Theorem altxpeq2 36930

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 3319	. . . 4 ⊢ (𝐴 = 𝐵 → (∃𝑦 ∈ 𝐴 𝑧 = ⟪𝑥, 𝑦⟫ ↔ ∃𝑦 ∈ 𝐵 𝑧 = ⟪𝑥, 𝑦⟫))
2	1	rexbidv 3190	. . 3 ⊢ (𝐴 = 𝐵 → (∃𝑥 ∈ 𝐶 ∃𝑦 ∈ 𝐴 𝑧 = ⟪𝑥, 𝑦⟫ ↔ ∃𝑥 ∈ 𝐶 ∃𝑦 ∈ 𝐵 𝑧 = ⟪𝑥, 𝑦⟫))
3	2	abbidv 2829	. 2 ⊢ (𝐴 = 𝐵 → {𝑧 ∣ ∃𝑥 ∈ 𝐶 ∃𝑦 ∈ 𝐴 𝑧 = ⟪𝑥, 𝑦⟫} = {𝑧 ∣ ∃𝑥 ∈ 𝐶 ∃𝑦 ∈ 𝐵 𝑧 = ⟪𝑥, 𝑦⟫})
4		df-altxp 36915	. 2 ⊢ (𝐶 ×× 𝐴) = {𝑧 ∣ ∃𝑥 ∈ 𝐶 ∃𝑦 ∈ 𝐴 𝑧 = ⟪𝑥, 𝑦⟫}
5		df-altxp 36915	. 2 ⊢ (𝐶 ×× 𝐵) = {𝑧 ∣ ∃𝑥 ∈ 𝐶 ∃𝑦 ∈ 𝐵 𝑧 = ⟪𝑥, 𝑦⟫}
6	3, 4, 5	3eqtr4g 2823	1 ⊢ (𝐴 = 𝐵 → (𝐶 ×× 𝐴) = (𝐶 ×× 𝐵))

This proof depends on syntax axioms:   → wi 4   = wceq 1570  {cab 2741  ∃wrex 3090  ⟪caltop 36912   ×× caltxp 36913

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-rex 3091  df-altxp 36915

This theorem is used by: (None)