Parker is sober. Her car won‟t break down unless Parker is drunk. But if he were
drunk, he would not drive. So, if Parker is driving, he is sober and Brains‟ plan will
work.
(W = Brain‟s plan will work; B = Lady Penelope‟s car will not break down;
S = Parker is sober; D = Parker will drive)
Making assumptions in the shortcut method.
In a few, exceptional cases, we arrive at situations where we are not forced to make
assignments but still do not have enough information to determine whether the argument is
valid or not. In such situations, we have to consider certain possible truth values in order to
get our answer.
Example: Determine whether or not the following SL-argument is valid:
P Q, P Q, Q P ╞═ P & Q
In this situation, no further
assignment is forced on us. But, we
know that „P‟ is either True or False.
If we can show that we get a
contradiction both on the assumption that it is true and on the assumption that it is false, then
P Q, P Q, Q P ╞═ P & Q
T T T F
1 2 3 4
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 15
we will have shown that it is not possible for the premises to be true and the conclusion to
befalse. So, we might set things out as follows.
Case 1: „P‟ is True.
Steps 9 and 8 contradict step 3. The
„@‟ beneath the final „P‟ signifies
that the assigned value is an
assumption. The number in
brackets under step 9 tells us
where the value was derived from. I could, for example, have determined the value under
the negation to be F, on the basis of steps 3 and 8. But in that case, step 9 would have
contradicted step 5.
The contradiction we have discovered tells us that the premises cannot be true while the
conclusion is false when „P‟ is True. Having ruled out this possibility, we now need to
consider the possibility that „P‟ is False.
Case 2: „P‟ is False.
Steps 7 and 9 are contradictory. In
other words, the premises cannot
be true while the conclusion is false
when „P‟ is False. Since „P‟ has to
be true or false, we have shown that it is impossible for the premises to be true while the
conclusion is false, period. Thus, the argument is VALID.
Another example: Determine whether or not the following SL-argument is valid:
A B, B A ╞═ A B.
First step: assign the premises True and the
conclusion False.
Unfortunately, nothing is forced on us.
So now we must consider different possibilities. Take „A‟, for instance. It has just two possible
truth values, True and False. We can check whether anything is forced on us by either
assumption.
Case 1: „A‟ is True
There is no contradiction. When „A‟ is True and „B‟ is
False, the premises are true and the conclusion is
false. So the argument is INVALID.
Because we have shown that the premises can be true while the conclusion is false, we do
not need to consider the other alternative, that „A‟ is False. We have done enough to
prove the argument INVALID.
P Q, P Q, Q P ╞═ P & Q
F T F T T T F T F F
6 1 7 2 9 3 8 @ 4 5
(5)
P Q, P Q, Q P ╞═ P & Q
T T T T T T F T F F
5 1 7 6 2 8 9 3 @ 4
A B, B A ╞═ A B
T T F
1 2 3
A B, B A ╞═ A B
T T F T T T T F F
5 1 7 8 2 6 @ 3 4
(4)
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 16
More exercises on the shortcut method
Determine whether the following arguments of SL are valid or invalid using the shortcut
method:
1. (A B) ╞═ A & B
2. P Q ╞═ (P & Q) & ( P & Q)
3. (L M) & (P M) ╞═ P L
4. M A, C H, A X, C M ╞═ X
5. L (A Q), L A, (L & Q) |= L Q
Derivations in Sentential Logic (1):
Some Simple Derivation Rules
Introduction
In the first half of the course we considered the logical language Sentential Logic (SL),
including the truth-table definitions of its connectives and accompanying account of SL-
validity. One motivation for such a language was to provide a simple framework for
determining the validity of natural-language arguments. In this section and the next, we will
consider a language which has precisely the same sentences as SL—so we shall continue to
refer to it as „SL‟—but where the connectives are defined by basic derivation (or inference)
rules. We can then prove that one sentence (the conclusion) is derivable from other sentences
(the premises) by showing that we can reach the conclusion from the premises by way of
simple steps sanctioned by the basic derivation rules.
Here is an example of a derivation (so you know what one looks like):
DERIVATION EXAMPLE 1
This is a derivation of „L & G‟ from the three
premises „(F G) H‟, „M & G‟ and „H
L‟. The numbers in the left-most column
enable one to refer to steps at various parts of
the argument. So e.g. the right-most column,
the justification column, at step 4 tells us that
step 4 was derived from step 2, using
derivation rule „&E‟ (to be explained shortly).
The vertical line is called a scope line—why we need it will become clearer in the next section.
The phrase „Prem.‟ in the justification column indicates that the sentence on that line was
given as a premise. Let us now consider the basic derivation rules used in the derivation.
Some basic derivation rules
There are two kinds of basic derivation rules, Introduction and Elimination Rules, for each
connective, which we will signify by suffixing the connective with „I‟ or „E‟, respectively. Here
are the rules we used in DERIVATION EXAMPLE 1.
1.
2.
3.
4.
5.
6.
7.
8.
(F G) H
M & G
H L
G
F G
H
L
L & G
Prem.
Prem.
Prem.
2, &E
4, I
1, 5, E
3, 6, E
4, 7, &I
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 17
m P
n Q
P & Q m, n &I
Q & P m, n, &I
&-Introduction (&I)
If sentences P and Q appear as steps in a derivation,
one may derive P & Q or Q & P. In the
justification column, one puts down the line-
numbers of the steps where P occurs and where Q
occurs and the name of the rule, „&I‟.
m P & Q
P m, &E
Q m, &E
&-Elimination (&E)
If P & Q appears as a step in a derivation, one may
derive P or Q. In the justification column, one puts
down the line-number of the line where P & Q
appears and the name of the rule, „&E‟.
Step 4 of DERIVATION EXAMPLE 1 uses the &-Elimination rule: given this rule, step 4
follows from step 2.
Step 8 of the derivation uses the &-Introduction rule: given this rule, step 8 follows from
steps 4 and 7.
Here are the other rules used in DERIVATION EXAMPLE 1:
m P
P Q m, I
Q P m, I
-Introduction (I)
If P appears as a step in a derivation, one may
derive P Q or Q P. In the justification column,
one puts down the line-number of the line where P
appears and the name of the rule, „I‟.
m P Q
n P
Q m, n E
or
m P Q
n Q
P m, n E
-Elimination (E)
If sentences P Q and P appear as steps in a
derivation, one may derive Q. If sentences P Q
and Q appear as steps in a derivation, one may
derive P. In the justification column one puts down
the line numbers of the steps where P Q occurs
and where P (or Q) occurs and the name of the
rule, „E‟.
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 18
m P Q
n P
Q m, n E
-Elimination (E)
If sentences P Q and P appear as steps in a
derivation, one may derive Q. In the justific-ation
column, one puts down the line-numbers of the
steps where P Q occurs and where P occurs and
the name of the rule, „E‟.
The -Elimination rule was used in step 7: from „H L‟ (step 3) and „ H‟ (step 6), one
may infer „L‟. Note: you need both steps 3 and 6 in order to derive ‘L’ (it does not follow
from step 3 by itself).
The -Elimination rule is used in step 6 of the derivation: the rule sanctions the derivation
of step 6, „ H‟, from steps 1 and 5, „(F G) H‟ and „F G‟. (Again, you need both
steps 1 and 5 to derive step 6.)
We will consider two more rules in this section: -Elimination and -Elimination.
m P Q
n P
Q m, n E
or
m P Q
n Q
P m, n E
-Elimination (E)
If sentences P Q and P appear as steps in a
derivation, one may derive Q. If sentences P Q
and Q appear as steps in a derivation, one may
derive P. In the justification column one puts down
the line numbers of the steps where P Q occurs
and where P (or Q) occurs and the name of the
rule, „E‟.
m P
P m, -E
-Elimination (E)
If sentence P appears as a step in a derivation,
one may derive P. In the justification column, one
puts down the line-numbers of the step where P
occurs and the name of the rule, „E‟.
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 19
Examples. Using the rules we have considered so far, construct derivations for the following
arguments („├—‟ is the derivability sign; it separates the premises from the conclusion):
(i) A B, C & B ├— A & C
1
2
3
4
5
6
7
A B
C & B
C
B
A
A
A & C
Prem.
Prem.
2, &E
2, &E
1,4 E
5, E
3,6 &I
(ii) (M Q) L, L X, Q ├— X
1
2
3
4
5
6
(M Q) L
L X
Q
M Q
L
X
Prem.
Prem.
Prem.
3, I
1,4 E
2,5 E
Exercises
Complete the derivations below by entering the appropriate justifications:
(a)
1 A (L M)
2 L & A
3 L
4 A
5 L M
6 L
7 M
8 A K
9 M & (A K)
Prem.
Prem.
_________________
_________________
_________________
_________________
_________________
_________________
_________________
1 R (P Q)
2 X Q
3 M & X
4 X
5 Q
6 P Q
7 R
8 M
9 M & R
Prem.
Prem.
Prem.
______________
______________
______________
______________
______________
______________
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 20
Strategies for constructing derivations
If an SL-argument is valid, then there will be a derivation for it; conversely, if the
argument‟s conclusion is derivable from its premises, then it will also be a valid argument. But,
unlike the truth-table or shortcut method, there is no sure-fire recipe for constructing
derivations. Failure to construct a derivation may simply be due to our failing to see how it
should be done. It is only by attempting (and thinking hard) about derivations that we can get
better at constructing them.
One way of approaching the construction of a derivation is to work from the bottom up.
Here is an example. Suppose we are asked to construct a derivation for the following
argument (from R.L. Simpson, Essentials of Symbolic Logic, p. 71):
B, B A, (A & B) R ├— R.
FIRST STAGE
Sketch the initial shape of the derivation. We don‟t know yet
how many steps will be needed, so, for the time being, we
use „z‟ in place of the last line-number.
Now, we ask ourselves: how could we have got „R‟ from
the premises? Evidently, only by using step 3. „R‟ does not
follow from step 3 by itself; but we know that if we had „A
& B‟ as a step in our derivation, then our rule -Elimination would permit us to derive „R‟.
So our derivation should have this shape:
SECOND STAGE
We put down „A & B‟ as the previous step and number the
line „y‟. We now know how step z is to be justified: it
follows from steps 3 and y, using E.
Now, how could „A & B‟ be derived? It does appear as the
antecedent of the conditional in step 3, but no rule allows us
to derive the left-hand-side of a conditional. So, we must
look elsewhere. There is only on way we could have got „A
& B‟ in this derivation—by &-Introduction; that is, we must be able to get „A‟ and „B‟ in
earlier steps and used &-Introduction on these:
THIRD STAGE
We already have „B‟ as a step (it is the first premise), so,
prior to deriving „A & B‟, we must have derived „A‟. We
know now that the derivation must look something like the
figure on the left.
Next question: how could we have got „A‟? It could only
have been derived from the „ A‟ which appears as the
right-hand-side of the bi-conditional in step 2, „B A‟:
we know that -Elimination allows us to derive „A‟ from
„ A‟. See next diagram.
FOURTH STAGE
1
2
3
Z
B
B A
(A & B) R
R
Prem.
Prem.
Prem.
?
1
2
3
y
Z
B
B A
(A & B) R
A & B
R
Prem.
Prem.
Prem.
?
3,y E
1
2
3
X
Y
Z
B
B A
(A & B) R
A
A & B
R
Prem.
Prem.
Prem.
?
1,X &I
3,y E
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 21
Next: how could „ A‟ be derived? It does not follow
from step 2 by itself. Our rule for -Elimination tells us
that we can derive one side of a bi-conditional (double
arrow) only if we have the other side as a step. But we do:
namely, „B‟ in step 1. Hence, we do not need any further
intermediate step to derive „ A‟— it follows from steps 1
and 2, using -Elimination. The justification for step w is
simply: „1,2 E‟.
FINAL STAGE
Remember to replace the letters used in place of line-
numbers with numbers, and to alter the justifications in the
right-hand column accordingly. (There is no need to show
the various stages of your construction—just give the final,
complete derivation as in the diagram)
A useful tip if you are stuck: use any elimination rules you can! This is a „pot-luck‟
method and may not always work, but often it will indicate how the remainder of the
derivation could be constructed. Here is an example. Construct a derivation for the following
argument: M & R, (Q M) R ├— Q W.
STAGE 1
Write down the premises and, to remind you of
what it is you are trying to derive, the conclusion.
STAGE 2
We know that if an &-statement is a step, we are
allowed, by &-elimination, to derive both sides—so
do so, and see if any other eliminations suggest
themselves. As it turns out, we can use E and
E.
1
2
3
W
X
y
Z
B
B A
(A & B) R
A
A
A & B
R
Prem.
Prem.
Prem.
?
1,W E
1,X &I
3,y E
1
2
3
4
5
6
7
B
B A
(A & B) R
A
A
A & B
R
Prem.
Prem.
Prem.
1,2 E
4, E
1,6 &I
3,6 E
1
2
z
M & R
(Q M) R
Q W
Prem.
Prem.
?
1
2
3
4
z
M & R
(Q M) R
M
R
Q W
Prem.
Prem.
1, &E
1, &E
?
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 22
STAGE 3
Step 5 follows from step 3, using -Elimination;
and Step 6, follows from steps 2 and 4. NOTE: you
need both steps 2 and 4 to derive step 6.
STAGE 4 (FINAL ANSWER)
Step 6 is the disjunction „Q M‟. We can derive
either side of the disjunct if we have the explicit
negation of the other side as a step (this is what the
-Elimination rule tells us). Thus, we may derive
„Q‟ from steps 3 and 6. NOTE: it would be incorrect
to derive it from steps 5 and 6; step 5, because the
explicit negation of „ M‟ is not „M‟ but „ M‟.
Once we have got „Q‟ (step 6), we may derive „Q
W‟, using -Introduction (even though „W‟
appears nowhere else in the derivation!). So, the
derivation is complete.
Exercises
1. Locate the errors in the derivation below (explain why they are errors):
2. Construct derivations for each of the following arguments using any of the rules we have
considered in this section:
(a) K & ( L K) ├─ L
(b) R M, (R K) Z, R ├─ Z & M
(c) P R, R & X, Y P ├─ Y
(d) P (M Z), (A P) & (M Z) ├─ A
(e) (F & G) H, H F, (F & G) ├─ F G
1
2
3
4
5
6
z
M & R
(Q M) R
M
R
M
Q M
Q W
Prem.
Prem.
1, & E
1, &E
3, E
2,4 E
?
1 M & R Prem.
2 (Q M) R Prem.
3 M 1 &E
4 R 1 &E
5 M 3 E
6 Q M 2,4 E
7 Q 3,6 E
8 Q W 7 I
1
2
3
4
5
6
A
B (A C)
A C
B
C
C B
Prem.
Prem.
2, E
2, E
3 E
5, I
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 23
Derivations in Sentential Logic (2): Assumptions
Assumptions
Some derivations require one to make assumptions at some stage. For example, in some
cases, the best (or only!) way of showing that a certain proposition, p, is false is to show that
an absurd or contradictory result follows if one assumes the truth of p. Conversely, we could
show a proposition to be true by showing that an absurd or contradictory result follows from
the assumption that it is false.
In fact, this is precisely the idea behind the shortcut method we used in Sentential Logic.
The method begins by assuming the premises to be true and the conclusion to be false; if this
results in a contradiction, we concluded that the argument was valid—that the conclusion had
to be true if the premisses were.
Reductio ad absurdum: this is the name given to the style of argument just mentioned, i.e.
an argument for a conclusion by way of show that assuming otherwise leads to absurdity or
contradiction. The phrase indirect proof is also used, though generally restricted to cases
when one argues for a conclusion by way of showing that assuming otherwise leads to
contradiction.
Two of our derivation rules for Sentential Logic, -Introduction and -Introduction,
proceed from assumptions.
-Introduction.
Let us approach this rule by asking: when are we entitled to infer an arrow-sentence of the
form „P Q‟? Earlier, we took such a sentence to affirm something like: If P is true, then so
is Q. Our rule for arrow-introduction (below) basically says that if on the assumption that P is
true we can derive Q using any of the derivation rules, then we are entitled to derive „P Q‟.
m P Ass.
n Q
P Q m, n I
-Introduction (I)
If one can derive Q by assuming (or under the
assumption of) P, then one may derive P Q. In
the justification column, one puts down the line
numbers of the steps at which P is assumed and at
which Q is derived, and the name of the rule, „I‟.
Here is an example of a derivation using this rule. To show:
M P, P Q |— M Q
1. M P Prem.
2. P Q Prem.
3. M Ass.
4. P 1,3 E
5. Q 2,4 E
6. M Q 3,5 I
Explanation
In order to prove the arrow-statement „
M Q‟, we assume the (whole) left-hand
side of the arrow (in this case, „ M‟) and
derive the right-hand side (in this case
„Q‟).
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 24
Whenever we make an assumption in a derivation we introduce another scope line (moving
over one place to the right, as it were). This is what we have done at step 3 above. We do so
because we must distinguish between what follows from our original premises and what
follows from them given that the assumption is also true. When we discharge (finished with)
the assumption, we move back one scope line.
Another example. Construct a derivation for:
A, (A & B) Q |— (B & C) (C & Q)
1 A Prem. Explanation.
In order to prove an arrow-statement (see step
10), we assume the left-hand side (the
antecedent) of the arrow statement and show
that the right-hand side (the consequent) is
derivable. In this proof the antecedent has been
assumed at step 4. Notice that we have added
another scope line a bit to the right of the main
one—to indicate that these steps are what
follow given the assumption. The consequent
has been shown to be derivable under the
assumption at step 9. This entitles us to derive
the arrow statement at step 10.
2 (A & B) Q Prem.
3 A 1, E
4 B & C Ass.
5 B 4, &E
6 A & B 3,5 &I
7 Q 2,6
8 C 4, &E
9 C & Q 7,8 &I
10 (B & C) (C & Q) 4,9 I
Exercises
Construct derivations for each of the following arguments using any of the rules we
have considered so far:
(1) (P L) M, K M |— L ( M &
(2) ( X & Y) Z, Y X |— (H & Y) Z
(3) L, (M X) L |— (L M) X
(4) A (B C) |— (A & B) (C Z)
-Introduction. This rule just reflects the point made earlier that one can establish a
conclusion by showing that assuming otherwise leads to a contradiction.
m P Ass.
n Q
o Q
P m, n, o I
-Introduction (I)
If one can derive a sentence, Q, and its explicit
negation, Q, assuming P, then one may derive
P. In the justification column, one puts down the
line numbers of the steps at which P is assumed,
and at which Q and Q occur, and the name of the
rule, „I‟.
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 25
Here‟s an example using this rule, a derivation for: F (G & H) |— (H & F)
1 F (G & H) Prem. Explanation.
We have assumed the precise „opposite‟
of the conclusion at step 2. At steps 5
and 6 we have derived a statement and
its explicit negation. So we are entitled
to conclude that the original assumption
was false: i.e. we can derive the
negation of the assumption.
2 H & F Ass.
3 F 2, &E
4 G & H 1,3 E
5 H 4, &E
6 2, &E
7 (H & F)
Another example. A derivation for: (A B), (A B) D |— D & A
1 (A B) Prem.
2 (A B) D Prem.
3 D 1,2 E
4 A Ass.
5 4, I
6 1, R
7 A
8 D & A
Step 6 uses a new, but quite trivial, rule, Reiteration. This rule basically allows one to
reiterate a step later in the same derivation or any sub-derivation. We need it in this case
because our rule for -Introduction requires us to have the contradictory steps (P and P)
appearing under the assumption (that is, in the same scope line as the assumption).
m P
P m, R
m P
P m, R
Reiteration (R)
If sentence P appears as a step in
a derivation, one may derive P
later in the derivation or in any
sub-derivation.
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 26
Exercises
Construct derivations for each of the following arguments using any of the
rules we have considered so far:
(1) P & Q |— P Q)
(2) (F G) M, F M |— F
(3) A B) C |— (C & A) B
(4) G (H & Y), ( F K) G, F ├— K Y
(5) J ( J & D), D F, F J ├— J
The power of -Introduction
Derivation (5) in Exercises 2 above reveals how -introduction may be used to derive
conclusions that are NOT negation-statements. For, from “ J” we may derive “J” by use of
the -Elimination rule. Thus, we can construct a derivation for the following argument:
J ( J & D), D F, F J ├— J
namely, by deriving “ J” as we did for (5) by using -Introduction, and then deriving “J” by
using -Elimination. In fact, we cannot derive “J” from the above premises unless we use
-Introduction.
Double assumptions
A derivation may involve two or more assumptions, one after the other. Consider e.g.
(P Q) |— P & Q
To derive the conclusion one has to derive “ P” by way of -Introduction, then derive “
Q” by way of -Introduction, and, finally derive the conclusion by way of &-Introduction.
Exercises: Construct a derivation for the argument above. Then construct a derivation for:
P & Q |— (P Q)
This does not require two assumptions, but by constructing both derivations you have shown
that “ (P Q)” and “ P & Q” are logically equivalent!
Introduction to Predicate Logic
The inadequacy of Sentential Logic
The shortcut method provides a quick and effective way of determining whether an
argument of Sentential Logic (SL) is valid or not. But there are many arguments of
ordinary language where it seems the translation into SL is inadequate. Consider e.g. the
following argument we have come across:
Premises:
Conclusion:
All vegetarians are healthy.
Tracy is a vegetarian.
Hence, Tracy is healthy. Valid
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 27
Because none of the statements involved contain sentence connectives, each of them must be
treated as simple (i.e. atomic) sentences of SL. So, we get something like the following SL-
invalid argument on translation: V, T ╞═ H.
Another example:
Premises:
Conclusion:
Superman can fly.
Batman cannot fly.
Hence, Superman is not Batman. Valid
This argument would get translated into SL as: S, B |= I (S = Superman can fly; B =
Batman can fly; I = Superman is Batman). This too comes out invalid using the shortcut
method.
What gives? The problem is that we need more than sentence connectives to capture the
„shape‟ of sentences. In the latter argument, for example, we need to capture the fact that one
and the same property (of being into fly) is invoked in the two premises; in the first argument,
we need to capture the properties of being a vegetarian and being healthy. So, we need a
more sophisticated logical language that breaks sentences down into simpler components.
Enter, Predicate Logic.
Predicate Logic (PL):
Constants, Predicates and Quantifiers
VOCABULARY CATEGORY USE
a, b, c, ..., w Constants / Names Designating objects
A, B, C, ..., Z Predicate letters Ascribing properties
x, y, z Variables Quantifying over objects
(x) Existential Quantifier There is an x such that ...
(x) Universal Quantifier Every x is such that …
, &, , , Logical connectives Same as Sentential Logic
Constants (names) are used when a particular thing is being referred to, and the
predicate letters are used to ascribe properties to that thing. So, e.g., we may symbolize
Harry is bald as „Bh‟, where „Bx‟ stands for „x is bald‟ and „h‟ stands for Harry. If Harry is
dull, so is Betty may be translated into Predicate Logic (PL) as: Dh Db (where „Dx‟ stands
for „x is dull‟ and „b‟ for Betty).
NOTE: it would be wrong to translate Politicians are dull as „Dp‟ (with „p‟ standing
for politicians). For this statement is not about a particular individual but about all individuals
with a certain property. (We‟ll see how to symbolize this kind of statement next term.)
Predicate letters are also used to signify relations between individuals. E.g. „Txy‟ could
stand for „x is taller than y‟; „Bxyz‟ for „x is between y and z‟. (Note: the former is called a
two-place predicate, because it has two „place-holders‟; „Bxyz‟ is a three-place predicate, and
so on.)
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 28
An n-place predicate letter followed by n names (constants) is called an atomic sentence of
PL. One can construct complex sentences from atomic sentences with the logical connectives
of Sentential Logic, the translations of the connectives being the same as before.
SENTENCE TRANSLATION
I am taller than Jenny. Tmj (m = me)
Kevin is not older than Marion. Okm
Jenny is taller than both Kevin and Marion. Tjk & Tjm
Marion is older than Kev if Jenny is. Ojk Omk
Brighton is not between Oxford and Glasgow. Bbog
Exercises
1. Translate the following sentences of Predicate Logic into English:
(a) Fba & Fma (Fxy = x is y‟s friend; b = Ben; a =Alice; m = me)
(b) (Tma & Tbm) Tba (Txy = x is taller than y)
(c) Sp Ad
(Sx: x will survive; Ax: x arrives soon; p = the patient; d = the doctor)
(d) (Sb Sa) & (Sb & Sa)
2. Translate the following English sentences into Predicate Logic (using the above key:
(a) Tracy is a vegetarian if her brother is.
(Vx = x is vegetarian; t = Tracy; b = Tracy‟s brother)
(b) Neither Tracy nor Alice are friends of mine. (Use previous definitions)
(c) Ben won‟t survive unless Alice arrives soon. (Use previous definitions)
(d) I‟ll be Alice‟s friend only if she isn‟t Tracy‟s friend.
(e) Alice and Tracy are the same height. (Using Txy = x is taller than y).
(f) I‟ll be in trouble if Tracy believes either Ben or Alice.
(Tx = x is in trouble; Bxy = x believes y)
Existential quantifiers are used to affirm the existence of objects possessing certain
properties. „(x)Fx‟ means:
There is an individual (or object) which is (an) F.
So e.g. we may translate „Unicorns exist‟ or „There are unicorns‟ into PL as „(x)Ux‟, where
„Ux‟ means „x is a unicorn‟.
NOTE: We could just as well have used: „(y)Fy‟ or „(z)Fz‟. The variables „x‟, „y‟, and
„z‟ do not stand for any specific individuals. So „Fx‟, for example, is not a grammatical
sentence. Neither is „(y)Fy & Gy‟, because the „y‟ in „ Gy‟ is not bound (governed) by
the quantifier; „(y)(Fy and Gy)‟, on the other hand, is fine: both tokens of „y‟ are bound by
the existential quantifier. Variables can only occur in sentences if they are bound (governed)
by a quantifier.
NOTE: PL does not discriminate between the following claims:
There are Fs; there is at least one F; there are many Fs; something is F
All x)Fx‟.
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 29
Statements of the form, „Nothing is F‟ simply deny that something is F. So they may be
translated as „ (x)Fx‟.
Statements of the form, „Some Fs are G‟ or „Some F is G‟ are logically equivalent to
„Something is both F and G‟. So they may be translated as „(y)(Fy & Gy)‟.
„No F is G‟ simply denies that some F is G. So it may be translated as „ (z)(Fz & Gz)‟.
Examples
SENTENCE TRANSLATION
There are many dogs. (x)Dx
Absolute cads exist! (y)Ay
Someone is not an absolute cad. (x) Ax
There are no dragons. (y)Dy
If someone is drunk, Alan is. (x)Dx Da
Something is bothering Alice. (x)Bxa
Nothing bothers me. (x)Bxm
Some cats are black. (z)(Cz &Bz)
Some cats are not black. (z)(Cz & Bz)
A dog is chasing Harry. (x)(Dx & Cxh)
Martha is chasing someone. (x)Cmx
Martha isn‟t chasing anyone. (x)Cmx
No dogs chase Alice. (x)(Dx & Cxa)
Translation Exercises involving existential quantification
Use the following key for questions 1 and 2:
KEY
Cx : x is crazy a : Archie
Ex : x is an elephant b : Blanche
Sx : x smokes m : Me
Vx : x is a vegetarian t : that room
Ixy : x is in y u : this university
Mxy : x is married to y
1. Translate the following PL-sentences into natural English (do your best with (i) and (j)!):
a. Iat & Iau
b. Mab (Sa Sb)
c. (y)Iyt
d. (x)(Mxb & Vx)
e. (z)(Ez & Cz)
f. (y)Mya Em
g. (y)Mby (
h. (x)(Ixt & Sx)
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 30
i. (y)((Cy & Ey) & Iyu)
j. (x) (y)Mxy
2. Translate the following sentences into sentences of Predicate Logic using existential
quantifiers (where necessary!):
a. I‟m not married to Archie.
b. Blanche is a vegetarian only if she is crazy.
c. Nobody‟s crazy.
d. I don‟t smoke, but many do.
e. Archie is married. (Clue: Archie is married if he‟s married to someone)
f. Some elephants are vegetarians.
g. Neither Blanche nor Archie are in this university.
h. Someone in that room is not a vegetarian.
i. No one is crazy if Archie isn‟t.
j. Unless I‟m crazy, there‟s an elephant in that room.
Predicate Logic: Universal Quantifiers and Domains Of Quantification
Domains of Quantification
Consider the following pairs of sentences:
(a1) Some students missed today‟s class.
(a2) Some students want to be academics.
(b1) Everyone enjoyed the party.
(b2) Everyone is to blame for the state of our society.
(a1) is presumably true if and only if there are students officially taking the course in
question that missed the class; so the fact that there are other students (e.g. students in other
universities) who did not attend the class does not automatically make (a1) true. (a2), on the
other hand, is presumably true if there are students somewhere—at any rate, not necessarily
those officially taking the course alluded to in (a1)—who want to be academics. So, the group
of individuals under discussion in (a1) is different from those under discussion in (a2). Thus, the
domain of quantification (or universe of discourse) is said to be different in the two cases.
Likewise, the quantifier „everyone‟ ranges over one set of individuals in (b1)—namely, the
people who came to the party—and a different set of individuals in (b2)—namely, the people
of this country (or the world). The domains of quantification are the respective sets of
individuals.
The upshot of all this is that whenever quantifiers are used in ordinary language, a domain
of quantification is generally assumed or taken for granted. When we translate into Predicate
Logic, however, we should, strictly speaking, make the domain explicit—especially when the
domain varies from question to question (or sentence to sentence). But in this course, we will
specify domains only when the context is ambiguous.
Universal quantifiers: (x), (x)
Universal quantifiers are used when we wish to affirm that everything (or everything of a
certain kind) possesses a certain property. „(x)Fx‟ means: For each and every x, x is F. So
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 31
e.g. we may translate „Everyone is moral‟ as „(x)Mx‟, where „Mx‟ means „x is moral‟;
„Everyone likes Archie‟ may be translated as „(y)Lya‟; and so on.
Note: the following statements are equivalent:
Every F is G; all Fs are G; whatever (whoever) is F is G
A statement of the form, „No one is (an) F‟ is equivalent to „Everyone is non-F‟ and can be
translated as „(x) Fx‟. (Ditto for: „Nothing is F‟).
Warning!!: „Everyone is not a smoker‟ (or „all are not smokers‟) generally does not mean:
„everyone is a non-smoker‟. Rather, it is more likely to be used to deny that everyone is a
smoker. So it should be translated as: (x)Sx. (Ditto for other statements of the same form.)
Universal quantifiers are also used to affirm that everything with a certain property has
another property. But it would be incorrect to translate All dogs bark, for example, as:
(x)(Dx & Bx). This PL-sentence affirms that everything is a barking dog! The correct
translation of „Every dog barks‟ (or „All dogs bark‟) is of the form: (x)(Dx Bx). Here‟s
why.
Suppose you know that every dog barks. What would you thereby know about any
randomly selected object from the world? You would know simply that: if the chosen object is
a dog, then it barks. So, it follows from the claim that every dog barks, that for any object, x,
if x is a dog, x barks; symbolically: (x)(Dx Bx).
Next, suppose it is true of each and every object in the world that: if it is a dog, it barks.
Then, clearly, it must be true that any given dog barks; in other words, it must be true that
every dog barks.
Hence, the correct translation of „Every dog barks‟ is: (x)(Dx Bx). More generally,
statements of the form: every F is G, any F is G, each F is G, all Fs are G, whatever is F is G
are all translated as (x)(Fx Gx).
Statements of the form: Only Fs are G are equivalent to Every G is F. So the PL-
translation is: (x)(Gx Fx)
We saw earlier that statements of the form No Fs are G can be translated as:
(x)(Fx & Gx). But this is equivalent to: (x)(Fx Gx). Either translation will do.
SENTENCE TRANSLATION
Everyone is hungry (y)Hy
Nothing is sacred (x) Sx
Every journalist drinks (x)(Jx Dx)
All yacht-owners love the Pacific Ocean (z)(Yz Lzp)
(key: Yx = x is a yacht-owner; Lxy = x loves y; p = the Pacific Ocean)
Any cow eats grass (y)(Cy Ey)
All bricklayers are not sexist (x)(Bx Sx)
(see earlier warning!)
No horse reads (x)(Hx Rx)
OR (x)(Hx & Rx)
Only horses run at Ascot (z)(Rz Hz)
Whoever hit Alan is vicious (y)(Hya Vy)
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 32
Translation Exercises
1. Using the key below, translate the following PL-sentences into English:
KEY
Cx : x supports City b : Bill
Lx : x likes football e : Eric
Sx : x is a snob m : Martina
Tx : x likes tennis p : Penny
Ux : x supports United
Bxy : x bores y Fxy : x is a fan of y
a. Lb Fbe
b. (x)Tx & (y)Ly
c. (y)(Fym & Bym)
d. (z)Fbz Fbb
e. (x)(Cx Bxm)
f. (y) Byp
g. (z)(Uz Cz)
h. (x)(y)Byx
i. (x)((Tx & Lx) Sx)
2. Using the following key, translate the following sentences into Predicate Logic:
KEY
Cx : x is crazy a : Archie
Ex : x is an elephant b : Blanche
Sx : x smokes m : Me
Vx : x is a vegetarian t : that room
Ixy : x is in y u : this university
Mxy : x is married to y
a. Everyone is crazy.
b. All elephants are vegetarians.
c. No elephant smokes.
d. Whoever Blanche is married to is a vegetarian.
e. Everyone in that room is a non-smoker.
f. Not everyone in this university is in that room.
g. Every vegetarian smoker is crazy.
h. Only elephants are in that room.
i. Everyone in that room is not a vegetarian.
j. If some elephants are vegetarians all are.
k. Someone is in this university if anyone is in that room.
l. If anyone is in that room, they are in this university.
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 33
Predicate Logic with Identity You may find the following useful for this section: S. Guttenplan, The Languages of Logic,
pp. 196-204 (Library code: [BC 71 Gut])
Adding a two-place identity predicate, „x ═ y‟ (read: x is identical with y) to Predicate
Logic (PL) increases the language‟s expressive power considerably. To begin with, we need
the predicate to express the identity or distinctness of individuals. E.g. “Superman is Clark
Kent” might be expressed as “s ═ c”, and “Bruce Wayne is not Robin” might be expressed as
as “b r” (where “x y” just abbreviates “ (x = y)”). But there are other significant uses.
Here are some examples.
(1) Carla didn’t go to the party but everyone else did.
How should (1) be expressed in Predicate Logic? (Take „Px‟ to stand for „x went to the party‟
and „c‟ for Carla). Here is a wrong answer:
(1WRONG) Pc & (x)Px
(1WRONG) affirms that: Carla didn’t go to the party and (but) everyone went. This is wrong
because it is contradictory: if Carla didn‟t go to the party, it cannot be true that everyone
went! Sentence (1), however, clearly is not contradictory. What we want to express is the
proposition that: Carla did not go to the party but everyone other than (or: everyone who is
not) Carla did go. This is captured by
(1RIGHT) Pc & (x)(x c Px)
(1RIGHT) also captures the content of “Everyone but [except, other than] Carla went to the
party” (though I would also accept „(x)(x c Px)‟ as a translation of “Everyone other than
Carla went to the party”).
(2) Only Phil drinks milk (No one but Phil drinks milk)
(Key: p = Phil, Mx = x drinks milk)
To see how (2) should be translated, it is useful to compare it with something like: Only cats
drink milk. This, we have seen, affirms the same thing as (M)
(M) Everything which drinks milk is a cat
(x)(Mx Cx)
So, (2) should be read as affirming that everything which drinks milk is (identical with) Phil:
(2a) (x)(Mx x = p)
NOTE: Just as (M) does not entail that there are cats who drink milk, (2a) does not entail that
Phil drinks milk. If this seems inadequate, i.e. if one does take (2) to entail that Phil drinks
milk, then (2) may be expressed as (2b) or (2c):
(2b) Mp & (x)(Mx x = p)
(2c) (x)(Mx x = p)
[It is the double-arrow in (2c) which ensures that if something is Phil, it drinks milk—i.e. that
Phil drinks milk.]
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 34
Another example: “Only Phil and Sue drink milk” is translated as:
(3a) (x)(Mx (x = p x = s))
(3a) is in fact compatible with neither Phil nor Sue being milk-drinkers. If one took the original
sentence to imply that they are, then one should translate it with the double-arrow, i.e. as:
(3b) (x)(Mx (x = p x = s))
(3) There are at least two Latvians (There is more than one Latvian)
(Key: Lx = x is a Latvian)
Here is a wrong translation of (3):
(3WRONG) (x)(y)(Lx & Ly)
The occurrence of an existential quantifier in a sentence simply signifies the existence of a
certain kind of thing: that there is at least one thing of that kind. Thus, (3WRONG) just affirms the
existence of Latvians „twice over‟, as it were: There is at least one Latvian and there is at
least one Latvian. To express the fact that there is more than one, we need to specify that
there are Latvians x and y which are distinct:
(3RIGHT) (x)(y)((Lx & Ly) & x y)
One would express the claim that there are at least four Latvians in the following way:
(x1)(x2)(x3)(x4)(Lx1 & … & Lx4 & x1 x2 & x1 x3 … & x3 x4)
(4) There is exactly one U.S. President (There is one, and only one, U.S. President;
there is only one U.S. President) (Key: Px = x is U.S. President)
(4) affirms that there is a U.S. President that s/he is the only U.S. President. This may be
expressed in Predicate Logic as either (4a) or (4b):
(4a) (x)(Px & (y)(Py x = y))
(4b) (x)(y)(Py x = y)
[Compare these with (2b) and (2c), respectively (on the previous page), which may be
regarded as expressing that Phil is the only (or the one and only) milk-drinker.]
Here is how one would symbolize “There are exactly (only) two tigers”:
(x)(y)((Tx & Ty & (z)(Tz (z = x z = y))) & x y)
Or: (x)(y)((z)(Tz (z = x z = y)) & x y)
(5) There is no more than one unicorn. (Ux = x is a unicorn)
At most one unicorn exists.
(5) does not affirm the existence of a unicorn—it does not state there is one unicorn but no
more. It simply denies that there are at least two unicorns. So one way of symbolizing it is
thus:
(5a) (x)(y)((Ux & Uy) & x y)
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 35
Another way of expressing the same fact is to maintain that if x and y are unicorns (for any x
and y), x and y are identical:
(5b) (x)(y)((Ux & Uy) x = y)
Exercises
1. Translate the following sentences of Predicate Logic into English:
a. (x)(x r Bxm)
b. (y)(Dy y ═ t) (t = Tibbles)
c. (y)(Byt y = r)
d. (x)(Dx & (y)(Dy & y x))
2. Translate the following sentences into Predicate Logic with Identity:
a. At least two dogs bit Tibbles. (Dx; Bxy; t)
b. Rover bit everyone but me. (Bxy; r; m)
c. Rover bit me, and so did someone else.
d. At most one dog bit me.
3. Give a simpler Predicate Logic translation of: (y)(Bym & y = t)
Exercises 4
Use the following key for questions 1 and 2 (the asterisked questions involve identity):
KEY
Ax : x is an actor (actress) a : Akira
Dx : x is a director b : Boris
Hx : x is happy d : Doris
Ox : x has won an Oscar j : Juanita
Bxy : x is a better actor than y m : Me
Fxy : x is a fan of y s : Sanjay
Mxy : x has met y
Wxy : x has worked with y
1. Translate the following sentences of Predicate Logic into English:
a. Fab Fas
b. (Wda Wdb) & Wdm
c. (y)(Ay & Hy) & (y)(Ay & Hy)
d. (z)(Wza Fmz)
e. (x)((Dx & Wxj) Ox)
f. (x)(Wdx & Bxd)
g. (x)(Ax & Dx) (y)(Ay Hy)
h. (x)(Dx (y)(Ay & Fyx))
i. (y)(Wby Oy)
*j. b=s (x)(y)((Fxm & Fym) x=y)
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 36
*k. (z)((Dz & Az) s=z)
*l (x)((Mbx & Fxb) & (y)(Fyb x=y))
2. Translate the following English sentences into Predicate Logic:
a. I‟ve met Akira—he is a director.
b. I won‟t be happy until I‟ve met Doris.
c. Anyone who has worked with Juanita is a fan of hers.
d. I‟ve met an Oscar-winner!
e. No one is a better actor than Boris.
f. Doris has a fan who is a better actress than her!
g. All Oscar-winning directors are happy.
h. If Sanjay is happy, I‟ve met a happy actor.
i. Boris has met an actress and a director she is working with.
j. If anyone is a better actor than Akira, Juanita is.
k. My fans will be unhappy only if I do not win the Oscar.
*l. Everyone but Sanjay has worked with Doris.
*m. I‟ve only worked with Boris and Akira.
*n. Sanjay is the only happy actor.
Leibniz’s Law(s)
The Indiscernibility of Identicals: a = b (Fa Fb)
This is a general principle concerning identity that is often simply called Leibniz’s Law. It
maintains that if a and b are identical, then whatever is true of a is true of b, and vice versa
(for any a and b). So e.g. if Clark Kent is Superman, then Clark Kent can fly if and only if
Superman can.
The Identity of Indiscernibles: (Fa Fb) a = b
This is another principle Leibniz held, but it is rather more contentious. It maintains that if
everything true of a is also true of b, and vice versa, then a is identical with b.
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 37
Predicate Logic Derivations: Two Simple Rules + Examples + Exercises
Predicate Logic Derivation Rules: -Introduction and -Elimination
m
(v)* m, I
-Introduction ()
From a PL-sentence containing occurrences of a
name, c, one may derive (v)*, where v is a
variable (e.g. x, y, or z) that does not appear in ,
and * is the result of replacing one or more (or
all) occurrences of c in with v.
m (v)
* m, E
-Elimination (E)
From a PL-sentence of the form „(v)‟, one may
derive *, where * is the result of replacing every
occurrence of v in with a particular name, c.
Some sample derivations: please complete the justification columns
1. (x)(Fx & Gx), Ft Ht |— Ht
1 (x)(Fx & Gx) Prem
2 Ft Ht Prem
3 Ft & Gt
4 Ft
5 Ht
2. Fb, (x)(Fx Gx) |— (z)(Fz & Gz)
1 Fb Prem
2 (x)(Fx Gx) Prem
3 Fb Gb
4 Gb
5 Fb & Gb
6 (z)(Fz & Gz)
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 38
3. (x)Fxb (x) Fxb, Fab |— Fhb
1 (x)Fxb (x) Fxb Prem
2 Fab Prem
3 (x)Fxb Ass.
4 Fab
5 Fab
6 (x)Fxb
7 (x) Fxb
8 Fhb
4. Ft Gt, (z)(Gz Hz) (x)Gx |— Ft Gm
1 Ft Gt Prem
2 z)(Gz Hz) (x)Gx Prem
3 Ft
4 Gt
5 Gt Ht
6 (z)(Gz Hz)
7 (x)Gx
8 Gm
9 Ft Gm
Construct derivations for the following PL-arguments:
5. (x)(Mx Hx), (y) Hy |— Mk & Hk
6. Abm & Fmb, (x)Abx (x)Fbx |— (y)(Fby & Fyb)
(Clue: derive “Fbm & Fmb” before the conclusion)
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 39
Predicate Logic Derivations: Two More Rules + Examples + Exercises
Predicate Logic Derivation Rules: -Introduction and -Elimination
m
(v)* m I
-Introduction ()
From a PL-sentence containing occurrences of a
name, c, that does not appear in the premises or
any governing assumptions, one may derive (v)*,
where v is a variable (e.g. x, y, or z) that does not
appear in , and * is the result of replacing every
occurrence of c in with v.
m (v)
n * Ass.
o α
α m,n,o E
-Elimination (E)
One may derive a statement α from (v) if one can
derive α under the assumption of *, where * is the
result of replacing every occurrence of v in with a
name, c, that does not appear anywhere earlier in
the derivation or in α.
Some sample derivations
1. (x)(Fx & Gx) |— (y)Gy
1 (x)(Fx & Gx) Prem
2 Fa & Ga 1 E
3 Ga 2 &E
4 (y)Gy 3 I
2. (x)(Vx Hx), (x)(Hx Tx) |— (x)(Vx Tx)
1 (x)(Vx Hx) Prem
2 (x)(Hx Tx) Prem
3 Vb Ass.
4 Vb Hb 1 E
5 Hb Tb 2 E
6 Hb 3,4 E
7 Tb 5,6 E
8 Vb Tb 3,7 I
9 (x)(Vx Tx) 8 I
Dept of Philosophy, WITS SYMBOLIC LOGIC Page 40
3.y)Gy |— (z)Gz
1 y)Gy Prem
2 Gm Ass.
3 y)Gy 2 I
4 y)Gy 1 R
5 Gm 2,3,4 I
6 (z) Gz 5 I
4. Ft, (x)(Fx Gx)|— (z)(Fz Gz)
1 Ft Prem
2 (x)(Fx Gx) Prem
3 Ft Gt 2 E
4 Gt 1,3 E
5 Ft & Gt 1,4 &I
6 (z)(Fz Gz) 5 I
5. (x)(Txa Txb), (x)xa |— (x)Txb
1 (x)(Txa Txb) Prem
2 (x)xa Prem
3 Tka Tkb Ass.
4 ka 2 E
5 Tkb 3,4 E
6 (x)Txb 5 I
7 (x)Txb 1,3,6 E
Exercises
Construct derivations for the following arguments of Predicate Logic:
1. (y)Ay, (z)Bz |— (x)(Ax & Bx)
2. (y)Cy Cb |— (x)(Cx Cb)
3. (x)(Lx tx & Bx)), Lb & Db |— (y)Mty
4. Ft Gt, (z)Gz (x)Gz ├— Ft Gm
5. (x)(Fx Gx), (y)Fy ├— (x)(Fx & Gx)
6. (y)(Ty & Qay), (y)Qay Qaa├— (y)Qyy
End of course. Good luck in the exam!
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