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	1	com12	|- ( A = B -> ( ph -> ps ) )
4	2	com12	|- ( A =/= B -> ( ph -> ps ) )
5	3 4	pm2.61ine	|- ( ph -> ps )