Introductory Logic PHI 120 Presentation: "Truth Tables – Sequents" This PowerPoint Presentation contains a large number of slides, a good many of which are nearly identical. If you print this Presentation, I recommend six or nine slides per page.
Feb 25, 2016
Introductory LogicPHI 120
Presentation: "Truth Tables – Sequents"
This PowerPoint Presentation contains a large number of slides, a good many of which are nearly identical. If you print
this Presentation, I recommend six or nine slides per page.
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