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69 votes
69 votes

Which one of the following well-formed formulae in predicate calculus is NOT valid ?

  1. $(\forall _{x} p(x) \implies \forall _{x} q(x)) \implies (\exists _{x} \neg p(x) \vee \forall _{x} q(x))$
  2. $(\exists x p(x) \vee \exists x q (x)) \implies \exists x (p(x) \vee q (x))$
  3. $\exists x (p(x) \wedge q(x)) \implies (\exists x p(x) \wedge \exists x q(x))$
  4. $\forall x (p(x) \vee q(x)) \implies (\forall x p(x) \vee \forall x q(x))$
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6 Answers

4 votes
4 votes

We can also prove this by making T or F combination & the trick we've to apply is T->F which is F i.e not valid.

for check validity it's easy to prove not valid by 1 combination(T->F) rather than proving valid by 3 combinations(T->T, F->T, F->F).

option A

  X P(X) Q(X)
X1 T T
X2 T T

So, using this combination we can't get T -> F, i.e option A is valid.

always choose a combination such that LHS of  implication will be T & using that combination check whether RHS will be F or not.

option B

X P(X) Q(X)
X1 T F
X2 F T

We can't get T -> F

option C

X P(X) Q(X)
X1 T T
X2 F F

NO T -> F

OPTION D

X P(X) Q(X)
X1 T F
X2 F T

So, here we get T -> F  i.e it's not a valid one

4 votes
4 votes

option A is not well explained in most all the answers . so i will try explaining that in detail.

Answer:

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