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Metamath Proof Explorer

Theorem pm2.61ne

Description: Deduction eliminating an inequality in an antecedent. (Contributed by NM, 24-May-2006) (Proof shortened by Andrew Salmon, 25-May-2011) (Proof shortened by Wolf Lammen, 25-Nov-2019)

	Ref	Expression
Hypotheses	pm2.61ne.1	\|- ( A = B -> ( ps <-> ch ) )
	pm2.61ne.2	\|- ( ( ph /\ A =/= B ) -> ps )
	pm2.61ne.3	\|- ( ph -> ch )
Assertion	pm2.61ne	\|- ( ph -> ps )

Proof

Step	Hyp	Ref	Expression
1		pm2.61ne.1	\|- ( A = B -> ( ps <-> ch ) )
2		pm2.61ne.2	\|- ( ( ph /\ A =/= B ) -> ps )
3		pm2.61ne.3	\|- ( ph -> ch )
4	3 1	imbitrrid	\|- ( A = B -> ( ph -> ps ) )
5	2	expcom	\|- ( A =/= B -> ( ph -> ps ) )
6	4 5	pm2.61ine	\|- ( ph -> ps )