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Theorem altxpeq2 33520
 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 3397 . . . 4 (𝐴 = 𝐵 → (∃𝑦𝐴 𝑧 = ⟪𝑥, 𝑦⟫ ↔ ∃𝑦𝐵 𝑧 = ⟪𝑥, 𝑦⟫))
21rexbidv 3289 . . 3 (𝐴 = 𝐵 → (∃𝑥𝐶𝑦𝐴 𝑧 = ⟪𝑥, 𝑦⟫ ↔ ∃𝑥𝐶𝑦𝐵 𝑧 = ⟪𝑥, 𝑦⟫))
32abbidv 2888 . 2 (𝐴 = 𝐵 → {𝑧 ∣ ∃𝑥𝐶𝑦𝐴 𝑧 = ⟪𝑥, 𝑦⟫} = {𝑧 ∣ ∃𝑥𝐶𝑦𝐵 𝑧 = ⟪𝑥, 𝑦⟫})
4 df-altxp 33505 . 2 (𝐶 ×× 𝐴) = {𝑧 ∣ ∃𝑥𝐶𝑦𝐴 𝑧 = ⟪𝑥, 𝑦⟫}
5 df-altxp 33505 . 2 (𝐶 ×× 𝐵) = {𝑧 ∣ ∃𝑥𝐶𝑦𝐵 𝑧 = ⟪𝑥, 𝑦⟫}
63, 4, 53eqtr4g 2884 1 (𝐴 = 𝐵 → (𝐶 ×× 𝐴) = (𝐶 ×× 𝐵))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   = wceq 1538  {cab 2802  ∃wrex 3134  ⟪caltop 33502   ×× caltxp 33503 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1971  ax-7 2016  ax-8 2117  ax-9 2125  ax-ext 2796 This theorem depends on definitions:  df-bi 210  df-an 400  df-ex 1782  df-sb 2071  df-clab 2803  df-cleq 2817  df-clel 2896  df-rex 3139  df-altxp 33505 This theorem is referenced by: (None)
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