# 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 )`