MPE Home Metamath Proof Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >  3netr4g Structured version   Unicode version

Theorem 3netr4g 2630
Description: Substitution of equality into both sides of an inequality. (Contributed by NM, 14-Jun-2012.)
Hypotheses
Ref Expression
3netr4g.1  |-  ( ph  ->  A  =/=  B )
3netr4g.2  |-  C  =  A
3netr4g.3  |-  D  =  B
Assertion
Ref Expression
3netr4g  |-  ( ph  ->  C  =/=  D )

Proof of Theorem 3netr4g
StepHypRef Expression
1 3netr4g.1 . 2  |-  ( ph  ->  A  =/=  B )
2 3netr4g.2 . . 3  |-  C  =  A
3 3netr4g.3 . . 3  |-  D  =  B
42, 3neeq12i 2613 . 2  |-  ( C  =/=  D  <->  A  =/=  B )
51, 4sylibr 204 1  |-  ( ph  ->  C  =/=  D )
Colors of variables: wff set class
Syntax hints:    -> wi 4    = wceq 1652    =/= wne 2599
This theorem is referenced by:  aalioulem2  20250  mapdpglem18  32487
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1555  ax-5 1566  ax-17 1626  ax-9 1666  ax-8 1687  ax-11 1761  ax-ext 2417
This theorem depends on definitions:  df-bi 178  df-ex 1551  df-cleq 2429  df-ne 2601
  Copyright terms: Public domain W3C validator