Contingency or Not ?

Question

~ $\forall$ x [  P(x) -> (Q(x) v P(x) )  ]

Comments

@Anu sir,

it is, $\exists x(0)$

now predicate is exactly 0(always), which does not depend on value of 'x'

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@anu sir there is no x to check its value.

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oh sory ashwin and nitish my bad. got it

Vs here o is constant. whaterver be x it is not change.

Answer 1

$\sim ∀ x$ $[P(x) \rightarrow (Q(x) \vee P(x))]$

this can also be written as

$\sim$ ∀x$[(P(x) \rightarrow (Q(x) )\vee (P(x)\rightarrow P(x) )]$

$P(x)\rightarrow P(x)$ will alwaysb be true,

$\sim$ ∀x$[(P(x) \rightarrow (Q(x) )\vee True]$ 

$\sim$ ∀x[True]

"This is not the case that for all "x" the expression is True"

We know that, $\sim$∀$x P(x)$ $\equiv ∃x \sim P(x)$

$∃x \sim[True]$

$∃x [False]$

Value of predicate which is false does not depends on what is the value of x. Hence the expression is a Contradiction.