Theorem pm2.61da3ne

Description: Deduction eliminating three inequalities in an antecedent. (Contributed by NM, 15-Jun-2013) (Proof shortened by Wolf Lammen, 25-Nov-2019)

	Ref	Expression
Hypotheses	pm2.61da3ne.1	|- ( ( ph /\ A = B ) -> ps )
	pm2.61da3ne.2	|- ( ( ph /\ C = D ) -> ps )
	pm2.61da3ne.3	|- ( ( ph /\ E = F ) -> ps )
	pm2.61da3ne.4	|- ( ( ph /\ ( A =/= B /\ C =/= D /\ E =/= F ) ) -> ps )
Assertion	pm2.61da3ne	|- ( ph -> ps )

Proof

Step	Hyp	Ref	Expression
1		pm2.61da3ne.1	|- ( ( ph /\ A = B ) -> ps )
2		pm2.61da3ne.2	|- ( ( ph /\ C = D ) -> ps )
3		pm2.61da3ne.3	|- ( ( ph /\ E = F ) -> ps )
4		pm2.61da3ne.4	|- ( ( ph /\ ( A =/= B /\ C =/= D /\ E =/= F ) ) -> ps )
5	1	a1d	|- ( ( ph /\ A = B ) -> ( ( C =/= D /\ E =/= F ) -> ps ) )
6	4	3exp2	|- ( ph -> ( A =/= B -> ( C =/= D -> ( E =/= F -> ps ) ) ) )
7	6	imp4b	|- ( ( ph /\ A =/= B ) -> ( ( C =/= D /\ E =/= F ) -> ps ) )
8	5 7	pm2.61dane	|- ( ph -> ( ( C =/= D /\ E =/= F ) -> ps ) )
9	8	imp	|- ( ( ph /\ ( C =/= D /\ E =/= F ) ) -> ps )
10	2 3 9	pm2.61da2ne	|- ( ph -> ps )