Theorem pm2.61dne

Description: Deduction eliminating an inequality in an antecedent. (Contributed by NM, 1-Jun-2007) (Proof shortened by Andrew Salmon, 25-May-2011)

	Ref	Expression
Hypotheses	pm2.61dne.1	|- ( ph -> ( A = B -> ps ) )
	pm2.61dne.2	|- ( ph -> ( A =/= B -> ps ) )
Assertion	pm2.61dne	|- ( ph -> ps )

Proof

Step	Hyp	Ref	Expression
1		pm2.61dne.1	|- ( ph -> ( A = B -> ps ) )
2		pm2.61dne.2	|- ( ph -> ( A =/= B -> ps ) )
3		nne	|- ( -. A =/= B <-> A = B )
4	3 1	syl5bi	|- ( ph -> ( -. A =/= B -> ps ) )
5	2 4	pm2.61d	|- ( ph -> ps )