Theorem nanbi1d 1301

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

Hypothesis

Ref	Expression
nanbid.1	|- (ph -> (ps <-> ch))

Assertion

Ref	Expression
nanbi1d	|- (ph -> ((ps -/\ th) <-> (ch -/\ th)))

Proof of Theorem nanbi1d

Step	Hyp	Ref	Expression
1		nanbid.1	. 2 |- (ph -> (ps <-> ch))
2		nanbi1 1295	. 2 |- ((ps <-> ch) -> ((ps -/\ th) <-> (ch -/\ th)))
3	1, 2	syl 15	1 |- (ph -> ((ps -/\ th) <-> (ch -/\ th)))

Syntax hints:   -> wi 4   <-> wb 176   -/\ 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:  nineq1  3235