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Theorem altxpeq1 32593
 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 3322 . . 3 (𝐴 = 𝐵 → (∃𝑥𝐴𝑦𝐶 𝑧 = ⟪𝑥, 𝑦⟫ ↔ ∃𝑥𝐵𝑦𝐶 𝑧 = ⟪𝑥, 𝑦⟫))
21abbidv 2918 . 2 (𝐴 = 𝐵 → {𝑧 ∣ ∃𝑥𝐴𝑦𝐶 𝑧 = ⟪𝑥, 𝑦⟫} = {𝑧 ∣ ∃𝑥𝐵𝑦𝐶 𝑧 = ⟪𝑥, 𝑦⟫})
3 df-altxp 32579 . 2 (𝐴 ×× 𝐶) = {𝑧 ∣ ∃𝑥𝐴𝑦𝐶 𝑧 = ⟪𝑥, 𝑦⟫}
4 df-altxp 32579 . 2 (𝐵 ×× 𝐶) = {𝑧 ∣ ∃𝑥𝐵𝑦𝐶 𝑧 = ⟪𝑥, 𝑦⟫}
52, 3, 43eqtr4g 2858 1 (𝐴 = 𝐵 → (𝐴 ×× 𝐶) = (𝐵 ×× 𝐶))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   = wceq 1653  {cab 2785  ∃wrex 3090  ⟪caltop 32576   ×× caltxp 32577 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1891  ax-4 1905  ax-5 2006  ax-6 2072  ax-7 2107  ax-9 2166  ax-10 2185  ax-11 2200  ax-12 2213  ax-ext 2777 This theorem depends on definitions:  df-bi 199  df-an 386  df-or 875  df-tru 1657  df-ex 1876  df-nf 1880  df-sb 2065  df-clab 2786  df-cleq 2792  df-clel 2795  df-nfc 2930  df-rex 3095  df-altxp 32579 This theorem is referenced by: (None)
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