Theorem altxpeq2 36217

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.

Step	Hyp	Ref	Expression
1		rexeq 3295	. . . 4 ⊢ (𝐴 = 𝐵 → (∃𝑦 ∈ 𝐴 𝑧 = ⟪𝑥, 𝑦⟫ ↔ ∃𝑦 ∈ 𝐵 𝑧 = ⟪𝑥, 𝑦⟫))
2	1	rexbidv 3165	. . 3 ⊢ (𝐴 = 𝐵 → (∃𝑥 ∈ 𝐶 ∃𝑦 ∈ 𝐴 𝑧 = ⟪𝑥, 𝑦⟫ ↔ ∃𝑥 ∈ 𝐶 ∃𝑦 ∈ 𝐵 𝑧 = ⟪𝑥, 𝑦⟫))
3	2	abbidv 2807	. 2 ⊢ (𝐴 = 𝐵 → {𝑧 ∣ ∃𝑥 ∈ 𝐶 ∃𝑦 ∈ 𝐴 𝑧 = ⟪𝑥, 𝑦⟫} = {𝑧 ∣ ∃𝑥 ∈ 𝐶 ∃𝑦 ∈ 𝐵 𝑧 = ⟪𝑥, 𝑦⟫})
4		df-altxp 36202	. 2 ⊢ (𝐶 ×× 𝐴) = {𝑧 ∣ ∃𝑥 ∈ 𝐶 ∃𝑦 ∈ 𝐴 𝑧 = ⟪𝑥, 𝑦⟫}
5		df-altxp 36202	. 2 ⊢ (𝐶 ×× 𝐵) = {𝑧 ∣ ∃𝑥 ∈ 𝐶 ∃𝑦 ∈ 𝐵 𝑧 = ⟪𝑥, 𝑦⟫}
6	3, 4, 5	3eqtr4g 2801	1 ⊢ (𝐴 = 𝐵 → (𝐶 ×× 𝐴) = (𝐶 ×× 𝐵))

Syntax hints:   → wi 4   = wceq 1548  {cab 2719  ∃wrex 3065  ⟪caltop 36199   ×× caltxp 36200

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1975  ax-7 2016  ax-9 2131  ax-ext 2713

This theorem depends on definitions:  df-bi 209  df-an 398  df-ex 1788  df-sb 2075  df-clab 2720  df-cleq 2733  df-rex 3066  df-altxp 36202

This theorem is referenced by: (None)