# GATE | Gate IT 2008 | Question 20

Which of the following first order formula is logically valid? Here α(x) is a first order formula with x as a free variable, and β is a first order formula with no free variable.
(A) [β→(∃x,α(x))]→[∀x,β→α(x)]
(B) [∃x,β→α(x)]→[β→(∀x,α(x))]
(C) [(∃x,α(x))→β]→[∀x,α(x)→β]
(D) [(∀x,α(x))→β]→[∀x,α(x)→β]

Explanation:

A formula is logically valid (or simply valid) if it is true in every interpretation. These formulas play a role similar to tautologies in propositional logic. A formula is VALID if no instance of it is false. So, it is enough to give any instance which gives false and prove that our formula is not valid

Choice of this question:
option (a)
[β→(∃x,α(x))] → [∀x,β→α(x)]
LHS: If β is true, then there exists an x for which α(x) is true.
RHS: For all x, if β is true then α(x) is true. This is same as saying if β is true then for all x, α(x) is true. (β⟹∀x,α(x))

So,
RHS⟹LHS and LHS⟹̸ RHS.

Option (b)
[∃x,β→α(x)] → [β→(∀x,α(x))]
LHS: There exists an x such that if β is true then α(x) is true.
RHS: If β is true then for all x, α(x) is true.

So, RHS⟹LHS and LHS⟹̸ RHS.

Option (c)
[(∃x,α(x))→β] → [∀x,α(x)→β]
LHS: If there is an x such that α(x) is true, then β is true.
RHS: For all x, if α(x) is true, then β is true.

Here, both LHS and RHS are same because β is a formula with no free variable which is independent of x. (if β is true for one x, it is true for every x and vice versa).

So, RHS⟹LHS and LHS⟹RHS.

Option (d)
[(∀x,α(x))→β]→[∀x,α(x)→β]
RHS: For every x, if α(x) is true then β is true.

So, RHS⟹LHS and LHS⟹̸ RHS
So, option c is correct one. because for option c, LHS and RHS being equivalent, even if the implication is reversed, it remains valid.

This solution is contributed by Nitika Bansal.

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