Which of the following predicate logic is true?

Question

$$S_1: \forall x (f(x) \rightarrow g(x)) \Leftrightarrow \forall x f(x) \rightarrow \forall x g(x)$$

$$S_2: \exists x (f(x) \rightarrow g(x)) \Leftrightarrow \forall x f(x) \rightarrow \exists x g(x)$$

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I got $S_2$ true but for $S_1$, I think only reverse implication exists but not the forward implication.

Am I doing something wrong here ?

Answer 1

Actually, for the first one the forward implication does hold, and the reverse does not:

Proof for forward:

1. $\forall x (f(x) \rightarrow g(x))$ (Premise)

2. $\qquad \forall x f(x)$ (Assumption)

3. $\qquad \qquad a$ (Consider any 'a')

4. $\qquad \qquad f(a) \rightarrow g(a)$ ($\forall$ Elim 1)

5. $\qquad \qquad f(a)$ ($\forall$ Elim 2)

6. $\qquad \qquad g(a)$ ($\rightarrow$ Elim 2)

7. $\qquad \forall x g(x)$ ($\forall$ Intro 3-6)

8. $\forall x f(x) \rightarrow \forall x g(x)$ ($\rightarrow$ Intro 2-7)

Counterexample for reverse implication:

Domain: Natural Numbers

$f(x)$: x is an even number

$g(x)$ : x is an odd number

Then $\forall x f(x) \rightarrow \forall x g(x)$ is true (since the antecedent is false), but $\forall x (f(x) \rightarrow g(x))$ is false.

For the second one:

$\exists x (f(x) \rightarrow g(x)) \Leftrightarrow$ (Implication)

$\exists x (\neg f(x) \lor g(x)) \Leftrightarrow$ (Distribution $\exists$ over $\lor$)

$\exists x \neg f(x) \lor \exists x g(x) \Leftrightarrow$ (Quantifier Negation)

$\neg \forall x f(x) \lor \exists x g(x) \Leftrightarrow$ (Implication)

$\forall x f(x) \rightarrow \exists x g(x)$