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Theorem xpeq2 4287
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 2083 . . . 4 (A = B → (y Ay B))
21anbi2d 440 . . 3 (A = B → ((x 𝐶 y A) ↔ (x 𝐶 y B)))
32opabbidv 3797 . 2 (A = B → {⟨x, y⟩ ∣ (x 𝐶 y A)} = {⟨x, y⟩ ∣ (x 𝐶 y B)})
4 df-xp 4278 . 2 (𝐶 × A) = {⟨x, y⟩ ∣ (x 𝐶 y A)}
5 df-xp 4278 . 2 (𝐶 × B) = {⟨x, y⟩ ∣ (x 𝐶 y B)}
63, 4, 53eqtr4g 2079 1 (A = B → (𝐶 × A) = (𝐶 × B))
Colors of variables: wff set class
Syntax hints:  wi 4   wa 97   = wceq 1228   wcel 1374  {copab 3791   × cxp 4270
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1316  ax-7 1317  ax-gen 1318  ax-ie1 1363  ax-ie2 1364  ax-8 1376  ax-11 1378  ax-4 1381  ax-17 1400  ax-i9 1404  ax-ial 1409  ax-i5r 1410  ax-ext 2004
This theorem depends on definitions:  df-bi 110  df-tru 1231  df-nf 1330  df-sb 1628  df-clab 2009  df-cleq 2015  df-clel 2018  df-opab 3793  df-xp 4278
This theorem is referenced by:  xpeq12  4291  xpeq2i  4293  xpeq2d  4296  xpeq0r  4673  xpdisj2  4675
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