Theorem bj-stand 14640

Description: The conjunction of two stable formulas is stable. Deduction form of bj-stan 14639. Its proof is shorter (when counting all steps, including syntactic steps), so one could prove it first and then bj-stan 14639 from it, the usual way. (Contributed by BJ, 24-Nov-2023.) (Proof modification is discouraged.)

Hypotheses

Ref	Expression
bj-stand.1	|- ( ph -> STAB ps )
bj-stand.2	|- ( ph -> STAB ch )

Assertion

Ref	Expression
bj-stand	|- ( ph -> STAB ( ps /\ ch ) )

Proof of Theorem bj-stand

Step	Hyp	Ref	Expression
1		bj-nnan 14628	. . 3 |- ( -. -. ( ps /\ ch ) -> ( -. -. ps /\ -. -. ch ) )
2		bj-stand.1	. . . . 5 |- ( ph -> STAB ps )
3		df-stab 831	. . . . 5 |- (STAB ps <-> ( -. -. ps -> ps ) )
4	2, 3	sylib 122	. . . 4 |- ( ph -> ( -. -. ps -> ps ) )
5		bj-stand.2	. . . . 5 |- ( ph -> STAB ch )
6		df-stab 831	. . . . 5 |- (STAB ch <-> ( -. -. ch -> ch ) )
7	5, 6	sylib 122	. . . 4 |- ( ph -> ( -. -. ch -> ch ) )
8	4, 7	anim12d 335	. . 3 |- ( ph -> ( ( -. -. ps /\ -. -. ch ) -> ( ps /\ ch ) ) )
9	1, 8	syl5 32	. 2 |- ( ph -> ( -. -. ( ps /\ ch ) -> ( ps /\ ch ) ) )
10		df-stab 831	. 2 |- (STAB ( ps /\ ch ) <-> ( -. -. ( ps /\ ch ) -> ( ps /\ ch ) ) )
11	9, 10	sylibr 134	1 |- ( ph -> STAB ( ps /\ ch ) )

Syntax hints:   -. wn 3   -> wi 4   /\ wa 104  STAB wstab 830

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 614  ax-in2 615

This theorem depends on definitions:  df-bi 117  df-stab 831

This theorem is referenced by: (None)