Answer: (1) Contradiction
(2) Tautology
(3) Neither Tautology nor Contradiction
(4) Tautology
Step-by-step explanation:
Given from the option that ,
(1) (P ↔ Q) ↔ ((P ∧ ¬Q) ∨ (¬P ∧ Q))
(2) (P → Q) V (Q →P)
(3) (¬R→((P V Q) → Q)) V R
(4) ((P→Q) ∧ (Q→R)) → (P→R)
From the first option (1), we have that;
(1) (P ↔ Q) ↔ ((P ∧ ¬Q) ∨ (¬P ∧ Q))
P Q ¬P ¬Q P ↔ Q P ∧ ¬Q ¬P ∧Q (P∧¬Q) V (¬P∧Q) result
T T F T F F F F F
T F F T F T F T F
F T T F F F T T F
F F T T T F F F F
From the result in the last column, this is a contradiction.
From option (2) we have that;
(2) (P → Q) V (Q →P)
The resulting table becomes
P Q P →Q Q → P result
T T T T T
T F F T T
F T T F T
F F T T T
It is seen from the table that the result of the last column has all T which implies that it is a Tautology.
From option (3) we have that;
(3) (¬R→((P V Q) → Q)) V R
The resulting table becomes
P Q R ¬R PVQ (PVQ)→Q ¬R→(PVQ→Q) Result
T T T F T T T T
T T F T T T T T
T F T F T F T T
T F F T T F F F
F T T F T T T T
F T F T T T T T
F F T F F T T T
F F F T F T T T
There is a little difference in the last column when compared to others,
Here we can see that the last column contains all T but just one F, this implies that it is neither a Tautology nor Contradiction.
From option (4) we have that;
(4) ((P→Q) ∧ (Q→R)) → (P→R)
The resulting table becomes
P Q R P →Q Q →R (P→Q)∧(Q→R) P→R result
T T T T T T T T
T T F T F F F T
T F T F T F T T
T F F F T F F T
F T T T T T T T
F T F T F F T T
F F T T T T T T
F F F T T T T T
Here we can see that the last column contains all T which implies that it is a Tautology.
i hope this helps, cheers.
