GATE CSE 2021 | Set 2 | Question: 15

Question

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

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

• A. $S$ is neither a tautology nor a contradiction
• B. $S$ is a tautology
• C. $S$ is a contradiction
• D. The antecedent of $S$ is logically equivalent to the consequent of $S$

Comments

nice

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can you do with by taking p=t/f

 

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Antecedent of S is equivalent to Consequent of S.

Hence Option D is right.

S is tautology then option A is correct

Option B,D

 

Answer 1

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

Comments

Here can i say that "If the (A)Antecedent of S is logically equivalent to (C)Consequent of S", ie "A ≡ C" or "A→C", which would always be an tautology.

And then If i prove that S is tautology, it should be good.

 

Is this an another approach? Or Am i missing a point somewhere?

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@Dhaya_Sagar, Can you simplify your doubt ?

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Yes it is tautology and both are equivaoent

Answer 2

Options $(b)$ and $(d)$.

Refer to the image below.

 

Comments

Shouldn’t it be a tautology?

The left hand side and right hand side are saying the same thing.

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Correct. Edited.

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Very interesting approach. Thanks for sharing!!

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on what basis do you assign the values here?

Answer 3

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.

Comments

your approach is little longer

Answer 4

 

…..….….….….…..…..

Answer 5

option b and are correct ..see the solution 

Answer 6

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

Consequent of S :-  (P $\wedge$ Q)’ V (Q → R) $\equiv$ (P’ V Q’) V (Q’ V R) $\equiv$ (P’ V Q’ V R) $\equiv$ ((P $\wedge$ Q)’ V R) $\equiv$ ((P $\wedge$ Q) → R)

=> S : ((P $\wedge$ Q) → R) → ((P $\wedge$ Q) → R) $\equiv$ True.