Theorem biadan

Description: An implication is equivalent to the equivalence of some implied equivalence and some other equivalence involving a conjunction. A utility lemma as illustrated in biadanii and elelb . (Contributed by BJ, 4-Mar-2023) (Proof shortened by Wolf Lammen, 8-Mar-2023)

	Ref	Expression
Assertion	biadan	|- ( ( ph -> ps ) <-> ( ( ps -> ( ph <-> ch ) ) <-> ( ph <-> ( ps /\ ch ) ) ) )

Proof

Step	Hyp	Ref	Expression
1		pm4.71r	|- ( ( ph -> ps ) <-> ( ph <-> ( ps /\ ph ) ) )
2		bicom	|- ( ( ph <-> ( ps /\ ph ) ) <-> ( ( ps /\ ph ) <-> ph ) )
3		bicom	|- ( ( ph <-> ( ps /\ ch ) ) <-> ( ( ps /\ ch ) <-> ph ) )
4		pm5.32	|- ( ( ps -> ( ph <-> ch ) ) <-> ( ( ps /\ ph ) <-> ( ps /\ ch ) ) )
5	3 4	bibi12i	|- ( ( ( ph <-> ( ps /\ ch ) ) <-> ( ps -> ( ph <-> ch ) ) ) <-> ( ( ( ps /\ ch ) <-> ph ) <-> ( ( ps /\ ph ) <-> ( ps /\ ch ) ) ) )
6		bicom	|- ( ( ( ps -> ( ph <-> ch ) ) <-> ( ph <-> ( ps /\ ch ) ) ) <-> ( ( ph <-> ( ps /\ ch ) ) <-> ( ps -> ( ph <-> ch ) ) ) )
7		biluk	|- ( ( ( ps /\ ph ) <-> ph ) <-> ( ( ( ps /\ ch ) <-> ph ) <-> ( ( ps /\ ph ) <-> ( ps /\ ch ) ) ) )
8	5 6 7	3bitr4ri	|- ( ( ( ps /\ ph ) <-> ph ) <-> ( ( ps -> ( ph <-> ch ) ) <-> ( ph <-> ( ps /\ ch ) ) ) )
9	1 2 8	3bitri	|- ( ( ph -> ps ) <-> ( ( ps -> ( ph <-> ch ) ) <-> ( ph <-> ( ps /\ ch ) ) ) )