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Choose the correct choice(s) regarding the following proportional logic assertion $S$:

$$S: (( P \wedge Q) \rightarrow R) \rightarrow (( P \wedge Q) \rightarrow (Q \rightarrow R))$$

- $S$ is neither a tautology nor a contradiction
- $S$ is a tautology
- $S$ is a contradiction
- The antecedent of $S$ is logically equivalent to the consequent of $S$

19 votes

Best answer

Antecedent of $S: (P \wedge Q) \to R$

$\qquad \equiv \neg ( P\wedge Q) \vee R$

$\qquad \equiv \neg P \vee \neg Q \vee R$

Consequent of $S : (P \wedge Q) \to (Q \to R)$

$\qquad\equiv (P \wedge Q) \to (\neg Q \vee R)$

$\qquad\equiv \neg( P\wedge Q) \vee (\neg Q \vee R)$

$\qquad\equiv \neg P \vee \neg Q \vee (\neg Q \vee R)$

$\qquad\equiv \neg P \vee \neg Q \vee R$

Antecedent of $S$ is equivalent to Consequent of $S.$ Hence Option D is right.

$A \to A$ is a Tautology. Hence options A and C are wrong and option B is right.

Correct Answer: B;D

$\qquad \equiv \neg ( P\wedge Q) \vee R$

$\qquad \equiv \neg P \vee \neg Q \vee R$

Consequent of $S : (P \wedge Q) \to (Q \to R)$

$\qquad\equiv (P \wedge Q) \to (\neg Q \vee R)$

$\qquad\equiv \neg( P\wedge Q) \vee (\neg Q \vee R)$

$\qquad\equiv \neg P \vee \neg Q \vee (\neg Q \vee R)$

$\qquad\equiv \neg P \vee \neg Q \vee R$

Antecedent of $S$ is equivalent to Consequent of $S.$ Hence Option D is right.

$A \to A$ is a Tautology. Hence options A and C are wrong and option B is right.

Correct Answer: B;D

1 vote

A **tautology** is a proposition that is always true for every value of its propositional variables.

A **contradiction** is a proposition that is always false for every value of its propositional variables.

**Logically equivalent**: Compound propositions with the same truth value in all possible cases are logically equivalent.

in another way, the compound propositions $p$ and $q$ are called logically equivalent if $p\leftrightarrow q$ is a tautology.

$P$ |
$Q$ |
$R$ | $P\land Q$ | $((P\land Q)\rightarrow R)$ | $(Q\rightarrow R)$ | $((P\land Q)\rightarrow(Q\rightarrow R))$ | $((P\land Q)\rightarrow R)\rightarrow((P\land Q)\rightarrow(Q\rightarrow R))$ |

$T$ | $T$ | $T$ | $T$ | $T$ | $T$ | $T$ | $T$ |

$T$ | $T$ | $F$ | $T$ | $F$ | $F$ | $F$ |
$T$ |

$T$ | $F$ | $T$ | $F$ | $T$ | $T$ | $T$ |
$T$ |

$T$ | $F$ | $F$ | $F$ |
$T$ |
$T$ | $T$ | $T$ |

$F$ | $T$ | $T$ | $F$ | $T$ | $T$ | $T$ | $T$ |

$F$ | $T$ | $F$ | $F$ | $T$ | $F$ | $T$ | $T$ |

$F$ | $F$ | $T$ | $F$ | $T$ | $T$ | $T$ | $T$ |

$F$ | $F$ | $F$ | $F$ | $T$ | $T$ | $T$ | $T$ |

It’s clearly visible from the above truth table that $S$ is a tautology as all its values are true. so option $B$ is true and $A$ is false.

$S$ is not a contradiction because all its values are true. so option $C$ is false.

The antecedent of $S$ is logically equivalent to the consequent of $S$, this option is true.

From the above truth table, we can see that $((P\land Q)\rightarrow R)$ $\equiv$ $((P\land Q)\rightarrow(Q\rightarrow R))$ and their truth tables are the same.

Option $B,D$ are correct.