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Theorem xpeq2 4799
Description: Equality theorem for cross product. (Contributed by NM, 5-Jul-1994.)
Assertion
Ref Expression
xpeq2 (A = B → (C × A) = (C × B))

Proof of Theorem xpeq2
Dummy variables x y are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 eleq2 2414 . . . 4 (A = B → (y Ay B))
21anbi2d 684 . . 3 (A = B → ((x C y A) ↔ (x C y B)))
32opabbidv 4625 . 2 (A = B → {x, y (x C y A)} = {x, y (x C y B)})
4 df-xp 4784 . 2 (C × A) = {x, y (x C y A)}
5 df-xp 4784 . 2 (C × B) = {x, y (x C y B)}
63, 4, 53eqtr4g 2410 1 (A = B → (C × A) = (C × B))
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  xpeq2i  4805  xpeq2d  4808  xpnz  5045  xpdisj2  5048  dmxpss  5052  rnxpid  5054  xpcan  5057  ovcross  5845  pmvalg  6010  xpcomeng  6053
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