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

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

  1. $S$ is neither a tautology nor a contradiction
  2. $S$ is a tautology
  3. $S$ is a contradiction
  4. The antecedent of $S$ is logically equivalent to the consequent of $S$
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5 Answers

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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
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2 Comments

and A→¬A, is contingency, right?
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yes
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4 votes
4 votes

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

Refer to the image below.

 

edited by

3 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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1 vote
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.

1 comment

your approach is little longer
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1 vote
1 vote

 

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