1

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.

3

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.

2

Here is a proof using the Fitch software:

enter image description here

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