Introductory LogicPHI 120

Presentation: "Truth Tables – Sequents"

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Homework1. Study Allen/Hand Logic Primer– Sec. 1.1, p. 1-2: “validity”– Sec. 2.2, p. 43-4, “validity” & “invalidating assignment

2. Complete Ex. 2.1, p. 42: i-x

Turn to page 40 in The Logic Primeralso take out TTs handout

Truth TablesTruth Value of Sentences

Section 2.1(quick review)

PTF

Atomic sentence

Simple

Truth TablesComplex Sentences

See bottom of Truth Tables Handout

Φ ~ ΦTF

~Φ• False?

~Φ• False – if the statement being negated (Φ) is True

Φ ~ ΦT FF T

Φ & Ψ• False?

Φ Ψ Φ & ΨT TT FF TF F

Φ & Ψ• False – if one or both conjuncts are False

Φ Ψ Φ & ΨT TT F FF T FF F F

Φ & Ψ• False – if one or both conjuncts are False

Φ Ψ Φ & ΨT T TT F FF T FF F F

Φ v Ψ• False?

Φ Ψ Φ v ΨT TT FF TF F

Φ v Ψ• False – only if both disjuncts are False

Φ Ψ Φ v ΨT TT FF TF F F

Φ v Ψ• False – only if both disjuncts are False

Φ Ψ Φ v ΨT T TT F TF T TF F F

Φ -> Ψ• False?

Φ Ψ Φ -> ΨT TT FF TF F

Φ -> Ψ• False – if antecedent is True and consequent is False

Φ Ψ Φ -> ΨT TT F FF TF F

Φ -> Ψ• False – if antecedent is True and consequent is False

Φ Ψ Φ -> ΨT T TT F FF T TF F T

Φ <-> Ψ• False?

Φ Ψ Φ <-> ΨT TT FF TF F

Φ <-> Ψ• False – if the two conditions have a different truth value

Φ Ψ Φ <-> ΨT TT F FF T FF F

Φ <-> Ψ• False – if the two conditions have a different truth value

Φ Ψ Φ <-> ΨT T TT F FF T FF F T

Φ v ~Φ

(P & ~Q) v ~(P & ~Q)Identify the main connective.

How many atomic sentences are in this WFF?

Note the binary structure

(P & ~Q) v ~(P & ~Q)Φ v ~Φ

P Q (P & ~ Q) v ~ (P & ~ Q)

Determine the number of rows for the WFF or the sequent as a whole

P Q (P & ~ Q) v ~ (P & ~ Q)

1 2 3 4 5 6 7 8 9 10 11 12

Determine the number of rows for the WFF or the sequent as a whole

(P & ~Q) v ~(P & ~Q)

TT Method in a Nutshell

Determine truth-values of:

1. atomic statements2. negations of atomics

3. inside parentheses4. negation of the parentheses

5. any remaining connectives

(P & ~Q) v ~(P & ~Q)Φ v ~Φ

P Q (P & ~ Q) v ~ (P & ~ Q)

1 2 3 4 5 6 7 8 9 10 11 12

Step 3 on HandoutFill in left main column first.

(P & ~Q) v ~(P & ~Q)Φ v ~Φ

P Q (P & ~ Q) v ~ (P & ~ Q)TTFF1 2 3 4 5 6 7 8 9 10 11 12

Step 3 on HandoutFill in left main column first.

(P & ~Q) v ~(P & ~Q)Φ v ~Φ

P Q (P & ~ Q) v ~ (P & ~ Q)T TT FF TF F1 2 3 4 5 6 7 8 9 10 11 12

Step 3 on HandoutFill in left main column first.

(P & ~Q) v ~(P & ~Q)Φ v ~Φ

P Q (P & ~ Q) v ~ (P & ~ Q)T TT FF TF F1 2 3 4 5 6 7 8 9 10 11 12

Step 4 on HandoutAssign truth-values for negation of simple statements

(P & ~Q) v ~(P & ~Q)Φ v ~Φ

P Q (P & ~ Q) v ~ (P & ~ Q)T T FT F TF T FF F T1 2 3 4 5 6 7 8 9 10 11 12

Step 4 on HandoutAssign truth-values for negation of simple statements

(P & ~Q) v ~(P & ~Q)Φ v ~Φ

P Q (P & ~ Q) v ~ (P & ~ Q)T T F FT F T TF T F FF F T T1 2 3 4 5 6 7 8 9 10 11 12

Step 4 on HandoutAssign truth-values for negation of simple statements

(P & ~Q) v ~(P & ~Q)Φ v ~Φ

P Q (P & ~ Q) v ~ (P & ~ Q)T T F FT F T TF T F FF F T T1 2 3 4 5 6 7 8 9 10 11 12

Step 5 on HandoutAssign truth-values for innermost binary connectives

When is a conjunction (an “&” statement) false?

(P & ~Q) v ~(P & ~Q)Φ v ~Φ

P Q (P & ~ Q) v ~ (P & ~ Q)T T F F FT F T TF T F FF F T T1 2 3 4 5 6 7 8 9 10 11 12

When is a conjunction (an “&” statement) false?

(P & ~Q) v ~(P & ~Q)Φ v ~Φ

P Q (P & ~ Q) v ~ (P & ~ Q)T T F F FT F T TF T F FF F T T1 2 3 4 5 6 7 8 9 10 11 12

When is a conjunction (an “&” statement) false?

(P & ~Q) v ~(P & ~Q)Φ v ~Φ

P Q (P & ~ Q) v ~ (P & ~ Q)T T F F FT F T TF T F F FF F T T1 2 3 4 5 6 7 8 9 10 11 12

When is a conjunction (an “&” statement) false?

(P & ~Q) v ~(P & ~Q)Φ v ~Φ

P Q (P & ~ Q) v ~ (P & ~ Q)T T F F FT F T TF T F F FF F F T T1 2 3 4 5 6 7 8 9 10 11 12

When is a conjunction (an “&” statement) false?

(P & ~Q) v ~(P & ~Q)Φ v ~Φ

P Q (P & ~ Q) v ~ (P & ~ Q)T T F F FT F T T TF T F F FF F F T T1 2 3 4 5 6 7 8 9 10 11 12

Step 5 on HandoutAssign truth-values for innermost binary connectives

(P & ~Q) v ~(P & ~Q)Φ v ~Φ

P Q (P & ~ Q) v ~ (P & ~ Q)T T F F F FT F T T T TF T F F F FF F F T F T1 2 3 4 5 6 7 8 9 10 11 12

(P & ~Q) v ~(P & ~Q)Φ v ~Φ

P Q (P & ~ Q) v ~ (P & ~ Q)T T F F F FT F T T T TF T F F F FF F F T F T1 2 3 4 5 6 7 8 9 10 11 12

Step 6a on HandoutAssign truth-values for negation of compounds

(P & ~Q) v ~(P & ~Q)Φ v ~Φ

P Q (P & ~ Q) v ~ (P & ~ Q)T T F F T F FT F T T F T TF T F F T F FF F F T T F T1 2 3 4 5 6 7 8 9 10 11 12

Step 6a on HandoutAssign truth-values for negation of compounds

(P & ~Q) v ~(P & ~Q)Φ v ~Φ

P Q (P & ~ Q) v ~ (P & ~ Q)T T F F T F FT F T T F T TF T F F T F FF F F T T F T1 2 3 4 5 6 7 8 9 10 11 12

Step 6b on HandoutAssign truth-values for remaining

When is a disjunction (a “v” statement) false?

(P & ~Q) v ~(P & ~Q)Φ v ~Φ

P Q (P & ~ Q) v ~ (P & ~ Q)T T F F T F FT F T T F T TF T F F T F FF F F T T F T1 2 3 4 5 6 7 8 9 10 11 12

Step 6b on HandoutAssign truth-values for remaining

(P & ~Q) v ~(P & ~Q)Φ v ~Φ

P Q (P & ~ Q) v ~ (P & ~ Q)T T F F T F FT F T T F T TF T F F T F FF F F T T F T1 2 3 4 5 6 7 8 9 10 11 12

When is a disjunction (a “v” statement) false?

(P & ~Q) v ~(P & ~Q)Φ v ~Φ

P Q (P & ~ Q) v ~ (P & ~ Q)T T F F T F FT F T T F T TF T F F T F FF F F T T F T1 2 3 4 5 6 7 8 9 10 11 12

When is a disjunction (a “v” statement) false?

(P & ~Q) v ~(P & ~Q)Φ v ~Φ

P Q (P & ~ Q) v ~ (P & ~ Q)T T F F T F FT F T T F T TF T F F T F FF F F T T F T1 2 3 4 5 6 7 8 9 10 11 12

When is a disjunction (a “v” statement) false?

(P & ~Q) v ~(P & ~Q)Φ v ~Φ

P Q (P & ~ Q) v ~ (P & ~ Q)T T F F T F FT F T T F T TF T F F T F FF F F T T F T1 2 3 4 5 6 7 8 9 10 11 12

When is a disjunction (a “v” statement) false?

(P & ~Q) v ~(P & ~Q)Φ v ~Φ

P Q (P & ~ Q) v ~ (P & ~ Q)T T F F T T F FT F T T T F T TF T F F T T F FF F F T T T F T1 2 3 4 5 6 7 8 9 10 11 12

The values under the governing connective are all T’s.

CLASSIFYING SENTENCESTTs: Sentences

p. 47-8: “tautology,” “inconsistency & contingent”

P Q (P & ~ Q) v ~ (P & ~ Q)T T F F T F FT F T T F T TF T F F T F FF F F T T F T1 2 3 4 5 6 7 8 9 10 11 12

Tautologies• Only Ts under main operator• Necessarily true

Look Under the Main Connective Φ v Ψ

P Q (P & ~ Q) v ~ (P & ~ Q)T T F F T T F FT F T T T F T TF T F F T T F FF F F T T T F T1 2 3 4 5 6 7 8 9 10 11 12

Tautologies• Only Ts under main operator• Necessarily true

Look Under the Main Connective Φ v Ψ

P Q ~ ((P & ~ Q) v ~ (P & ~ Q))T T F F T T F FT F T T T F T TF T F F T T F FF F F T T T F T1 2 3 4 5 6 7 8 9 10 11 12 13

Inconsistencies• Only Fs under main operator• Necessarily false

Look Under the Main Connective

~Φ

P Q ~ ((P & ~ Q) v ~ (P & ~ Q))T T F F F T T F FT F F T T T F T TF T F F F T T F FF F F F T T T F T1 2 3 4 5 6 7 8 9 10 11 12 13

Inconsistencies• Only Fs under main operator• Necessarily false

Look Under the Main Connective

~Φ

P Q P & ~ QT T FT F TF T FF F T1 2 3 4 5 6

Contingencies• At least one T and one F under main operator• Sometime true, sometime false

Look Under the Main Connective Φ & Ψ

P Q P & ~ QT T F FT F T TF T F FF F F T1 2 3 4 5 6

Contingencies• At least one T and one F under main operator• Sometime true, sometime false

Look Under the Main Connective Φ & Ψ

Truth TablesSection 2.2

Sequents

“turnstile”

(conclusion indicator)

P -> Q, Q ⊢ P Premise(s) ⊢ Conclusion

TESTING FOR VALIDITYTTs: Sequents

Testing for Validity I

• The Invalidating Assignment– Conclusion: False– All Premises: True

Φ -> Ψ, Ψ ⊢ Φ– The TT will contain an invalidating assignment

(Invalid form: “Affirming the consequent”)

“Affirming the Consequent”P Q P -> Q , Q ⊢ P

Φ -> Ψ , Ψ ⊢ Φ

“Affirming the Consequent”P Q P -> Q , Q ⊢ P

“Affirming the Consequent”P Q P -> Q , Q ⊢ P

TT Method in a Nutshell

Determine truth-values of:

1. atomic statements2. negations of atomics

3. inside parentheses4. negation of the parentheses

5. any remaining connectives

“Affirming the Consequent”P Q P -> Q , Q ⊢ PTTFF

“Affirming the Consequent”P Q P -> Q , Q ⊢ PT TT FF TF F

“Affirming the Consequent”P Q P -> Q , Q ⊢ PT TT FF TF F

Always circle the governing connective in each sentence.

“Affirming the Consequent”P Q P -> Q , Q ⊢ PT TT F FF TF F

“Affirming the Consequent”P Q P -> Q , Q ⊢ PT T TT F FF T TF F T

“Affirming the Consequent”P Q P -> Q , Q ⊢ PT T TT F FF T T T FF F T F F

“Affirming the Consequent”P Q P -> Q , Q ⊢ PT T TT F FF T T T FF F T

InvalidIf invalidating assignment, then argument is:

“Affirming the Consequent”P Q P -> Q , Q ⊢ PT T TT F FF T TF F T

Circle the invalidating assignment!

Homework1. Study Allen/Hand Logic Primer– Sec. 1.1, p. 1-2: “validity”– Sec. 2.2, p. 43-4, “validity” & “invalidating assignment

2. Complete Ex. 2.1, p. 42: i-x