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42 votes
42 votes

Which one of these first-order logic formulae is valid?

  1. $\forall x\left(P\left(x\right) \implies Q\left(x\right)\right) \implies \left(∀xP\left(x\right)\implies \forall xQ\left(x\right)\right)$
  2. $\exists x\left(P\left(x\right) \vee Q\left(x\right)\right)\implies \left(\exists xP\left(x\right)\implies \exists xQ\left(x\right)\right)$
  3. $\exists x\left(P\left(x\right) \wedge Q\left(x\right)\right) \iff \left(\exists xP\left(x\right) \wedge \exists xQ\left(x\right)\right)$
  4. $\forall x \exists y P\left(x, y\right)\implies \exists y \forall x P\left(x, y\right)$
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5 Answers

2 votes
2 votes

1. If there exist x for which P is true then for all such x Q will be true. 

Hence is P is true for all x then Q is true for all x (Correct)

2. there exist x for which either of P or Q is true. Then if for all x if P is true then for all x Q is true.

Not Necessary  :  consider a case when Q is false for all x , and P is true for all x.

3. here operator is <====> , we can consider any of LHS or RHS first. 

lets consider RHS first If there some x for which P is true and there is some other value of x (Note we are not using same value of x for both) for which Q is true . Then it is not necessary  that  there exist common value of x for which P and Q both are true.

4.    

 

For every value in X there exist some value in Y to which it is related to.

But there does not exist any value for Y which is related to all values of X

Answer:

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