In predicate logic, all $x$ in set $X$ who have quality A also have quality B

Question

How do I convert the premise that

all $x$ in set $X$ for which $A$ is true $\implies B$ is also true

the main part where I am having trouble finding examples is "all $x$ in set $X$ for which $A$ is true"...I can do there exists $x$ in $X$ for which $A$ is true, but am confused as to what is equivalent to the subset of $X$ for which $A$ is true in predicate logic...

I guess what I am asking is if I have to specify the truth set: $\{x \in X\mid A(x)\}$ in the predicate logic expression.

Answer 1

all x in set X for which A is true -> B is also true

Isn't this just?: $\qquad\forall x\in X\;\big(A(x) \to B(x)\big)$

Which is short for: $\quad\forall x \;\Big(\big(x\in X\wedge A(x)\big)\to B(x)\Big)$

Though some may prefer the equivalent: $\;\forall x\Big(x\in X \to \big(A(x)\to B(x)\big)\Big)$

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I can do there exists x in X for which A is true,

To be sure, this would be: $\quad \exists x\in X\big(A(x)\to B(x)\big)$

Which is also expressed as: $\quad \exists x \Big( x\in X \wedge \big(A(x)\to B(x)\big)\Big)$