I have the following syllogism:
all cars are red
no motorcycle is red
thus
no car is a motorcycle
I tried to pass it to the logical predicate form:
$\forall x(A(x)\supset B(x))$
$\neg \exists x(C(x)\land B(x)) $
thus
$\neg \exists x(A(x)\land C(x))$
I think this syllogism is true, but I don't know how to prove it formally.
I would really appreciate any hints or feedback.
Suppose $C(x)$ is "$x$ is a car", $R(x)$ is "$x$ is red" and $M(x)$ is "$x$ is a motorcycle". We have $$S1: (\forall x)(C(x)\Rightarrow R(x))$$ (all cars are red) and $\neg (\exists x)(M(x) \wedge R(x) )$ (no motorcycle is red). The second is equivalent to $$S2: (\forall x)(M(x)\Rightarrow \neg R(x))$$ (if $x$ is a motorcycle then it is not red).
[Here we distribute the $\neg$ over, and use the standard definition of $\vee$ as $(a\vee b) \equiv (\neg a \Rightarrow b)$. Another slight subtlety: we are using excluded middle.]
So using the two statements, we have $(\exists x) (C(x)\wedge M(x))\Rightarrow (\exists x)(R(x)\wedge \neg R(x)) \Rightarrow \texttt{False}.$ But for any proposition $P$, $(\neg P)$ is by definition $(P\Rightarrow \texttt{False})$ so we have $\neg (\exists x)(M(x)\wedge C(x))$.