predicate logic - counterexample $(A \models \phi \implies A \models \psi) \implies A \models \phi \rightarrow \psi$

Question

It's predicate logic and I need to find a counterexample to disprove the follwowing claim

$(A \models \phi \implies A \models \psi) \implies A \models \phi \rightarrow \psi$

Answer 1

I don't think there is a counterexample.

Suppose $A \models \phi \rightarrow \psi$ is false. This can only happen if $A \models \phi$ and $A \not\models \psi$. Hence $(A \models \phi \Rightarrow A \models \psi)$ is false.

Answer 2

Some people define a formula (with free occurrences of variables) to be true in the structure $A$ if the sentence obtained by universally quantifying these variables is true in $A$. Then annoying things can happen, which is why I prefer not to do it.

For an informal example, let $A$ be the natural numbers, let $\phi(x)$ be the formula that says $x$ is even, and let $\psi(x)$ be the formula that says $x$ is odd. (It is not hard to write down the appropriate formulas.) Then it is not the case that $\phi$ is true in $A$. It follows that $A \models \phi \implies A \models \psi$ is true. But $ A \models \phi \rightarrow \psi$ is false, since it is not the case that for all $x$, if $x$ is even then $x$ is odd.

Remark: Expressing commutativity of addition as $x+y=y+x$ rather than $\forall x\forall y(x+y=y+x)$ is a useful abbreviation. However, building that abbreviation into the logic introduces complications, as illustrated by the example above.