Theorem nanbi1 1455

Description: Introduce a right anti-conjunct to both sides of a logical equivalence. (Contributed by SF, 2-Jan-2018.)

Assertion

Ref	Expression
nanbi1	|- ( ( ph <-> ps ) -> ( ( ph -/\ ch ) <-> ( ps -/\ ch ) ) )

Proof of Theorem nanbi1

Step	Hyp	Ref	Expression
1		anbi1 743	. . 3 |- ( ( ph <-> ps ) -> ( ( ph /\ ch ) <-> ( ps /\ ch ) ) )
2	1	notbid 308	. 2 |- ( ( ph <-> ps ) -> ( -. ( ph /\ ch ) <-> -. ( ps /\ ch ) ) )
3		df-nan 1448	. 2 |- ( ( ph -/\ ch ) <-> -. ( ph /\ ch ) )
4		df-nan 1448	. 2 |- ( ( ps -/\ ch ) <-> -. ( ps /\ ch ) )
5	2, 3, 4	3bitr4g 303	1 |- ( ( ph <-> ps ) -> ( ( ph -/\ ch ) <-> ( ps -/\ ch ) ) )

Syntax hints:   -. wn 3   -> wi 4   <-> wb 196   /\ wa 384   -/\ wnan 1447

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 197  df-an 386  df-nan 1448

This theorem is referenced by:  nanbi2  1456  nanbi12  1457  nanbi1i  1458  nanbi1d  1461  nabi1  32389