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Metamath Proof Explorer

Theorem bibiad

Description: Eliminate an hypothesis th in a biconditional. (Contributed by Thierry Arnoux, 4-May-2025)

Ref Expression
Hypotheses bibiad.1 
|- ( ( ph /\ ps ) -> th )
bibiad.2 
|- ( ( ph /\ ch ) -> th )
bibiad.3 
|- ( ( ph /\ th ) -> ( ps <-> ch ) )
Assertion bibiad 
|- ( ph -> ( ps <-> ch ) )

Proof

Step Hyp Ref Expression
1 bibiad.1 
 |-  ( ( ph /\ ps ) -> th )
2 bibiad.2 
 |-  ( ( ph /\ ch ) -> th )
3 bibiad.3 
 |-  ( ( ph /\ th ) -> ( ps <-> ch ) )
4 simpl 
 |-  ( ( ph /\ ps ) -> ph )
5 simpr 
 |-  ( ( ph /\ ps ) -> ps )
6 3 biimpa 
 |-  ( ( ( ph /\ th ) /\ ps ) -> ch )
7 4 1 5 6 syl21anc 
 |-  ( ( ph /\ ps ) -> ch )
8 simpl 
 |-  ( ( ph /\ ch ) -> ph )
9 simpr 
 |-  ( ( ph /\ ch ) -> ch )
10 3 biimpar 
 |-  ( ( ( ph /\ th ) /\ ch ) -> ps )
11 8 2 9 10 syl21anc 
 |-  ( ( ph /\ ch ) -> ps )
12 7 11 impbida 
 |-  ( ph -> ( ps <-> ch ) )