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Chapter 13

Formal Proofs and Quantifiers

Now that we have learned the basic informal methods of proof for quantifiers,

we turn to the task of giving formal rules that correspond to them. Again, we

can to do this by having two rules for each quantifier.

Before getting down to the rules, though, we should emphasize that formal

proofs in the system F contain only sentences, never wffs with free variables.

This is because we want every line of a proof to make a definite claim. Wffs

with free variables do not make claims, as we have noted. Some deductive

systems do allow proofs containing formulas with free variables, where such

variables are interpreted universally, but that is not how the system F works.

Section 13.1

Universal quantifier rules

The valid inference step of universal instantiation or elimination is easily for-

malized. Here is the schematic version of the rule:

Universal Elimination (∀ Elim):

∀x S(x)...

▷ S(c)

Here x stands for any variable, c stands for any individual constant (whether

or not it has been used elsewhere in the proof), and S(c) stands for the result

of replacing free occurrences of x in S(x) with c. If the language contains

function symbols, c can also be any complex term that contains no variables.

Next, let us formalize the more interesting methods of general conditional

proof and universal generalization. This requires that we decide how to rep-

resent the fact that a constant symbol, say c, has been introduced to stand

for an arbitrary object satisfying some condition, say P(c). We indicate this

by means of a subproof with assumption P(c), insisting that the constant c in

question occur only within that subproof. This will guarantee, for example,

that the constant does not appear in the premises of the overall proof.

351

352 / Formal Proofs and Quantifiers

To remind ourselves of this crucial restriction, we will introduce a new

graphical device, boxing the constant symbol in question and putting it inboxed constant

front of the assumption. We will think of the boxed constant as the formal

analog of the English phrase “Let c denote an arbitrary object satisfying P(c).”

General Conditional Proof (∀ Intro):

c P(c)

...Where c does not occur out-

side the subproof where it is

introduced.Q(c)

▷ ∀x (P(x)→ Q(x))

When we give the justification for universal introduction, we will cite the

subproof, as we do in the case of conditional introduction. The requirement

that c not occur outside the subproof in which it is introduced does not pre-

clude it occurring within subproofs of that subproof. A sentence in a subproof

of a subproof still counts as a sentence of the larger subproof.

As a special case of ∀ Intro we allow a subproof where there is no sentential

assumption at all, just the boxed constant on its own. This corresponds to

the method of universal generalization discussed earlier, where one assumes

that the constant in question stands for an arbitrary object in the domain of

discourse.

Universal Introduction (∀ Intro):

c



introduced.P(c)

▷ ∀x P(x)

As we have indicated, we don’t really need both forms of ∀ Intro. We use

both because the first is more natural while the second is more general.

Chapter 13

Universal quantifier rules / 353

Let’s illustrate how to use these rules by giving a formal proof mirroring

the informal proof given on page 335. We prove that the following argument

is valid:

∀x (P(x)→ Q(x))

∀z (Q(z)→ R(z))

∀x (P(x)→ R(x))

(This is a general form of the argument about all math majors being smart

given earlier.) Here is a completed proof:

1. ∀x (P(x)→ Q(x))

2. ∀z (Q(z)→ R(z))

3. d P(d)

4. P(d)→ Q(d) ∀ Elim: 1

5. Q(d) → Elim: 3, 4

6. Q(d)→ R(d) ∀ Elim: 2

7. R(d) → Elim: 5, 6

8. ∀x (P(x)→ R(x)) ∀ Intro: 3-7

Notice that the constant symbol d does not appear outside the subproof. It is

newly introduced at the beginning of that subproof, and occurs nowhere else

outside it. That is what allows the introduction of the universal quantifier in

the final step.

You try it. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

◀1. Open the file Universal 1. This file contains the argument proven above.

We’ll show you how to construct this proof in Fitch.

◀2. Start a new subproof immediately after the premises. Before typing any-

thing in, notice that there is a blue, downward pointing triangle to the left

of the blinking cursor. It looks just like the focus slider, but sort of stand-

ing on its head. Use your mouse to click down on this triangle. A menu

will pop up, allowing you to choose the constant(s) you want “boxed” in

this subproof. Choose d from the menu. (If you choose the wrong one, say

c, then choose it again to “unbox” it.)

◀3. After you have d in the constant box, enter the sentence P(d) as your

assumption. Then add a step and continue the subproof.

Section 13.1


▶ 4. You should now be able to complete the proof on your own. When you’re

done, save it as Proof Universal 1.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Congratulations

Default and generous uses of the ∀ rules

Both of the universal quantifier rules have default uses. If you cite a universaldefault uses of ∀ rules

sentence and apply ∀ Elim without entering a sentence, Fitch will replace the

universally quantified variable with its best guess of the name you intended. It

will either choose the alphabetically first name that does not already appear

in the sentence, or the first name that appears as a boxed constant in the

current subproof. For example, in steps 4 and 6 of the above proof, the default

mechanism would choose d, and so generate the correct instances.

If you know you want a different name substituted for the universally

quantified variable, you can indicate this by typing a colon (:), followed byindicating substitutions

the variable, followed by the greater-than sign (>), followed by the name

you want. In other words, if instead of a sentence you enter “: x > c”, Fitch

will instantiate ∀x P(x) as P(c), rather than picking its own default instance.

(Think of “: x > c” as saying “substitute c for x.”)

If you apply ∀ Intro to a subproof that starts with a boxed constant on its

own, without entering a sentence, Fitch will take the last sentence in the cited

subproof and universally quantify the name introduced at the beginning of the

subproof. If the cited subproof starts with a boxed constant and a sentence,

then Fitch will write the corresponding universal conditional, using the first

sentence and the last sentence of the proof to create the conditional.

You try it. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

▶ 1. Open the file Universal 2. Look at the goal to see what sentence we are

trying to prove. Then focus on each step in succession and check the step.

Before moving to the next step, make sure you understand why the step

checks out and, more important, why we are doing what we are doing at

that step. At the empty steps, try to predict which sentence Fitch will

provide as a default before you check the step.

▶ 2. When you are done, make sure you understand the completed proof. Save

the file as Proof Universal 2.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Congratulations

Chapter 13

Universal quantifier rules / 355

Fitch has generous uses of both ∀ rules. ∀ Elim will allow you to remove

several universal quantifiers from the front a sentence simultaneously. For

example, if you have proven ∀x∀y SameCol(x, y) you could infer SameCol(f, c)

in one step in Fitch. If you want to use the default mechanism to generate this

step, you can enter the substitutions “: x > f : y > c” before checking the step.

In a like manner, you can often prove a sentence starting with more than

one universal quantifier by means of a single application of ∀ Intro. You do

this by starting a subproof with the appropriate number of boxed constants.

If you then prove a sentence containing these constants you may end the

subproof and infer the result of universally quantifying each of these constants

using ∀ Intro. The default mechanism allows you to specify the variables to

be used in the generated sentence by indicating the desired substitutions, for

example “: a > z : b > w” will generate ∀z∀wR(w, z) when applied to R(b, a).

Notice the order used to specify substitutions: for ∀ Elim it will always be

“: variable > name,” while for ∀ Intro it must be “: name > variable.”

Add Support Steps can be used with the ∀ Intro rule. If the focus step

contains a universally quantified formula then the support will be a subproof

with a new constant at the assumption step. The last step of the subproof will

contain the appropriate instance of the universal formula. With the ∀ Elim

rule a single support step containing a universal formula will be added. The

support formula will be a universal generalization of the formula at the focus

step, with the first constant replaced by the variable of quantification.

Remember

The formal rule of ∀ Intro corresponds to the informal method of general

conditional proof, including the special case of universal generalization.

Exercises

13.1ö

If you skipped the You try it sections, go back and do them now. Submit the files Proof

Universal 1 and Proof Universal 2.

For each of the following arguments, decide whether or not it is valid. If it is, use Fitch to give a formal

proof. If it isn’t, use Tarski’s World to give a counterexample. In this chapter you are free to use Taut

Con to justify proof steps involving only propositional connectives.

Section 13.1


13.2ö

∀x (Cube(x)↔ Small(x))

∀x Cube(x)

∀x Small(x)

13.3ö

∀x Cube(x)∀x Small(x)

∀x (Cube(x) ∧ Small(x))

13.4ö

¬∀x Cube(x)

¬∀x (Cube(x) ∧ Small(x))

13.5ö

∀x∀y ((Cube(x) ∧ Dodec(y))

→ Larger(y, x))

∀x∀y (Larger(x, y)↔ LeftOf(x, y))


→ LeftOf(y, x))

13.6ö

∀x ((Cube(x) ∧ Large(x))

∨ (Tet(x) ∧ Small(x)))

∀x (Tet(x)→ BackOf(x, c))

∀x¬(Small(x) ∧ Large(x))

∀x (Small(x)→ BackOf(x, c))

(See Exercise 12.9. Notice that we have

included a logical truth as an additional

premise here.)

13.7ö


→ FrontOf(x, y))

∀x (Cube(x)→ ∀y (Dodec(y)

→ FrontOf(x, y)))

13.8ö


→ FrontOf(x, y)))


→ FrontOf(x, y))

13.9ö⋆


→ Larger(x, y))

∀x∀y ((Dodec(x) ∧ Tet(y))

→ Larger(x, y))

∀x∀y ((Cube(x) ∧ Tet(y))

→ Larger(x, y))

Section 13.2

Existential quantifier rules

Recall that in our discussion of informal proofs, existential introduction was a

simple proof step, whereas the elimination of ∃ was a subtle method of proof.

Thus, in presenting our formal system, we begin with the introduction rule.

Chapter 13

Existential quantifier rules / 357

Existential Introduction (∃ Intro):

S(c)...

▷ ∃x S(x)

Here too x stands for any variable, c stands for any individual constant (or

complex term without variables), and S(c) stands for the result of replacing

free occurrences of x in S(x) with c. Note that there may be other occurrences

of c in S(x) as well.

When we turn to the rule of existential elimination, we employ the same,

“boxed constant” device as with universal introduction. If we have proven

∃x S(x), then we introduce a new constant symbol, say c, along with the as-

sumption that the object denoted by c satisfies the formula S(x). If, from this

assumption, we can derive some sentence Q not containing the constant c,

then we can conclude that Q follows from the original premises.

Existential Elimination (∃ Elim):

∃x S(x)...

c S(c)



introduced.Q

▷ Q

Again we think of the notation at the beginning of the subproof as the formal

counterpart of the English “Let c be an arbitrary individual such that S(c).”

The rule of existential elimination is quite analogous to the rule of disjunc-

tion elimination, both formally and intuitively. With disjunction elimination, comparison with

∨ Elimwe have a disjunction and break into cases, one for each disjunct, and estab-

lish the same result in each case. With existential elimination, we can think

of having one case for each object in the domain of discourse. We are required

to show that, whichever object it is that satisfies the condition S(x), the same

result Q can be obtained. If we can do this, we may conclude Q.

To illustrate the two existential rules, we will give a formal counterpart to

the proof given on page 332.

Section 13.2


1. ∀x [Cube(x)→ Large(x)]

2. ∀x [Large(x)→ LeftOf(x, b)]

3. ∃x Cube(x)

4. e Cube(e)

5. Cube(e)→ Large(e) ∀ Elim: 1

6. Large(e) → Elim: 5, 4

7. Large(e)→ LeftOf(e, b) ∀ Elim: 2

8. LeftOf(e, b) → Elim: 7, 6

9. Large(e) ∧ LeftOf(e, b) ∧ Intro: 6, 8

10. ∃x (Large(x) ∧ LeftOf(x, b)) ∃ Intro: 9

11. ∃x (Large(x) ∧ LeftOf(x, b)) ∃ Elim: 3, 4-10

Default and generous uses of the ∃ rules

Defaults for the existential quantifier rules work similarly to those for the

universal quantifier. If you cite a sentence and apply ∃ Intro without typingdefault uses of ∃ rules

a sentence, Fitch will supply a sentence that existentially quantifies the al-

phabetically first name appearing in the cited sentence. When replacing the

name with a variable, Fitch will choose the first variable in the list of variables

that does not already appear in the cited sentence. If this isn’t the name or

variable you want used, you can specify the substitution yourself; for example

“: max > z” will replace max with z and add ∃z to the front of the result.

In a default application of ∃ Elim, Fitch will supply the last sentence in

the cited subproof, providing that sentence does not contain the temporary

name introduced at the beginning of the subproof.

You try it. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

▶ 1. Open the file Existential 1. Look at the goal to see the sentence we are

trying to prove. Then focus on each step in succession and check the step.

Before moving to the next step, make sure you understand why the step

checks out and, more important, why we are doing what we are doing at

that step.

▶ 2. At any empty steps, you should try to predict which sentence Fitch will

provide as a default before you check the step. Notice in particular step

eight, the one that contains “: a > y”. Can you guess what sentence would

Chapter 13

Existential quantifier rules / 359

have been supplied by Fitch had we not specified this substitution? You

could try it if you like.

◀3. When you are done, make sure you understand the completed proof. Save

the file as Proof Existential 1.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Congratulations

As with ∀, Fitch has generous uses of both ∃ rules. ∃ Intro will allow

you to add several existential quantifiers to the front a sentence. For example,

if you have proved SameCol(b, a) you could infer ∃y ∃z SameCol(y, z) in one

step in Fitch. In a like manner, you can use a sentence beginning with more

than one existential quantifier in a single application of ∃ Elim. You do this

by starting a subproof with the appropriate number of boxed constants. If

you then prove a sentence not containing these constants, you may end the

subproof and infer the result using ∃ Elim.

The Add Support Steps command cannot be used with either of the ∃rules.

Remember

The formal rule of ∃ Elim corresponds to the informal method of exis-

tential instantiation.

Exercises

13.10ö

If you skipped the You try it section, go back and do it now. Submit the file Proof Existential 1.

For each of the following arguments, decide whether or not it is valid. If it is, use Fitch to give a formal

proof. If it isn’t, use Tarski’s World to give a counterexample. Remember that in this chapter you are

free to use Taut Con to justify proof steps involving only propositional connectives.

13.11ö

∀x (Cube(x) ∨ Tet(x))

∃x¬Cube(x)

∃x¬Tet(x)

13.12ö

∀x (Cube(x) ∨ Tet(x))

∃x¬Cube(x)

∃xTet(x)

13.13ö

∀y [Cube(y) ∨ Dodec(y)]

∀x [Cube(x)→ Large(x)]

∃x¬Large(x)

∃xDodec(x)

13.14ö

∀x (Cube(x)↔ Small(x))

∃x¬Cube(x)

∃x¬Small(x)

Section 13.2


13.15ö

∃x (Cube(x)→ Small(x))

∀x Cube(x)

∃x Small(x)

13.16ö

∃x∃y Adjoins(x, y)

∀x∀y (Adjoins(x, y)

→ ¬SameSize(x, y))

∃x∃y¬SameSize(y, x)

In our discussion of the informal methods, we observed that the method that introduces new constants

can interact to give defective proofs, if not used with care. The formal system F automatically prevents

these misapplications of the quantifier rules. The next two exercises are designed to show you how the

formal rules prevent these invalid steps by formalizing one of the fallacious informal proofs we gave

earlier.

13.17ö

Here is a formalization of the pseudo-proof given on page 340:

1. ∀x ∃y SameCol(x, y)

2. c

3. ∃y SameCol(c, y) ∀ Elim: 1

4. d SameCol(c, d)

5. SameCol(c, d) Reit: 4

6. SameCol(c, d) ∃ Elim: 3, 4–5

7. ∀x SameCol(x, d) ∀ Intro: 2–6

8. ∃y ∀x SameCol(x, y) ∃ Intro: 7

1. Write this “proof” in a file using Fitch and check it out. You will discover that step

6 is incorrect; it violates the restriction on existential elimination that requires the

constant d to appear only in the subproof where it is introduced. Notice that the other

steps all check out, so if we could make that move, then the rest of the proof would be

fine.

2. Construct a counterexample to the argument to show that no proof is possible.

Submit both files.

13.18ö

Let’s contrast the faulty proof from the preceding exercise with a genuine proof

that ∀x∃y R(x, y) follows from ∃y ∀xR(x, y). Use Fitch to create the following proof.

Chapter 13

Strategy and tactics / 361

1. ∃y ∀x SameCol(x, y)

2. d ∀x SameCol(x, d)

3. c

4. SameCol(c, d) ∀ Elim: 2

5. ∃y SameCol(c, y) ∃ Intro: 4

6. ∀x∃y SameCol(x, y) ∀ Intro: 3–5

7. ∀x ∃y SameCol(x, y) ∃ Elim: 1, 2–6

Notice that in this proof, unlike the one in the previous exercise, both constant symbols c

and d are properly sequestered within the subproofs where they are introduced. Therefore the

quantifier rules have been applied properly. Submit your proof.

Section 13.3

Strategy and tactics

We have seen some rather simple examples of proofs using the new rules. In

more interesting examples, however, the job of finding a proof can get pretty

challenging. So a few words on how to approach these proofs will be helpful.

We have given you a general maxim and two strategies for finding sen-

tential proofs. The maxim—to consider what the various sentences mean—is consider meaning

even more important with the quantifiers. Only if you remember what they

mean and how the formal methods mirror common-sense informal methods

will you be able to do any but the most boring of exercises.

Our first strategy was to try to come up with an informal proof of the informal proof as guide

goal sentence from the premises, and use it to try to figure out how your

formal proof will proceed. This strategy, too, is even more important in proofs

involving quantifiers, but it is a bit harder to apply. The key skill in applying

the strategy is the ability to identify the formal rules implicit in your informal

reasoning. This takes a bit of practice. Let’s work through an example, to see

some of the things you should be looking out for.

Suppose we want to prove that the following argument is valid:

∃x (Tet(x) ∧ Small(x))

∀x (Small(x)→ LeftOf(x, b))

∃x LeftOf(x, b)

Section 13.3


Obviously, the conclusion follows from the given sentences. But ask yourself

how you would prove it, say, to your stubborn roommate, the one who likes

to play devil’s advocate. You might argue as follows:

Look, Bozo, we’re told that there is a small tetrahedron. So we know

that it is small, right? But we’re also told that anything that’s small

is left of b. So if it’s small, it’s got to be left of b, too. So something’s

left of b, namely the small tetrahedron.

Now we don’t recommend calling your roommate “Bozo,” so ignore that bit.

The important thing to notice here is the implicit use of three of our quantifier

rules: ∃ Elim, ∀ Elim, and ∃ Intro. Do you see them?

What indicates the use of ∃ Elim is the “it” appearing in the second

sentence. What we are doing there is introducing a temporary name (in this

case, the pronoun “it”) and using it to refer to a small tetrahedron. That

corresponds to starting the subproof needed for an application of ∃ Elim.

So after the second sentence of our informal proof, we can already see the

following steps in our reasoning (using “c” for “it”):

1. ∃x (Tet(x) ∧ Small(x))

2. ∀x (Small(x)→ LeftOf(x, b))

3. c Tet(c) ∧ Small(c)

4. Small(c) ∧ Elim: 3...

5. ∃x LeftOf(x, b) ??

6. ∃x LeftOf(x, b) ∃ Elim: 3-5

In general, the key to recognizing ∃ Elim is to watch out for any reference to

an object whose existence is guaranteed by an existential claim. The reference

might use a pronoun (it, he, she), as in our example, or it might use a definite

noun phrase (the small tetrahedron), or finally it might use an actual name

(let n be a small tetrahedron). Any of these are signs that the reasoning is

proceeding via existential elimination.

The third and fourth sentences of our informal argument are where the

implicit use of ∀ Elim shows up. There we apply the claim about all small

things to the small tetrahedron we are calling “it.” This gives us a couple

more steps in our formal proof:

Chapter 13





4. Small(c) ∧ Elim: 3

5. Small(c)→ LeftOf(c, b) ∀ Elim: 2

6. LeftOf(c, b) → Elim: 4, 5...

8. ∃x LeftOf(x, b) ?

9. ∃x LeftOf(x, b) ∃ Elim: 3-8

The distinctive mark of universal elimination is just the application of a gen-

eral claim to a specific individual. For example, we might also have said at

this point: “So the small tetrahedron [there’s the specific individual] must be

left of b.”

The implicit use of ∃ Intro appears in the last sentence of the informal

reasoning, where we conclude that something is left of b, based on the fact that

“it,” the small tetrahedron, is left of b. In our formal proof, this application

of ∃ Intro will be done within the subproof, giving us a sentence that we can

export out of the subproof since it doesn’t contain the temporary name c.




4. Small(c) ∧ Elim: 3

5. Small(c)→ LeftOf(c, b) ∀ Elim: 2

6. LeftOf(c, b) → Elim: 4, 5

7. ∃x LeftOf(x, b) ∃ Intro: 6

8. ∃x LeftOf(x, b) ∃ Elim: 1, 3-7

One thing that’s a bit tricky is that in informal reasoning we often leave out

simple steps like ∃ Intro, since they are so obvious. Thus in our example, we

might have left out the last sentence completely. After all, once we conclude

that the small tetrahedron is left of b, it hardly seems necessary to point out

that something is left of b. So you’ve got to watch out for these omitted steps.

Section 13.3


This completes our formal proof. To a trained eye, the proof matches the

informal reasoning exactly. But you shouldn’t feel discouraged if you would

have missed it on your own. It takes a lot of practice to recognize the steps

implicit in our own reasoning, but it is practice that in the end makes us more

careful and able reasoners.

The second strategy that we stressed is that of working backwards: startingworking backward

from the goal sentence and inserting steps or subproofs that would enable us

to infer that goal. It turns out that of the four new quantifier rules, only ∀Intro really lends itself to this technique.

Suppose your goal sentence is of the form ∀x (P(x)→ Q(x)). After survey-

ing your given sentences to see whether there is any immediate way to infer

this conclusion, it is almost always a good idea to start a subproof in which

you introduce an arbitrary name, say c, and assume P(c). Then add a step to

the subproof and enter the sentence Q(c), leaving the rule unspecified. Next,

end the subproof and infer∀x (P(x)→ Q(x)) by ∀ Intro, citing the subproof

in support. When you check this partial proof, an X will appear next to the

sentence Q(c), indicating that your new goal is to prove this sentence.

Remember

1. Always be clear about the meaning of the sentences you are using.

2. A good strategy is to find an informal proof and then try to formalize

it.

3. Working backwards can be very useful in proving universal claims,

especially those of the form ∀x (P(x)→ Q(x)).

4. Working backwards is not useful in proving an existential claim ∃x S(x)

unless you can think of a particular instance S(c) of the claim that

follows from the premises.

5. If you get stuck, consider using proof by contradiction.

A worked example

We are going to work through a moderately difficult proof, step by step, using

what we have learned in this section. Consider the the following argument:

¬∀x P(x)

∃x¬P(x)

Chapter 13


This is one of four such inferences associated with the DeMorgan rules relating

quantifiers and negation. The fact that this inference can be validated in Fis one we will need in our proof of the Completeness Theorem for the system

F in the final chapter. (The other three DeMorgan rules will be given in the

review exercises at the end of this chapter. Understanding this example will

be a big help in doing those exercises.)

Before embarking on the proof, we mention that this inference is one of the

hallmarks of first-order logic. Notice that it allows us to assert the existence

of something having a property from a negative fact: that not everything has

the opposite property.

In the late nineteenth and early twentieth century, the validity of this sort

of inference was hotly debated in mathematical circles. While it seems obvious

to us now, it is because we have come to understand existence claims in a

somewhat different way than some (the so-called “intuitionists”) understood intuitionists

them. While the first-order understanding of ∃xQ(x) is as asserting that some

Q exists, the intuitionist took it as asserting something far stronger: that the

asserter had actually found a Q and proven it to be a Q. Under this stronger

reading, the DeMorgan principle under discussion would not be valid. This

point will be relevant to our proof.

Let us now turn to the proof. Following our strategy, we begin with an

informal proof, and then formalize it.

Proof: Since we are trying to prove an existential sentence, our first

thought would be to use existential introduction, say by proving

¬P(c) for some c. But if we think a bit about what our premise

means, we see there is no hope of proving of any particular thing

that it satisfies ¬P(x). From the fact that not everything satisfies

P(x), we aren’t going to prove of some specific c that ¬P(c). So this

is surely a dead end. (It is also why the intuitionist would not accept

the argument as valid, given his or her understanding of ∃.)

This leaves only one possible route to our desired conclusion: proof

by contradiction. Thus we will negate our desired conclusion and try

to obtain a contradiction. Thus, we assume ¬∃x¬P(x). How can we

hope to obtain a contradiction? Since our only premise is ¬∀x P(x),

the most promising line of attack would be to try for a proof of

∀x P(x) using universal generalization. Thus, let c be an arbitrary

individual in our domain of discourse. Our goal is to prove P(c).

How can we do this? Another proof by contradiction, for if P(c)

were not the case, then we would have ¬P(c), and hence ∃x¬P(x).

But this contradicts our assumption. Hence P(c) is the case. Since

Section 13.3


c was arbitrary, we get ∀x P(x). But this contradicts our premise.

Hence, we get our desired conclusion using the method of proof by

contradiction.

We now begin the process of turning our informal proof into a formal

proof.

You try it. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

▶ 1. Open Quantifier Strategy 1. This contains the skeleton of our proof:

1. ¬∀x P(x)

...

2. ∃x¬P(x)

▶ 2. The first step in our informal proof was to decide to try to give a proof by

contradiction. Formalize this idea by filling in the following:

1. ¬∀x P(x)

2. ¬∃x¬P(x)

...

4. ⊥ ⊥ Intro: ?, ?

5. ∃x¬P(x) ¬ Intro: 2–4

This step will check out because of the generous nature of Fitch’s ¬ Intro

rule, which lets us strip off as well as add a negation.

▶ 3. We next decided to try to contradict ¬∀x P(x) by proving ∀x P(x) using

universal generalization. Formalize this as follows:

1. ¬∀x P(x)

2. ¬∃x¬P(x)

3. c

...

5. P(c) ?

6. ∀x P(x) ∀ Intro: 3–5

7. ⊥ ⊥ Intro: 6, 1

8. ∃x¬P(x) ¬ Intro: 2–7

Chapter 13


◀4. Recall how we proved P(c). We said that if P(c) were not the case, then

we would have ¬P(c), and hence ∃x¬P(x). But this contradicted the as-

sumption at step 2. Formalize this reasoning by filling in the rest of the

proof.

1. ¬∀x P(x)

2. ¬∃x¬P(x)

3. c

4. ¬P(c)

5. ∃x¬P(x) ∃ Intro: 4

6. ⊥ ⊥ Intro: 5, 2

7. ¬¬P(c) ¬ Intro: 4–6

8. P(c) ¬ Elim: 7

9. ∀x P(x) ∀ Intro: 3–8

10. ⊥ ⊥ Intro: 9, 1

11. ∃x¬P(x) ¬ Intro: 2–10

◀5. This completes our formal proof of ∃x¬P(x) from the premise ¬∀x P(x).

Verify your proof and save it as Proof Quantifier Strategy 1.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Congratulations

Exercises

13.19ö

If you skipped the You try it section, go back and do it now. Submit the file Proof Quantifier

Strategy 1.

Recall that in Exercises 12.1–12.3 on page 336, you were asked to give logical analyses of purported

proofs of some arguments involving nonsense predicates. In the following exercises, we return to these

arguments. If the argument is valid, submit a formal proof. If it is invalid, turn in an informal coun-

terexample. If you submit a formal proof, be sure to use the Exercise file supplied with Fitch. In order

to keep your hand in at using the propositional rules, we ask you not to use Taut Con in these proofs.

13.20ö|.

∀x [(Brillig(x) ∨ Tove(x))→ (Mimsy(x) ∧ Gyre(x))]

∀y [(Slithy(y) ∨Mimsy(y))→ Tove(y)]

∃x Slithy(x)

∃x [Slithy(x) ∧Mimsy(x)]

(See Exercise 12.1 on p. 336.)

Section 13.3


13.21ö|.

∀x [Brillig(x)→ (Mimsy(x) ∧ Slithy(x))]

∀y [(Slithy(y) ∨Mimsy(y))→ Tove(y)]

∀x [Tove(x)→ (Outgrabe(x, b) ∧ Brillig(x))]

∀z [Brillig(z)↔ Mimsy(z)]

(See Exercise 12.2 on p. 337.)

13.22ö|.

∀x [(Brillig(x) ∧ Tove(x))→ Mimsy(x)]

∀y [(Tove(y) ∨Mimsy(y))→ Slithy(y)]

∃xBrillig(x) ∧ ∃xTove(x)

∃z Slithy(z)

(See Exercise 12.3, p. 337)

Some of the following arguments are valid, some are not. For each, either use Fitch to give a formal

proof or use Tarski’s World to construct a counterexample. In giving proofs, feel free to use Taut Con

if it helps.

13.23ö

∀y [Cube(y) ∨ Dodec(y)]

∀x [Cube(x)→ Large(x)]

∃x¬Large(x)

∃xDodec(x)

13.24ö

∃x (Cube(x) ∧ Small(x))

∃x Cube(x) ∧ ∃x Small(x)

13.25ö

∃x Cube(x) ∧ ∃x Small(x)

∃x (Cube(x) ∧ Small(x))

13.26ö

∀x (Cube(x)→ Small(x))

∀x (Adjoins(x, b)→ Small(x))

∀x ((Cube(x) ∨ Small(x))

→ Adjoins(x, b))

13.27ö

∀x (Cube(x)→ Small(x))

∀x (¬Adjoins(x, b)→ ¬Small(x))

∀x ((Cube(x) ∨ Small(x))→ Adjoins(x, b))

For each of the following, use Fitch to give a formal proof of the argument. These look simple but some

of them are a bit tricky. Don’t forget to first figure out an informal proof. Use Taut Con whenever it

is convenient but do not use FO Con.

13.28ö⋆

∀x∀y Likes(x, y)

∀x∃y Likes(x, y)

13.29ö⋆

∀x (Small(x)→ Cube(x))

∃x¬Cube(x)→ ∃x Small(x)

∃x Cube(x)

13.30ö⋆

Likes(carl,max)

∀x [∃y (Likes(y, x) ∨ Likes(x, y))

→ Likes(x, x)]

∃x Likes(x, carl)

13.31ö⋆

∀x∀y [Likes(x, y)→ Likes(y, x)]

∃x∀y Likes(x, y)

∀x∃y Likes(x, y)

Chapter 13


The following valid arguments come in pairs. The validity of the first of the pair makes crucial use of the

meanings of the blocks language predicates, whereas the second adds one or more premises, making the

result a first-order valid argument. For the latter, give a proof that does not make use of Ana Con. For

the former, give a proof that uses Ana Con but only where the premises and conclusions of the citation

are literals (including ⊥). You may use Taut Con but do not use FO Con in any of the proofs.

13.32ö

¬∃x (Tet(x) ∧ Small(x))

∀x [Tet(x)→ (Large(x) ∨Medium(x))]

13.33ö

¬∃x (Tet(x) ∧ Small(x))

∀y (Small(y) ∨Medium(y) ∨ Large(y))

∀x [Tet(x)→ (Large(x) ∨Medium(x))]

13.34ö

∀x (Dodec(x)→ SameCol(x, a))

SameCol(a, c)

∀x (Dodec(x)→ SameCol(x, c))

13.35ö

∀x (Dodec(x)→ SameCol(x, a))

SameCol(a, c)

∀x∀y ∀z ((SameCol(x, y) ∧ SameCol(y, z))

→ SameCol(x, z))

∀x (Dodec(x)→ SameCol(x, c))

13.36ö

∀x (Dodec(x)→ LeftOf(x, a))

∀x (Tet(x)→ RightOf(x, a))

∀x (SameCol(x, a)→ Cube(x))

13.37ö

∀x (Dodec(x)→ LeftOf(x, a))

∀x (Tet(x)→ RightOf(x, a))

∀x∀y (LeftOf(x, y)→ ¬SameCol(x, y))

∀x∀y (RightOf(x, y)→ ¬SameCol(x, y))

∀x (Cube(x) ∨ Dodec(x) ∨ Tet(x))

∀x (SameCol(x, a)→ Cube(x))

13.38ö


→ Larger(x, y)))

∀x (Dodec(x)→ ∀y (Tet(y)→ Larger(x, y)))

∃xDodec(x)

∀x (Cube(x)→ ∀y (Tet(y)→ Larger(x, y)))

(Compare this with Exercise 13.9. The

crucial difference is the presence of the

third premise, not the difference in form

of the first two premises.)

13.39ö


→ Larger(x, y)))

∀x (Dodec(x)→ ∀y (Tet(y)

→ Larger(x, y)))

∃xDodec(x)∀x∀y ∀z ((Larger(x, y) ∧ Larger(y, z))

→ Larger(x, z))

∀x (Cube(x)→ ∀y (Tet(y)

→ Larger(x, y)))

Section 13.3


Section 13.4

Soundness and completeness

In Chapter 8 we raised the question of whether the deductive system FT was

sound and complete with respect to tautological consequence. The same issues

arise with the full system F , which contains the rules for the quantifiers and

identity, in addition to the rules for the truth-functional connectives. Here,

the target consequence relation is the notion of first-order consequence, rather

than tautological consequence.

The soundness question asks whether anything we can prove in F fromsoundness of Fpremises P1, . . . ,Pn is indeed a first-order consequence of the premises. The

completeness question asks the converse: whether every first-order consequencecompleteness of Fof a set of sentences can be proven from that set using the rules of F .

It turns out that both of these questions can be answered in the affirma-

tive. Before actually proving this, however, we need to add more precision

to the notion of first-order consequence, and this presupposes tools from set

theory that we will introduce in Chapters 15 and 16. We state and prove

the soundness theorem for first-order logic in Chapter 18. The completeness

theorem for first-order logic is the main topic of Chapter 19.

Section 13.5

Some review exercises

In this section we present more problems to help you solidify your understand-

ing of the methods of reasoning involving quantifiers. We also present some

more interesting problems from a theoretical point of view.

Exercises

Some of the following arguments are valid, some are not. For each, either use Fitch to give a formal

proof or use Tarski’s World to construct a counterexample. In giving proofs, feel free to use Taut Con

if it helps.

13.40ö

∃x Cube(x) ∧ Small(d)

∃x (Cube(x) ∧ Small(d))

13.41ö

∀x (Cube(x) ∨ Small(x))

∀x Cube(x) ∨ ∀x Small(x)

Chapter 13

Some review exercises / 371

13.42ö

∀x Cube(x) ∨ ∀x Small(x)

∀x (Cube(x) ∨ Small(x))

Each of the following is a valid argument of a type discussed in Section 10.3. Use Fitch to give a proof

of its validity. You may use Taut Con freely in these proofs.

13.43ö

¬∀x Cube(x)

∃x¬Cube(x)

13.44ö

¬∃x Cube(x)

∀x¬Cube(x)

13.45ö

∀x¬Cube(x)

¬∃x Cube(x)

13.46ö

(Change of bound variables)

∀x Cube(x)

∀y Cube(y)

13.47ö

(Change of bound variables)

∃xTet(x)

∃y Tet(y)

13.48ö

(Null quantification)

Cube(b)↔ ∀x Cube(b)

13.49ö

∃x P(x)

∀x∀y ((P(x) ∧ P(y))→ x = y)

∃x (P(x) ∧ ∀y (P(y)→ y = x))

13.50ö

∃x (P(x) ∧ ∀y (P(y)→ y = x))

∀x∀y ((P(x) ∧ P(y))→ x = y)

13.51ö⋆⋆

∃x (P(x)→ ∀y P(y))

[Hint: Review your answer to Exer-

cise 12.22 where you should have given

an informal proof of something of this

form.]

13.52ö

¬∃x∀y [E(x, y)↔ ¬E(y, y)]

This result might be called Russell’s

Theorem. It is connected with the fa-

mous result known as Russell’s Paradox,

which is discussed in Section 15.9. In

fact, it was upon discovering this that

Russell invented the Barber Paradox, to

explain his result to a general public.

13.53ö|.

Is ∃x∃y¬LeftOf(x, y) a first-order consequence of ∃x¬LeftOf(x, x)? If so, give a formal proof.

If not, give a reinterpretation of LeftOf and an example where the premise is true and the

conclusion is false.

The next exercises are intended to help you review the difference between first-order satisfiability and

true logical possibility. All involve the four sentences in the file Padoa’s Sentences. Open that file now.

Section 13.5


13.54ö⋆

Any three of the sentences in Padoa’s Sentences form a satisfiable set. There are four sets of three

sentences, so to show this, build four worlds, World 13.54.123, World 13.54.124, World 13.54.134,

and World 13.54.234,where the four sets are true. (Thus, for example, sentences 1, 2 and 4 should

be true in World 13.54.124.)

13.55.⋆

Give an informal proof that the four sentences in Padoa’s Sentences taken together are incon-

sistent.

13.56.⋆

Is the set of sentences in Padoa’s Sentences first-order satisfiable, that is, satisfiable with some

reinterpretation of the predicates other than identity? [Hint: Imagine a world where one of the

blocks is a sphere.]

13.57.⋆

Reinterpret the predicates Tet and Dodec in such a way that sentence 3 of Padoa’s Sentences

comes out true in World 13.54.124. Since this is the only sentence that uses these predicates,

it follows that all four sentences would, with this reinterpretation, be true in this world. (This

shows that the set is first-order satisfiable.)

13.58ö|.⋆⋆

(Logical truth versus non-logical truth in all worlds) A distinction Tarski’s World helps us to

understand is the difference between sentences that are logically true and sentences that are,

for reasons that have nothing to do with logic, true in all worlds. The notion of logical truth

has to do with a sentence being true simply in virtue of the meaning of the sentence, and so

no matter how the world is. However, some sentences are true in all worlds, not because of

the meaning of the sentence or its parts, but because of, say, laws governing the world. We

can think of the constraints imposed by the innards of Tarski’s World as analogues of physical

laws governing how the world can be. For example, the sentence which asserts that there are at

most 12 objects happens to hold in all the worlds that we can construct with Tarski’s World.

However, it is not a logical truth.

Open Post’s Sentences. Classify each sentence in one of the following ways: (A) a logical

truth, (B) true in all worlds that can be depicted using Tarski’s World, but not a logical truth,

or (C) falsifiable in some world that can be depicted by Tarski’s World. For each sentence of

type (C), build a world in which it is false, and save it as World 13.58.x, where x is the number

of the sentence. For each sentence of type (B), use a pencil and paper to depict a world in

which it is false. (In doing this exercise, assume that Medium simply means neither small nor

large, which seems plausible. However, it is not plausible to assume that Cube means neither a

dodecahedron nor tetrahedron, so you should not assume anything like this.)

Chapter 13

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