Theorem bifald 33888

Description: Infer the equivalence to a contradiction from a negation, in deduction form. (Contributed by Giovanni Mascellani, 15-Sep-2017.)

Hypothesis

Ref	Expression
bifald.1	|- ( ph -> -. ps )

Assertion

Ref	Expression
bifald	|- ( ph -> ( ps <-> F. ) )

Proof of Theorem bifald

Step	Hyp	Ref	Expression
1		bifald.1	. 2 |- ( ph -> -. ps )
2		id 22	. . 3 |- ( ps -> ps )
3		falim 1498	. . 3 |- ( F. -> ps )
4	2, 3	pm5.21ni 367	. 2 |- ( -. ps -> ( ps <-> F. ) )
5	1, 4	syl 17	1 |- ( ph -> ( ps <-> F. ) )

Syntax hints:   -. wn 3   -> wi 4   <-> wb 196   F. wfal 1488

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 197  df-tru 1486  df-fal 1489

This theorem is referenced by: (None)