I am doing some practice sentential derivation proofs for an upcoming test and have attempt the following proof many, many times without success.
~(A ≡ B) ├ (~A ≡ B)
The logic system I am using is the Sentential Logic system which has the following inference rules: Reiteration, Conjunction Introduction/Elimination, Conditional Introduction/Elimination, Negation Introduction/Elimination, Disjunction Introduction/Elimination, and Biconditional Introduction/Elimination.
The bi-conditional is boring because you have to split into two parts:
1) ¬[(A→B) ∧ (B→A)] --- premise
2) A --- assumed [a]
3) B→A --- from 2)
4) B --- assumed [b]
5) A→B --- from 4)
6) (A→B) ∧ (B→A) --- from 3) and 5)
7) contradiction ! with 1)
8) ¬A --- from 2) and 7), discharging [a]
9) B → ¬A --- from 4) and 8), discharging [b].
In the same way, we have to derive: ¬A → B.
10) ¬A --- assumed [c]
11) ¬B --- assumed [d]
12) A --- assumed [e]
13) contradiction! with 10)
14) B --- from 13)
15) A→B --- from 12) and 14), discharging [e]
16) B --- assumed [f]
17) contradiction! with 11)
18) A --- from 17)
19) B→A --- from 16) and 18) discharging [f]
20) (A→B) ∧ (B→A) --- from 15) and 19)
21) contradiction! with 1)
22) B --- from 11) by Double Negation, discharging [d]
23) ¬A→B --- from 10) and 22), discharging [c].
Now we conclude from 9) and 23) with:
24) ¬A ≡ B.
Here is a proof using the Fitch software:
