Theorem altxpeq1 33436

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 3408	. . 3 ⊢ (𝐴 = 𝐵 → (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐶 𝑧 = ⟪𝑥, 𝑦⟫ ↔ ∃𝑥 ∈ 𝐵 ∃𝑦 ∈ 𝐶 𝑧 = ⟪𝑥, 𝑦⟫))
2	1	abbidv 2887	. 2 ⊢ (𝐴 = 𝐵 → {𝑧 ∣ ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐶 𝑧 = ⟪𝑥, 𝑦⟫} = {𝑧 ∣ ∃𝑥 ∈ 𝐵 ∃𝑦 ∈ 𝐶 𝑧 = ⟪𝑥, 𝑦⟫})
3		df-altxp 33422	. 2 ⊢ (𝐴 ×× 𝐶) = {𝑧 ∣ ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐶 𝑧 = ⟪𝑥, 𝑦⟫}
4		df-altxp 33422	. 2 ⊢ (𝐵 ×× 𝐶) = {𝑧 ∣ ∃𝑥 ∈ 𝐵 ∃𝑦 ∈ 𝐶 𝑧 = ⟪𝑥, 𝑦⟫}
5	2, 3, 4	3eqtr4g 2883	1 ⊢ (𝐴 = 𝐵 → (𝐴 ×× 𝐶) = (𝐵 ×× 𝐶))

Syntax hints:   → wi 4   = wceq 1537  {cab 2801  ∃wrex 3141  ⟪caltop 33419   ×× caltxp 33420

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-ext 2795

This theorem depends on definitions:  df-bi 209  df-an 399  df-ex 1781  df-sb 2070  df-clab 2802  df-cleq 2816  df-clel 2895  df-rex 3146  df-altxp 33422

This theorem is referenced by: (None)