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Metamath Proof Explorer

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 )