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Theorem pm2.61da3ne 2630
Description: Deduction eliminating three inequalities in an antecedent. (Contributed by NM, 15-Jun-2013.)
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.1 . 2  |-  ( (
ph  /\  A  =  B )  ->  ps )
2 pm2.61da3ne.2 . 2  |-  ( (
ph  /\  C  =  D )  ->  ps )
3 pm2.61da3ne.3 . . . 4  |-  ( (
ph  /\  E  =  F )  ->  ps )
43adantlr 696 . . 3  |-  ( ( ( ph  /\  ( A  =/=  B  /\  C  =/=  D ) )  /\  E  =  F )  ->  ps )
5 simpll 731 . . . 4  |-  ( ( ( ph  /\  ( A  =/=  B  /\  C  =/=  D ) )  /\  E  =/=  F )  ->  ph )
6 simplrl 737 . . . 4  |-  ( ( ( ph  /\  ( A  =/=  B  /\  C  =/=  D ) )  /\  E  =/=  F )  ->  A  =/=  B )
7 simplrr 738 . . . 4  |-  ( ( ( ph  /\  ( A  =/=  B  /\  C  =/=  D ) )  /\  E  =/=  F )  ->  C  =/=  D )
8 simpr 448 . . . 4  |-  ( ( ( ph  /\  ( A  =/=  B  /\  C  =/=  D ) )  /\  E  =/=  F )  ->  E  =/=  F )
9 pm2.61da3ne.4 . . . 4  |-  ( (
ph  /\  ( A  =/=  B  /\  C  =/= 
D  /\  E  =/=  F ) )  ->  ps )
105, 6, 7, 8, 9syl13anc 1186 . . 3  |-  ( ( ( ph  /\  ( A  =/=  B  /\  C  =/=  D ) )  /\  E  =/=  F )  ->  ps )
114, 10pm2.61dane 2628 . 2  |-  ( (
ph  /\  ( A  =/=  B  /\  C  =/= 
D ) )  ->  ps )
121, 2, 11pm2.61da2ne 2629 1  |-  ( ph  ->  ps )
Colors of variables: wff set class
Syntax hints:    -> wi 4    /\ wa 359    /\ w3a 936    = wceq 1649    =/= wne 2550
This theorem is referenced by:  trljco  30854  dvh4dimN  31562
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8
This theorem depends on definitions:  df-bi 178  df-an 361  df-3an 938  df-ne 2552
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