0

Can I prove that $A \implies B$ implies $\neg B \implies \neg A$ in this particular way? I assume $A \implies B$ and $ \neg (\neg B \implies \neg A)$ and derive a contradiction. Is this ok?

Adam
  • 3,422
  • 1
  • 33
  • 50
  • Yes, ok! ...... – Bram28 Jun 09 '22 at 17:25
  • That's a valid strategy, but how do you derive a contradiction? It is likely that inside that derivation you have a simpler derivation already. – Magdiragdag Jun 09 '22 at 17:25
  • 1
    Alternative: show that ($A\implies B)\implies (\neg B \implies \neg A$) is a tautology. – insipidintegrator Jun 09 '22 at 17:27
  • Here is how I derive a contradiction: I rewrite A implies B as not(A) or B. Next I rewrite not( not(B) => not(A) ) as not(B or not(A) ) which is the same as not( not(A) or B ) and that is a contradiction. – Adam Jun 09 '22 at 17:36
  • That works, but if you have that kind of rewriting available, you might as well rewrite $A \to B$ first as $\lnot A \lor B$, then as $\lnot \lnot B \lor \lnot A$, then as $\lnot B \to \lnot A$. – Magdiragdag Jun 09 '22 at 18:03

0 Answers0