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Theorem altxpeq1 25820
Description: Equality for alternate cross products. (Contributed by Scott Fenton, 24-Mar-2012.)
Assertion
Ref Expression
altxpeq1  |-  ( A  =  B  ->  ( A  XX.  C )  =  ( B  XX.  C
) )

Proof of Theorem altxpeq1
Dummy variables  x  y  z are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 rexeq 2907 . . 3  |-  ( A  =  B  ->  ( E. x  e.  A  E. y  e.  C  z  =  << x ,  y >> 
<->  E. x  e.  B  E. y  e.  C  z  =  << x ,  y >> ) )
21abbidv 2552 . 2  |-  ( A  =  B  ->  { z  |  E. x  e.  A  E. y  e.  C  z  =  << x ,  y >> }  =  { z  |  E. x  e.  B  E. y  e.  C  z  =  << x ,  y
>> } )
3 df-altxp 25806 . 2  |-  ( A 
XX.  C )  =  { z  |  E. x  e.  A  E. y  e.  C  z  =  << x ,  y
>> }
4 df-altxp 25806 . 2  |-  ( B 
XX.  C )  =  { z  |  E. x  e.  B  E. y  e.  C  z  =  << x ,  y
>> }
52, 3, 43eqtr4g 2495 1  |-  ( A  =  B  ->  ( A  XX.  C )  =  ( B  XX.  C
) )
Colors of variables: wff set class
Syntax hints:    -> wi 4    = wceq 1653   {cab 2424   E.wrex 2708   <<caltop 25803    XX. caltxp 25804
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1556  ax-5 1567  ax-17 1627  ax-9 1667  ax-8 1688  ax-6 1745  ax-7 1750  ax-11 1762  ax-12 1951  ax-ext 2419
This theorem depends on definitions:  df-bi 179  df-or 361  df-an 362  df-tru 1329  df-ex 1552  df-nf 1555  df-sb 1660  df-clab 2425  df-cleq 2431  df-clel 2434  df-nfc 2563  df-rex 2713  df-altxp 25806
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