Theorem xpeq2d 4798

Description: Equality deduction for cross product. (Contributed by Jeff Madsen, 17-Jun-2010.)

Hypothesis

Ref	Expression
xpeq1d.1	|- ( ph -> A = B )

Assertion

Ref	Expression
xpeq2d	|- ( ph -> ( C X. A ) = ( C X. B ) )

Proof of Theorem xpeq2d

Step	Hyp	Ref	Expression
1		xpeq1d.1	. 2 |- ( ph -> A = B )
2		xpeq2 4789	. 2 |- ( A = B -> ( C X. A ) = ( C X. B ) )
3	1, 2	syl 14	1 |- ( ph -> ( C X. A ) = ( C X. B ) )

This proof depends on syntax axioms:   -> wi 4   = wceq 1402   X. cxp 4772

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-11 1559  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-opab 4193  df-xp 4780

This theorem is used by:  csbresg  5066  fconstg  5589  fvdiagfn  6975  mapsncnv  6977  xpsneng  7120  exp3val  10978  mulgval  13925  reldvg  15780  dvfvalap  15782