Theorem biadanid 614

Description: Deduction associated with biadani 612. Add a conjunction to an equivalence. (Contributed by Thierry Arnoux, 16-Jun-2024.)

Hypotheses

Ref	Expression
biadanid.1	|- ( ( ph /\ ps ) -> ch )
biadanid.2	|- ( ( ph /\ ch ) -> ( ps <-> th ) )

Assertion

Ref	Expression
biadanid	|- ( ph -> ( ps <-> ( ch /\ th ) ) )

Proof of Theorem biadanid

Step	Hyp	Ref	Expression
1		biadanid.1	. . 3 |- ( ( ph /\ ps ) -> ch )
2		biadanid.2	. . . . . 6 |- ( ( ph /\ ch ) -> ( ps <-> th ) )
3	2	biimpa 296	. . . . 5 |- ( ( ( ph /\ ch ) /\ ps ) -> th )
4	3	an32s 568	. . . 4 |- ( ( ( ph /\ ps ) /\ ch ) -> th )
5	1, 4	mpdan 421	. . 3 |- ( ( ph /\ ps ) -> th )
6	1, 5	jca 306	. 2 |- ( ( ph /\ ps ) -> ( ch /\ th ) )
7	2	biimpar 297	. . 3 |- ( ( ( ph /\ ch ) /\ th ) -> ps )
8	7	anasss 399	. 2 |- ( ( ph /\ ( ch /\ th ) ) -> ps )
9	6, 8	impbida 596	1 |- ( ph -> ( ps <-> ( ch /\ th ) ) )

Syntax hints:   -> wi 4   /\ wa 104   <-> wb 105

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108

This theorem depends on definitions:  df-bi 117

This theorem is referenced by:  dflidl2  14044  df2idl2  14065