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Theorem altxpeq1 31250
 Description: Equality for alternate Cartesian products. (Contributed by Scott Fenton, 24-Mar-2012.)
Assertion
Ref Expression
altxpeq1 (𝐴 = 𝐵 → (𝐴 ×× 𝐶) = (𝐵 ×× 𝐶))

Proof of Theorem altxpeq1
Dummy variables 𝑥 𝑦 𝑧 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 rexeq 3116 . . 3 (𝐴 = 𝐵 → (∃𝑥𝐴𝑦𝐶 𝑧 = ⟪𝑥, 𝑦⟫ ↔ ∃𝑥𝐵𝑦𝐶 𝑧 = ⟪𝑥, 𝑦⟫))
21abbidv 2728 . 2 (𝐴 = 𝐵 → {𝑧 ∣ ∃𝑥𝐴𝑦𝐶 𝑧 = ⟪𝑥, 𝑦⟫} = {𝑧 ∣ ∃𝑥𝐵𝑦𝐶 𝑧 = ⟪𝑥, 𝑦⟫})
3 df-altxp 31236 . 2 (𝐴 ×× 𝐶) = {𝑧 ∣ ∃𝑥𝐴𝑦𝐶 𝑧 = ⟪𝑥, 𝑦⟫}
4 df-altxp 31236 . 2 (𝐵 ×× 𝐶) = {𝑧 ∣ ∃𝑥𝐵𝑦𝐶 𝑧 = ⟪𝑥, 𝑦⟫}
52, 3, 43eqtr4g 2669 1 (𝐴 = 𝐵 → (𝐴 ×× 𝐶) = (𝐵 ×× 𝐶))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   = wceq 1475  {cab 2596  ∃wrex 2897  ⟪caltop 31233   ×× caltxp 31234 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590 This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-rex 2902  df-altxp 31236 This theorem is referenced by: (None)
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