Users' Mathboxes Mathbox for Scott Fenton < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >   Mathboxes  >  altxpeq2 Structured version   Visualization version   GIF version

Theorem altxpeq2 32585
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 3320 . . . 4 (𝐴 = 𝐵 → (∃𝑦𝐴 𝑧 = ⟪𝑥, 𝑦⟫ ↔ ∃𝑦𝐵 𝑧 = ⟪𝑥, 𝑦⟫))
21rexbidv 3231 . . 3 (𝐴 = 𝐵 → (∃𝑥𝐶𝑦𝐴 𝑧 = ⟪𝑥, 𝑦⟫ ↔ ∃𝑥𝐶𝑦𝐵 𝑧 = ⟪𝑥, 𝑦⟫))
32abbidv 2916 . 2 (𝐴 = 𝐵 → {𝑧 ∣ ∃𝑥𝐶𝑦𝐴 𝑧 = ⟪𝑥, 𝑦⟫} = {𝑧 ∣ ∃𝑥𝐶𝑦𝐵 𝑧 = ⟪𝑥, 𝑦⟫})
4 df-altxp 32570 . 2 (𝐶 ×× 𝐴) = {𝑧 ∣ ∃𝑥𝐶𝑦𝐴 𝑧 = ⟪𝑥, 𝑦⟫}
5 df-altxp 32570 . 2 (𝐶 ×× 𝐵) = {𝑧 ∣ ∃𝑥𝐶𝑦𝐵 𝑧 = ⟪𝑥, 𝑦⟫}
63, 4, 53eqtr4g 2856 1 (𝐴 = 𝐵 → (𝐶 ×× 𝐴) = (𝐶 ×× 𝐵))
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1653  {cab 2783  wrex 3088  caltop 32567   ×× caltxp 32568
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 2775
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 2784  df-cleq 2790  df-clel 2793  df-nfc 2928  df-rex 3093  df-altxp 32570
This theorem is referenced by: (None)
  Copyright terms: Public domain W3C validator