Metamath Proof Explorer


Theorem biadanid

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

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

Proof

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