Truth tables part 2

1. LOGICAL EQUIVALENCE
• To show that a set of sentences is logically equivalent:
You need to construct complete truth table
• To show that a set of sentences is NOT logically equivalent:
You need only one line – where they are different

2. • To show that the sentences are not logically equivalent:
A B A A V B
0 1
A B A A V B
0 0 1
A B A A V B
0 0 0 1
A B A A V B
0 1 0 0 1 1
We proved that they are not
logically equivalent!

3. CONSISTENCY
• To show that a set of sentences is consistent, you have to show that there is at least
one line when they are all 1
• To show that a set of sentences is inconsistent, you have to construct the whole truth
table

4. • Show that the set of sentences is consistent
(E ↔ H) ↔ E and H & ¬ E
E H (E ↔ H) ↔ E H & ¬ E
1 1
E H (E ↔ H) ↔ E H & ¬ E
1 1 1 1 1
E H (E ↔ H) ↔ E H & ¬ E
0 1 1 1 1 1 0
E H (E ↔ H) ↔ E H & ¬ E
0 1 0 0 1 1 0 1 1 1 0
It’s consistent!

5. FINALLY, ARGUMENTS
• To show that an argument is VALID, you have to construct the complete truth table
• In order to show that an argument is invalid, you have to show that there is a line when
the premises are true and the conclusion is false

6. • Show that the following argument is invalid, using short method:
¬ (B V D)
¬H
_____
B
B D H ¬ (B V D) ¬ H B
1 1 0
B D H ¬ (B V D) ¬ H B
1 0 1 0
B D
o
H ¬ (B V D) ¬ H B
0 0 1 0 0 0 1 0
B D H ¬ (B V D) ¬ H B
0 0 0 1 0 0 0 1 0 0
We proved that it’s
invalid!

7. • HW:
1. “forall x” – rest of exercises for Part E and D
2. Rest of team exercises on short method (#3 for Part A, B, C, D, E)

8. SHORT METHOD-HW ANSWERS
• Part A
• 1. ((F V H) V (¬F ↔H)
F H ( (F V H) V (¬ F ↔ H)
0
F H ( (F V H) V (¬ F ↔ H)
0 0 0
F H ( (F V H) V (¬ F ↔ H)
0 0 0 0 0 0 0
F H ( (F V H) V (¬ F ↔ H)
0 0 0 0 0 0 1 0 0 0
We have proven that
it’s not a tautology!

9. • (F V H) V ¬ (¬F → H)
F H ( (F V H) V ¬ (¬ F → H)
0
F H ( (F V H) V ¬ (¬ F → H)
0 0 0 1
F H ( (F V H) V ¬ (¬ F → H)
0 0 0 0 0 0 0 1 0 1 0
Contradiction
 we have to
construct truth
table

10. • ¬ A → [(B & A) → C]
A B C ¬ A → [(B & A) → C]
0
A B C ¬ A → [(B & A) → C]
0 1 0 0 0
A B C ¬ A → [(B & A) → C]
0 0 1 0 0 1 0 0
A B C ¬ A → [(B & A) → C]
0 0 1 0 0 1 0 0 0
Contradiction 
have to construct
complete truth
table

11. PART B
• A ↔ (B ↔ A)
A B A ↔ (B ↔ A)

12. • [(F & G) → (C & ¬C)] & F
C F G [ (F & G) → (C & ¬ C)] & F