Theorem altxpeq1 35602

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.

Step	Hyp	Ref	Expression
1		rexeq 3319	. . 3 ⊢ (𝐴 = 𝐵 → (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐶 𝑧 = ⟪𝑥, 𝑦⟫ ↔ ∃𝑥 ∈ 𝐵 ∃𝑦 ∈ 𝐶 𝑧 = ⟪𝑥, 𝑦⟫))
2	1	abbidv 2797	. 2 ⊢ (𝐴 = 𝐵 → {𝑧 ∣ ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐶 𝑧 = ⟪𝑥, 𝑦⟫} = {𝑧 ∣ ∃𝑥 ∈ 𝐵 ∃𝑦 ∈ 𝐶 𝑧 = ⟪𝑥, 𝑦⟫})
3		df-altxp 35588	. 2 ⊢ (𝐴 ×× 𝐶) = {𝑧 ∣ ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐶 𝑧 = ⟪𝑥, 𝑦⟫}
4		df-altxp 35588	. 2 ⊢ (𝐵 ×× 𝐶) = {𝑧 ∣ ∃𝑥 ∈ 𝐵 ∃𝑦 ∈ 𝐶 𝑧 = ⟪𝑥, 𝑦⟫}
5	2, 3, 4	3eqtr4g 2793	1 ⊢ (𝐴 = 𝐵 → (𝐴 ×× 𝐶) = (𝐵 ×× 𝐶))

Syntax hints:   → wi 4   = wceq 1533  {cab 2705  ∃wrex 3067  ⟪caltop 35585   ×× caltxp 35586

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-9 2108  ax-ext 2699

This theorem depends on definitions:  df-bi 206  df-an 395  df-ex 1774  df-sb 2060  df-clab 2706  df-cleq 2720  df-rex 3068  df-altxp 35588

This theorem is referenced by: (None)