Users' Mathboxes Mathbox for Scott Fenton < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >   Mathboxes  >  altxpeq1 Structured version   Visualization version   GIF version

Theorem altxpeq1 32055
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 3134 . . 3 (𝐴 = 𝐵 → (∃𝑥𝐴𝑦𝐶 𝑧 = ⟪𝑥, 𝑦⟫ ↔ ∃𝑥𝐵𝑦𝐶 𝑧 = ⟪𝑥, 𝑦⟫))
21abbidv 2739 . 2 (𝐴 = 𝐵 → {𝑧 ∣ ∃𝑥𝐴𝑦𝐶 𝑧 = ⟪𝑥, 𝑦⟫} = {𝑧 ∣ ∃𝑥𝐵𝑦𝐶 𝑧 = ⟪𝑥, 𝑦⟫})
3 df-altxp 32041 . 2 (𝐴 ×× 𝐶) = {𝑧 ∣ ∃𝑥𝐴𝑦𝐶 𝑧 = ⟪𝑥, 𝑦⟫}
4 df-altxp 32041 . 2 (𝐵 ×× 𝐶) = {𝑧 ∣ ∃𝑥𝐵𝑦𝐶 𝑧 = ⟪𝑥, 𝑦⟫}
52, 3, 43eqtr4g 2679 1 (𝐴 = 𝐵 → (𝐴 ×× 𝐶) = (𝐵 ×× 𝐶))
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1481  {cab 2606  wrex 2910  caltop 32038   ×× caltxp 32039
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1720  ax-4 1735  ax-5 1837  ax-6 1886  ax-7 1933  ax-9 1997  ax-10 2017  ax-11 2032  ax-12 2045  ax-13 2244  ax-ext 2600
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1484  df-ex 1703  df-nf 1708  df-sb 1879  df-clab 2607  df-cleq 2613  df-clel 2616  df-nfc 2751  df-rex 2915  df-altxp 32041
This theorem is referenced by: (None)
  Copyright terms: Public domain W3C validator