Tautology of contingency

Question

Comments

given answer is wrong.

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Put p=0 (as we want to check this implication for the case (1$\rightarrow0)\leftrightarrow$0)

p$\rightarrow$~q is always true

If we put r=1, the implication r$\rightarrow$q is dependent on the value of q

So, the conjunction of all the terms in the LHS of implication is not always true or false.

So, it is a contingency i.e.option (B)

Answer 1

Tatuology: all the value are true

Contradiction: all value are false

Contingency: some of the value are ture and some of the false(mix of tautology and contraditcion)

by using truth table:

from the last column, it's neither tautology nor contradiction, it's contingency

Answer 2

Option(B) is correct.

Comments

No, you are wrong. The given propositional formula is contingency

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Yes, you are right.Now I have edited it.