- #1
antonio85
- 5
- 0
How can I prove that:
If a formula [tex]A[/tex] is provable without use of substitution axioms, nonlogical axioms, equality and identity axioms, and the [tex]\exists[/tex]-introduction rule, than [tex]A[/tex] 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?
If a formula [tex]A[/tex] is provable without use of substitution axioms, nonlogical axioms, equality and identity axioms, and the [tex]\exists[/tex]-introduction rule, than [tex]A[/tex] 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?