Metamath Proof Explorer


Theorem aisfina

Description: Given a is equivalent to F. , there exists a proof for not a. (Contributed by Jarvin Udandy, 30-Aug-2016)

Ref Expression
Hypothesis aisfina.1
|- ( ph <-> F. )
Assertion aisfina
|- -. ph

Proof

Step Hyp Ref Expression
1 aisfina.1
 |-  ( ph <-> F. )
2 nbfal
 |-  ( -. ph <-> ( ph <-> F. ) )
3 1 2 mpbir
 |-  -. ph