Show that the following inference is not correct:
Suppose every day that is not rainy is not windy, and some day is windy. Then every day is rainy.
$$\forall x (\lnot R(x)\rightarrow \lnot W(x)) \land \exists x W(x)$$
is how I would write it. I'm unsure what I can use to show that the inference is not correct. Intuitively it's false but... Should I use natural deduction?..
The correct inference is simply that then there is some day that is rainy.
$$\forall x (\lnot R(x)\rightarrow \lnot W(x)) \land \exists x W(x)\quad \equiv \quad \forall x(W(x)\rightarrow R(x)) \land \exists x W(x)\tag{1}$$
Since on every day that is windy it is raining, then all we know that if on some day it is windy, then it is also rainy. The fact that on some days it is windy tells us nothing about about whether or not it is raining on other days which may not be windy.
Counter example:
So, suppose it was windy yesterday. (Hence, there is some day that is windy). And it rained yesterday. (That is, there is some day that is rainy). But today it is not rainy, nor windy (so it is not the case that all days are rainy. The fact that it not windy is alright, because it is already true that on some day it is windy, namely, yesterday).
Nothing in the counterexample violated the truth of the given statement $(1)$. But it does show that the inference that every day is rainy does not follow from $(1)$.