Theorem nanbi1 1295

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 687	. . 3 |- ((ph <-> ps) -> ((ph /\ ch) <-> (ps /\ ch)))
2	1	notbid 285	. 2 |- ((ph <-> ps) -> (-. (ph /\ ch) <-> -. (ps /\ ch)))
3		df-nan 1288	. 2 |- ((ph -/\ ch) <-> -. (ph /\ ch))
4		df-nan 1288	. 2 |- ((ps -/\ ch) <-> -. (ps /\ ch))
5	2, 3, 4	3bitr4g 279	1 |- ((ph <-> ps) -> ((ph -/\ ch) <-> (ps -/\ ch)))

Syntax hints:  -. wn 3   -> wi 4   <-> wb 176   /\ wa 358   -/\ wnan 1287

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 177  df-an 360  df-nan 1288

This theorem is referenced by:  nanbi2  1296  nanbi12  1297  nanbi1i  1298  nanbi1d  1301