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Question

The false statement in the following is


A
p(p) is contradiction
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B
(pq)(qp) is a contradiction
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C
(p)p is a tautology
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D
p(p) is a tautology
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Solution

The correct option is B $$\left(p \rightarrow q \right) \leftrightarrow \left( q \rightarrow p \right) $$ is a contradiction
(A)
If $$p$$ is true then $$\sim p$$ is false 
$$\implies p\wedge \left( \sim p \right) $$ is false
If $$p$$ is false then $$\sim p$$ is true
$$\implies p\wedge \left( \sim p \right) $$ is false
Hence, $$p\wedge \left( \sim p \right) $$ is a contradiction

(B)
If $$p$$ is true and $$q$$ is false then $$p\rightarrow q$$ is false and $$q\rightarrow p$$ is true
So, $$\left(p \rightarrow q \right) \leftrightarrow \left( q \rightarrow p \right) $$ is false
If $$p$$ is false and $$q$$ is true then $$p\rightarrow q$$ is true and $$q\rightarrow p$$ is false
So, $$\left(p \rightarrow q \right) \leftrightarrow \left( q \rightarrow p \right) $$ is false
If both $$p$$ and $$q$$ are same(i.e. true of false) then $$p\rightarrow q$$ is true.
So, $$\left( p \rightarrow q \right) \leftrightarrow \left( q \rightarrow p \right) $$ is true
Thus, the truth value of $$\left( p \rightarrow q \right) \leftrightarrow \left( q \rightarrow p \right) $$ is neither tautology nor contradiction.

(C)
$$\sim \left( \sim p \right) \leftrightarrow p$$ is same as $$p \leftrightarrow p$$ which is always true
Thus, $$\sim \left( \sim p \right) \leftrightarrow p$$ is a tautology

(D)
$$p\vee \left( \sim p \right) $$ is always true
Hence, $$p\vee \left( \sim p \right) $$ is a tautology
Hence, only B is false.

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