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Given propositions P, Q, and R, determine if the following propositions are tautologies, contradictions, or neither. (1) (P ↔ Q) ↔ ((P ∧ ¬Q) ∨ (

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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.

User Gordon Gustafson
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