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Theorem pm2.61da3ne 2769
Description: Deduction eliminating three inequalities in an antecedent. (Contributed by NM, 15-Jun-2013.) (Proof shortened by Wolf Lammen, 25-Nov-2019.)
Hypotheses
Ref Expression
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
Ref Expression
pm2.61da3ne  |-  ( ph  ->  ps )

Proof of Theorem pm2.61da3ne
StepHypRef Expression
1 pm2.61da3ne.2 . 2  |-  ( (
ph  /\  C  =  D )  ->  ps )
2 pm2.61da3ne.3 . 2  |-  ( (
ph  /\  E  =  F )  ->  ps )
3 pm2.61da3ne.1 . . . . 5  |-  ( (
ph  /\  A  =  B )  ->  ps )
43a1d 25 . . . 4  |-  ( (
ph  /\  A  =  B )  ->  (
( C  =/=  D  /\  E  =/=  F
)  ->  ps )
)
5 pm2.61da3ne.4 . . . . . 6  |-  ( (
ph  /\  ( A  =/=  B  /\  C  =/= 
D  /\  E  =/=  F ) )  ->  ps )
653exp2 1206 . . . . 5  |-  ( ph  ->  ( A  =/=  B  ->  ( C  =/=  D  ->  ( E  =/=  F  ->  ps ) ) ) )
76imp4b 590 . . . 4  |-  ( (
ph  /\  A  =/=  B )  ->  ( ( C  =/=  D  /\  E  =/=  F )  ->  ps ) )
84, 7pm2.61dane 2767 . . 3  |-  ( ph  ->  ( ( C  =/= 
D  /\  E  =/=  F )  ->  ps )
)
98imp 429 . 2  |-  ( (
ph  /\  ( C  =/=  D  /\  E  =/= 
F ) )  ->  ps )
101, 2, 9pm2.61da2ne 2768 1  |-  ( ph  ->  ps )
Colors of variables: wff setvar class
Syntax hints:    -> wi 4    /\ wa 369    /\ w3a 965    = wceq 1370    =/= wne 2645
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8
This theorem depends on definitions:  df-bi 185  df-an 371  df-3an 967  df-ne 2647
This theorem is referenced by:  trljco  34703  dvh4dimN  35411
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