I'm working on a question right now that's asking me the difference between "For every integer a, b, if for every integer x, ax+b is even ... " and "For every integer a, b, x if ax+b is even ..."
I already proved the first statement as true but I'm struggling to find a concrete difference between the two statements. If anybody can sort of explain the difference for me, I'd greatly appreciate that. Thanks!
Following Graham's comment and filtering out the $a,b$: $$\big(\forall x~\mathrm P(x)\big)\Rightarrow\mathrm Q\tag{1e}$$$$\forall x~\big(\mathrm P(x)\Rightarrow\mathrm Q\big)\tag{2e}$$
Sentence $(2\mathrm e)$ asserts that $\ldots,\big(P(-7)\Rightarrow Q\big),\ldots,\big(P(0)\Rightarrow Q\big),\ldots,\big(P(31)\Rightarrow Q\big),\ldots$ are all true ($x$ is varying over its entire domain); equivalently, that if some $P(x)$ is true, then $Q$ is true; i.e., $$\forall x~\big(\mathrm P(x)\Rightarrow\mathrm Q\big)\;\equiv\;\big(\exists x~\mathrm P(x)\big)\Rightarrow\mathrm Q.\tag{2e}$$
In comparison, sentence $(1\mathrm e)$ is a weaker assertion: it is applicable (usefully invoked) in fewer scenarios, since its hypothesis $\big(\forall x~\mathrm P(x)\big)$ is harder to satisfy than $(2\mathrm e)$'s hypothesis $\big(\exists x~\mathrm P(x)\big).$
For completeness: $$\big(\forall x~\mathrm P(x)\big)\Rightarrow\mathrm Q\;\equiv\;\exists x~\big(\mathrm P(x)\Rightarrow\mathrm Q\big).\tag{1e}$$