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14CHECKING VALIDITY TREES III

Paal [email protected]

https://sites.google.com/site/paalantonsen/teaching/logica

Formal Logic

The story so far. . .

I In this lecture we will extend the tree method so that we will be able tohandle arguments in predicate logic. This is a method for checking forproofs argument forms (i.e. whether the conclusion is provable from thepremises) by checking whether trees close.

There is a proof for 〈X , A〉 (X ` A) iff there is a closed tree for 〈X , ∼A〉.

A tree is closed if all its branches are closed, and open otherwise.

A branch is (a) closed if a formula A and a formula ∼A (its negation) occurin that branch, (b) open iff it is complete and not closed.

The story so far. . .

I We gave a summary of how to use the tree method:

HOW TO: CHECK WHETHER THERE IS A TREE PROOF

(a) Where 〈X ,A〉 is an argument, write down all the premises X andnegation of conclusion ∼A (the root).

(b) Then apply the rules for the connectives to produce branches on thetree. Each time you apply a rule on a formula, remember to mark thatformula as resolved with X.

(c) If a formula along with its negation occurs in a branch of the the tree,then mark that branch as closed with ×.

There are then two possibilities:

(d) If you thereby produce a tree where all branches are closed, you haveshown that there is a tree proof for 〈X ,A〉 (X ` A).

(e) If you thereby produce a tree where all the formulas are resolved and atleast one branch is open, you have shown that there isn’t a tree proof for〈X ,A〉 (X 0 A).

How do we handle formulas in predicate logic?

I The description of what counted as formulas of predicate logic included∼A, (A & B), (A ∨ B), (A ⊃ B), (A ≡ B). So we can keep the rules:

closure

A...∼A

X

conjunction

(A & B)

AB

∼(A & B)

∼A ∼B

disjunction

(A ∨ B)

A B

∼(A ∨ B)

∼A∼B

negation

∼∼A

A

conditional

(A ⊃ B)

∼A B

∼(A ⊃ B)

A∼B

biconditional

(A ≡ B)

AB

∼A∼B

∼(A ≡ B)

∼AB

A∼B


I The extended tree method needs rules to handle formulas of the form:

Existential (∃x)ANegated existential ∼(∃x)A

Universal (∀x)ANegated universal ∼(∀x)A

I The rules we need must satisfy their intuitive truth conditions:

(∃x)A is true iff A is true for some individual∼(∃x)A is true iff A is not true for any individual

(∀x)A is true iff A is true for every individual∼(∀x)A is true iff A is not true for every individual


(1) (∀x)∼A iff ∼(∃x)A(2) (∃x)∼A iff ∼(∀x)A

(1) (∀x)∼A is true inMiff for every object, ∼A is true inMiff for every object, A is false inM

∼(∃x)A is true inMiff there isn’t some object, s.t. A is true inMiff for every object, A is false inM

(2) (∃x)∼A is true inMiff there is some object, s.t. ∼A is true inMiff there is some object, s.t. A is false inM

∼(∀x)A is true inMiff not for every object, A is true inMiff there is some object, s.t. A is false inM

Trees and predicate logic: universal quantification

I We add the two following rules for the universal quantifier

(∀x)A

A(a/x)

for any name a

X∼(∀x)A

∼A(a/x)

where a is a new name.

We can use any name when resolving (∀x)A because we assume thatthe condition A is supposed to hold for any individual.

We do not check (X) this formula, so we can use it several times.

We use a new name, i.e. one that doesn’t occur in that branch, whenresolving ∼(∀x)A, because we just assume that not all individuals meetcondition A; nothing about which in particular.

Trees and predicate logic: example I

(∀x)(Px ⊃ Qx), ∼Qa; therefore ∼Pa

The rule said: resolve (∀x)A by extending branch with the formula A(a/x)for any name a.



(∀x)(Px ⊃ Qx)∼Qa∼∼Pa




(∀x)(Px ⊃ Qx) /a∼Qa∼∼Pa

Pa ⊃ Qa





XPa ⊃ Qa

∼Pa Qa





XPa ⊃ Qa

∼Pa

⊗

Qa

⊗

All the branches close, so the argument is classified as valid. and that isvery good indeed

Trees and predicate logic: example II

(∀x)Px, (∀x)(Px ⊃ Qx); therefore (∀x)Qx




(∀x)Px(∀x)(Px ⊃ Qx)∼(∀x)Qx




(∀x)Px(∀x)(Px ⊃ Qx)∼(∀x)Qx

The rule said: resolve ∼(∀x)A by extending branch with the formula∼A(a/x) where a is new.



(∀x)Px(∀x)(Px ⊃ Qx)X∼(∀x)Qx /a

∼Qa

The rule said: resolve ∼(∀x)A by extending branch with the formula∼A(a/x) where a is new.



(∀x)Px(∀x)(Px ⊃ Qx)X∼(∀x)Qx /a

∼Qa

The rule said: resolve ∀x)A by extending branch with the formula A(a/x)for any name a.



(∀x)Px /a(∀x)(Px ⊃ Qx)X∼(∀x)Qx /a

∼Qa

Pa




(∀x)Px /a(∀x)(Px ⊃ Qx) /aX∼(∀x)Qx /a

∼Qa

Pa

Pa ⊃ Qa





∼Qa

Pa

Pa ⊃ Qa

∼Pa Qa





∼Qa

Pa

Pa ⊃ Qa

∼Pa

⊗

Qa

⊗


Trees and predicate logic: existential quantification

I We add the two following rules for the existential quantifier

X(∃x)A

A(a/x)

where a is a new name

∼(∃x)A

∼A(a/x)

for any name a

We use a new name, i.e. one that doesn’t occur in that branch, whenresolving (∃x)A, because we just assume that some individual meetcondition A; nothing about which in particular.

We can use any name when resolving ∼(∃x)A because we assume thatthe condition A doesn’t hold for any individal.

We do not check (X) this formula, so we can use it several times.

Trees and predicate logic: example III

(∃x)Px ⊃ Qa, Pb; therefore Qa




(∃x)Px ⊃ QaPb∼Qa




X(∃x)Px ⊃ QaPb∼Qa

∼(∃x)Px Qa





∼(∃x)Px Qa

⊗





∼(∃x)Px Qa

⊗

The rule said: resolve ∼(∃x)A by extending branch with the formula∼A(a/x) for any name a.




∼(∃x)Px /b

∼Pb

Qa

⊗

The rule said: resolve ∼(∃x)A by extending branch with the formula∼A(a/x) for any name a.




∼(∃x)Px /b

∼Pb

⊗

Qa

⊗


Trees and predicate logic: example IV

(∃x)Px, (∀x)(Px ⊃ Qx); therefore (∃x)Qx




(∃x)Px(∀x)(Px ⊃ Qx)∼(∃x)Qx




(∃x)Px(∀x)(Px ⊃ Qx)∼(∃x)Qx

The rule said: resolve (∃x)A by extending branch with the formulaA(a/x), where a is a new name.



X(∃x)Px /a(∀x)(Px ⊃ Qx)∼(∃x)Qx

Pa




X(∃x)Px /a(∀x)(Px ⊃ Qx)∼(∃x)Qx

Pa

The rule said: resolve (∀x)A by extending branch with the formulaA(a/x), for any name a.



X(∃x)Px /a(∀x)(Px ⊃ Qx) /a∼(∃x)Qx

Pa

Pa ⊃ Qa




X(∃x)Px /a(∀x)(Px ⊃ Qx) /a∼(∃x)Qx

Pa

Pa ⊃ Qa

The rule said: resolve ∼(∃x)A by extending branch with the formula∼A(a/x), for any name a.



X(∃x)Px /a(∀x)(Px ⊃ Qx) /a∼(∃x)Qx /a

Pa

Pa ⊃ Qa

∼Qa





Pa

XPa ⊃ Qa

∼Qa

∼Pa Qa





Pa

XPa ⊃ Qa

∼Qa

∼Pa

⊗

Qa

⊗


Restating the new rules

I Tomorrow we will continue with the tree method for predicate logic, so itis useful to memorize the four new rules:

(∀x)A

A(a/x)

for any name a

X∼(∀x)A

∼A(a/x)

where a is a new name.

X(∃x)A

A(a/x)

where a is a new name

∼(∃x)A

∼A(a/x)

for any name a

A final example

(∀x)(Px ⊃ Rxx), (∃x)Px ; therefore (∃x)Rxx


A final example


(∀x)(Px ⊃ Rxx)(∃x)Px∼(∃x)Rxx


A final example


(∀x)(Px ⊃ Rxx)X (∃x)Px /a∼(∃x)Rxx

Pa


A final example


(∀x)(Px ⊃ Rxx) /aX (∃x)Px /a∼(∃x)Rxx

Pa

Pa ⊃ Raa


A final example



Pa

XPa ⊃ Raa

∼Pa Raa


A final example



Pa

XPa ⊃ Raa

∼Pa

⊗

Raa


A final example


(∀x)(Px ⊃ Rxx) /aX (∃x)Px /a∼(∃x)Rxx /a

Pa

XPa ⊃ Raa

∼Pa

⊗

Raa

∼Raa


A final example


(∀x)(Px ⊃ Rxx) /aX (∃x)Px /a∼(∃x)Rxx /a

Pa

XPa ⊃ Raa

∼Pa

⊗

Raa

∼Raa

⊗

All the branches close (i.e. the tree closes), so the argument is classifiedas valid by the tree method.

paal antonsen [email protected] a formal logic · this is a method for checking for ... (px ˙qx),...

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