0
$\begingroup$

Given $\lnot(A→B)$. Then $A˄\lnot B$. Assume $\lnot(A→\lnot B)$. Then $A˄\lnot\lnot B$. By DN, $A˄B$. Also $A˄\lnot B (\text{reit})$. Contradiction, therefore, $\lnot\lnot(A→\lnot B)$ by indirect proof, which by DN is $(A→\lnot B)$. So $\lnot (A→B) ⊢ A→\lnot B$.

$\endgroup$
1
  • 1
    $\begingroup$ Why do you doubt your reasoning? Do you know the "truth table" method that easily solves questions like this, especially with only two letters? $\endgroup$
    – GEdgar
    Commented Apr 1, 2023 at 8:46

1 Answer 1

Reset to default
0
$\begingroup$

In propositional logic the following statements are equivalent:

  • $p\vdash q$
  • $\vdash p\to q$

So if you are still in doubt then you can just check whether statement:$$\neg(A\to B)\to(A\to\neg B)$$is a tautology by means of a truth table (as suggested in the comments on your question).

$\endgroup$

You must log in to answer this question.