# What is a natural deduction proof from ~(A↔B) to ~(A→B)?

It feels intuitively correct, but I cannot work out how to prove it. I would appreciate any help.

• It is not intuitively correct; Two predicates not being equivalent does not prohibit one from implying the other. – Graham Kemp May 29 '19 at 22:59

The following truth table shows that ~(A↔B) → ~(A→B) is not a tautology:

If A is False and B is True then the antecedent is True but the consequent is False making the conditional False.

Because the truth table does not show a tautology, one should not be able to derive a natural deduction proof of the result.

Michael Rieppel. Truth Table Generator. https://mrieppel.net/prog/truthtable.html

• Thanks very much, that makes sense. – zzzz May 31 '19 at 7:07

You can't derive ~(A→B) from ~(A↔B).

Consider:

A = I'm in Paris. B = I'm in France.

~(A↔B) is true, because being in Paris is not equivalent to being in France (I could be in France but not in Paris). But ~(A→B) is false, because if I'm in Paris then necessarily I'm in France. So you can't derive ~(A→B) from ~(A↔B).

• Ah, ok perfect - thank you very much. – zzzz May 31 '19 at 7:08