1
$\begingroup$

I don't understand this logic statement. I don't think the context helps at all but I thought i'd include it anyway.

$G$ is abelian and $H$ is not abelian, then $G\ncong H$, is the same as:

$G$ abelian and $G\cong H$, then $H$ is abelian

$\endgroup$
  • 1
    $\begingroup$ A and B $\implies$ C is the same as A and not C $\implies$ not B $\endgroup$ – Mathmo123 Jun 29 '14 at 13:30
3
$\begingroup$

Context is irrelevant.

"If $G$ is abelian and $H$ is not abelian, then $G\ncong H$" has the form $(A\land \neg B)\to \neg C$.

And since $(A\land \neg B)\to \neg C\iff \neg A\lor B\lor \neg C\iff (A\land C)\to B$,

for appropriate interpretations of $A, B$ and $C$, you get what you want.

$\endgroup$
1
$\begingroup$

Let $P \equiv ``H\ \text{is not Abelian}"$ and $Q \equiv ``G \not\cong H"$.

Assume that $G$ is Abelian. Then the first statement says $P \Rightarrow Q$, and the second says $\sim Q \Rightarrow \sim P$. Of course, these two are equivalent.

Only, that's not exactly what's given, I simplified it. Let $A \equiv ``G\ \text{is Abelian}"$. What is given is $(A \wedge P) \Rightarrow Q$. This is the same as $(A \wedge \sim Q) \Rightarrow \sim P$, in exactly the same way as the simpler versions above.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.