They're not the same. In your solution, for example, $x$ could be a teacher; in that case, for all $y$, $S(x) \wedge F(y)$ is false, so the implication $(S(x) \wedge F(y)) \Rightarrow \
eg A(x, y)$ is always true.
They're not the same. In your solution, for example, $x$ could be a teacher; in that case, for all $y$, $S(x) \wedge F(y)$ is false, so the implication $(S(x) \wedge F(y)) \Rightarrow \
eg A(x, y)$ is always true.