¬(A↔B) conclusion ¬A↔B
I'm having trouble with the second part of this proof.
I think I managed the first part:
1 |¬(A↔B)$\,$ $\,$ A prem.
2 ||B $\,$ $\,$ A →intro
3 |||A $\,$ $\,$ A →intro
4 |||B
5 ||A→B $\,$ $\,$ →intro 3,4
6 |||A $\,$ $\,$ A RAA intro
7 ||||B $\,$ $\,$ A →intro
8 ||||A
9 |||B→A $\,$ $\,$ →intro 7,8
10|||A↔B $\,$ $\,$ ↔ intro 5,9
11||¬A $\,$ $\,$ RAA(6) 1,10
12|B→¬A $\,$ $\,$ →intro 2,11
I was going to do the same for ¬A→B, but I don't see how I
can conclude B when I start with ¬A.
If anyone could help me.
You can still use the basic method of doing this by a RAA: assume $\neg B$, try and get to $A \leftrightarrow B$ to get your contradiction, and use RAA to conclude $\neg \neg B$, and from that you can get $B$
Also, to get to $A \leftrightarrow B$ once you have assumed $\neg A $ and $\neg B$: same idea. Just going left to right: assume $A$. Now assume $\neg B$. Get a contradiction between $A$ and $\neg A$. So by RAA $\neg \neg B$ and thus $B$