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6 votes

Let $p$ and $q$ be two propositions. Consider the following two formulae in propositional logic.

- $S_1: (\neg p\wedge(p\vee q))\rightarrow q$
- $S_2: q\rightarrow(\neg p\wedge(p\vee q))$

Which one of the following choices is correct?

- Both $S_1$ and $S_2$ are tautologies.
- $S_1$ is a tautology but $S_2$ is not a tautology
- $S_1$ is not a tautology but $S_2$ is a tautology
- Neither $S_1$ nor $S_2$ is a tautology

7 votes

Best answer

$S_1 : \neg p \wedge ( p \vee q ) \to q$

If consequence is false and hypothesis is true, then we will get False in the truth table.

Lets assume $q$ is false. So consequence is FALSE.

Can it make Hypothesis TRUE?

Hypothesis: $\neg p \wedge ( p \vee q ) \equiv \neg p \wedge ( p \vee \text{FALSE} ) \equiv \neg p \wedge ( p ) \equiv \text{FALSE}.$

Hypothesis can’t be true, So we can’t get False in the Truth Table.

∴ $S_1$ is Tautology.

$S_2: q \to \neg p \wedge ( p \vee q )$

If hypothesis is true and consequence is false, then we will get False in the truth table.

Lets assume $q$ is True, So Hypothesis is TRUE.

Can it make Consequence FALSE ?

Consequence: $\neg p \wedge ( p \vee q ) \equiv \neg p \wedge ( p \vee \text{TRUE} ) \equiv \neg p \wedge ( \text{TRUE} ) \equiv \neg p$

Consequence can be false and so we can get False in the Truth Table.

∴ $S_2$ is not Tautology.

Correct Option: B

If consequence is false and hypothesis is true, then we will get False in the truth table.

Lets assume $q$ is false. So consequence is FALSE.

Can it make Hypothesis TRUE?

Hypothesis: $\neg p \wedge ( p \vee q ) \equiv \neg p \wedge ( p \vee \text{FALSE} ) \equiv \neg p \wedge ( p ) \equiv \text{FALSE}.$

Hypothesis can’t be true, So we can’t get False in the Truth Table.

∴ $S_1$ is Tautology.

$S_2: q \to \neg p \wedge ( p \vee q )$

If hypothesis is true and consequence is false, then we will get False in the truth table.

Lets assume $q$ is True, So Hypothesis is TRUE.

Can it make Consequence FALSE ?

Consequence: $\neg p \wedge ( p \vee q ) \equiv \neg p \wedge ( p \vee \text{TRUE} ) \equiv \neg p \wedge ( \text{TRUE} ) \equiv \neg p$

Consequence can be false and so we can get False in the Truth Table.

∴ $S_2$ is not Tautology.

Correct Option: B

4 votes

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

$S_1: (\neg p \land(p\lor q)) \rightarrow q$

$p$ | $q$ | $\neg p$ | $p\lor q$ | $(\neg p \land(p\lor q))$ | $(\neg p \land(p\lor q))\rightarrow q$ |

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

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

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

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

$\therefore S_1$ is tautology.

$S_2: q\rightarrow (\neg p \land(p\lor q))$

$p$ | $q$ | $\neg p$ | $p\lor q$ | $\neg p \land(p \lor q)$ | $q \rightarrow (\neg p \land(p \lor q))$ |

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

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

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

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

$\therefore S_2$ is not tautology.

Option $B$ is correct.