# A formula provable without is a tautology

1. Sep 15, 2010

### antonio85

How can I prove that:

If a formula $$A$$ is provable without use of substitution axioms, nonlogical axioms, equality and identity axioms, and the $$\exists$$-introduction rule, than $$A$$ is a tautology.

I try to act this way: consider a tautology A and show that using propositional axioms I get another tautology, and that if the hypothesis of a rule are tautologies then also the conclusion is a tautology. But I don't know wether it is the correct way or not.
Anyone can help me?

2. Sep 15, 2010

### Office_Shredder

Staff Emeritus
A good way to start might be by induction

If your proof is one step then what can A possibly be?

Then if it's true for all proofs of less than or equal to n steps... for an n+1 step proof how would you proceed?