Theorem xpeq2d 4628

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

Hypothesis

Ref	Expression
xpeq1d.1	⊢ (𝜑 → 𝐴 = 𝐵)

Assertion

Ref	Expression
xpeq2d	⊢ (𝜑 → (𝐶 × 𝐴) = (𝐶 × 𝐵))

Proof of Theorem xpeq2d

Step	Hyp	Ref	Expression
1		xpeq1d.1	. 2 ⊢ (𝜑 → 𝐴 = 𝐵)
2		xpeq2 4619	. 2 ⊢ (𝐴 = 𝐵 → (𝐶 × 𝐴) = (𝐶 × 𝐵))
3	1, 2	syl 14	1 ⊢ (𝜑 → (𝐶 × 𝐴) = (𝐶 × 𝐵))

Syntax hints:   → wi 4   = wceq 1343   × cxp 4602

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-11 1494  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-opab 4044  df-xp 4610

This theorem is referenced by:  csbresg  4887  fconstg  5384  fvdiagfn  6659  mapsncnv  6661  xpsneng  6788  exp3val  10457  reldvg  13288  dvfvalap  13290