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Theorem xpeq1 4798
 Description: Equality theorem for cross product. (Contributed by NM, 4-Jul-1994.)
Assertion
Ref Expression
xpeq1 (A = B → (A × C) = (B × C))

Proof of Theorem xpeq1
Dummy variables x y are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 eleq2 2414 . . . 4 (A = B → (x Ax B))
21anbi1d 685 . . 3 (A = B → ((x A y C) ↔ (x B y C)))
32opabbidv 4625 . 2 (A = B → {x, y (x A y C)} = {x, y (x B y C)})
4 df-xp 4784 . 2 (A × C) = {x, y (x A y C)}
5 df-xp 4784 . 2 (B × C) = {x, y (x B y C)}
63, 4, 53eqtr4g 2410 1 (A = B → (A × C) = (B × C))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ∧ wa 358   = wceq 1642   ∈ wcel 1710  {copab 4622   × cxp 4770 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334 This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-opab 4623  df-xp 4784 This theorem is referenced by:  xpeq12  4803  xpeq1i  4804  xpeq1d  4807  dmxpid  4924  reseq2  4929  xpnz  5045  xpdisj1  5047  xpcan2  5058  ovcross  5845  pmvalg  6010  xpsneng  6050  xpcomeng  6053
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