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Theorem altxpeq2 35976
Description: Equality for alternate Cartesian products. (Contributed by Scott Fenton, 24-Mar-2012.)
Assertion
Ref Expression
altxpeq2 (𝐴 = 𝐵 → (𝐶 ×× 𝐴) = (𝐶 ×× 𝐵))

Proof of Theorem altxpeq2
Dummy variables 𝑥 𝑦 𝑧 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 rexeq 3321 . . . 4 (𝐴 = 𝐵 → (∃𝑦𝐴 𝑧 = ⟪𝑥, 𝑦⟫ ↔ ∃𝑦𝐵 𝑧 = ⟪𝑥, 𝑦⟫))
21rexbidv 3178 . . 3 (𝐴 = 𝐵 → (∃𝑥𝐶𝑦𝐴 𝑧 = ⟪𝑥, 𝑦⟫ ↔ ∃𝑥𝐶𝑦𝐵 𝑧 = ⟪𝑥, 𝑦⟫))
32abbidv 2807 . 2 (𝐴 = 𝐵 → {𝑧 ∣ ∃𝑥𝐶𝑦𝐴 𝑧 = ⟪𝑥, 𝑦⟫} = {𝑧 ∣ ∃𝑥𝐶𝑦𝐵 𝑧 = ⟪𝑥, 𝑦⟫})
4 df-altxp 35961 . 2 (𝐶 ×× 𝐴) = {𝑧 ∣ ∃𝑥𝐶𝑦𝐴 𝑧 = ⟪𝑥, 𝑦⟫}
5 df-altxp 35961 . 2 (𝐶 ×× 𝐵) = {𝑧 ∣ ∃𝑥𝐶𝑦𝐵 𝑧 = ⟪𝑥, 𝑦⟫}
63, 4, 53eqtr4g 2801 1 (𝐴 = 𝐵 → (𝐶 ×× 𝐴) = (𝐶 ×× 𝐵))
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1539  {cab 2713  wrex 3069  caltop 35958   ×× caltxp 35959
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808  ax-5 1909  ax-6 1966  ax-7 2006  ax-9 2117  ax-ext 2707
This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1779  df-sb 2064  df-clab 2714  df-cleq 2728  df-rex 3070  df-altxp 35961
This theorem is referenced by: (None)
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