Using truth tables, determine whether the following formulas
are logically equivalent or not:

21. [Exercise 21] p ∧ q → p , p ∨ p → r
p q r p ∧ q p ∧ q → p p ∨ p p ∨ p → r (p ∧ q → p) ∧ (p ∨ p → r)
1 1 1 1 1 1 1 1
1 1 0 1 1 1 0 0
1 0 1 0 1 1 1 1
1 0 0 0 1 1 0 0
0 1 1 0 1 0 1 1
0 1 0 0 1 0 1 1
0 0 1 0 1 0 1 1
0 0 0 0 1 0 1 1

NOT LOGICALLY EQUIVALENT. COUNTEREXAMPLE: I(p)= 1, I(q)=1 and I(r)= 0.

22. [Exercise 22] p ∧ (q ∨ r) , (p ∧ q) ∨ (p ∧ r)
p q r q ∨ r p ∧ (q ∨ r) p ∧ q p ∧ r (p ∧ q) ∨ (p ∧ r) p ∧ (q ∨ r) ↔ (p ∧ q) ∨ (p ∧ r)
1 1 1 1 1 1 1 1 1
1 1 0 1 1 1 0 1 1
1 0 1 1 1 0 1 1 1
1 0 0 0 0 0 0 0 1
0 1 1 1 0 0 0 0 1
0 1 0 1 0 0 0 0 1
0 0 1 1 0 0 0 0 1
0 0 0 0 0 0 0 0 1

LOGICALLY EQUIVALENT. The formulas are logically equivalent (Distributive Law).

23. [Exercise 23] p ∨ (q → r) , p → ¬¬(q → ¬r)
p q r ¬r q → r p ∨ (q → r) q → ¬r ¬(q → ¬r) ¬¬(q → ¬r) p →¬¬(q → ¬r) p ∨ (q → r) ∧ p →¬¬(q → ¬r)
1 1 1 0 1 1 0 1 0 0 0
1 1 0 1 0 1 1 0 1 1 1
1 0 1 0 1 1 1 0 1 1 1
1 0 0 1 1 1 1 0 1 1 1
0 1 1 0 0 0 0 1 0 1 0
0 1 0 1 1 1 1 0 1 1 1
0 0 1 0 1 1 1 0 1 1 1
0 0 0 1 1 1 1 0 1 1 1

NOT LOGICALLY EQUIVALENT. COUNTEREXAMPLE: I(p)= 1, I(q)=1, I(r)=1

24. [Exercise 24] (p → q) ∧ (q → r) → (p → r) , p → (q → r)
p q r p → q q → r (p → q) ∧ (p → r) p → r (p → q) ∧ (p → r) → (p → r) (p → (q → r) (p → q) ∧ (p → r) ↔ (p → (q → r)
1 1 1 1 1 1 1 1 1 1
1 1 0 1 0 0 0 1 0 0
1 0 1 0 1 0 1 1 1 1
1 0 0 0 1 0 0 1 1 1
0 1 1 1 1 1 1 1 1 1
0 1 0 1 0 0 1 1 1 1
0 0 1 1 1 1 1 1 1 1
0 0 0 1 1 1 1 1 1 1

NOT LOGICALLY EQUIVALENT. COUNTEREXAMPLE: I(p)=1, I(q)= 1, I(r) = 0

25. [Exercise 25] ¬(p ∨ q) ↔ ¬p ∨ ¬r , ¬(p → q) ↔ p ∧ r
p q r p ∨ q ¬(p ∨ q) ¬p ∨ ¬r ¬(p ∨ q) ↔ ¬p ∨ ¬r p → q ¬(p → q) p ∧ r ¬(p → q) ↔ p ∧ r Final biconditional
1 1 1 1 0 0 1 1 0 1 0 0
1 1 0 1 0 1 0 1 0 0 1 0
1 0 1 1 0 0 1 0 1 1 1 1
1 0 0 1 0 1 0 0 1 0 0 1
0 1 1 1 0 1 0 1 0 0 1 0
0 1 0 1 0 1 0 1 0 0 1 0
0 0 1 0 1 1 1 1 0 0 1 1
0 0 0 0 1 1 1 1 0 0 1 1

NOT LOGICALLY EQUIVALENT. COUNTEREXAMPLE: I(p)=1, I(q)= 1, I(r) = 1

 

Next >>