0

It is necessary to prove the derivability: $\vdash((\lnot B \to B) \to \lnot \lnot B)$.
My steps:
$\lnot B \to B \vdash \lnot \lnot B$
1)$\lnot B \to B$ - hypothesis
2)$(((\lnot B) \to B)) \to (((\lnot B) \to B) \to \lnot \lnot B)$ (A3)
3)$(((\lnot B) \to \lnot \lnot B) \to \lnot \lnot B)$ (1,2 MP)
But then I don't understand what to do.

Rey Legar
  • 3
  • Can you list the precise axioms you're using? It seems it's a Hilbert calculus, but different textbooks may use different names for the axioms... – Johannes Kloos Oct 27 '21 at 21:51

1 Answers1

0

Your application of A3 seems problematic, you can do a more careful inspection to see is your step 2) really correct by invoking A3 of your system, which seems Hilbert $H_2$ axiom system:

A3: $(\neg A\to \neg B)\to ((\neg A\to B)\to A)$

One way to proceed is to use proof by negation, assume $\lnot B$, then using MP you can easily get $B$ and clearly you arrive at $\bot$, a contradiction. So you then can safely conclude $(\lnot B \to B) \to \lnot \lnot B$.

cinch
  • 1,648
  • 1
  • 4
  • 12
  • Sorry, I don't understand how to use the negation proof, could you explain? – Rey Legar Oct 26 '21 at 14:30
  • @ReyLegar See reference post in this site here. Hope it helps. It's a common proof method applicable to both classical and intuitionistic logics. – cinch Oct 26 '21 at 14:36