Theorem xpeq2 4360

Description: Equality theorem for cross product. (Contributed by NM, 5-Jul-1994.)

Assertion

Ref	Expression
xpeq2	⊢ (𝐴 = 𝐵 → (𝐶 × 𝐴) = (𝐶 × 𝐵))

Proof of Theorem xpeq2

Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		eleq2 2101	. . . 4 ⊢ (𝐴 = 𝐵 → (𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵))
2	1	anbi2d 437	. . 3 ⊢ (𝐴 = 𝐵 → ((𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐴) ↔ (𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐵)))
3	2	opabbidv 3823	. 2 ⊢ (𝐴 = 𝐵 → {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐴)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐵)})
4		df-xp 4351	. 2 ⊢ (𝐶 × 𝐴) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐴)}
5		df-xp 4351	. 2 ⊢ (𝐶 × 𝐵) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐵)}
6	3, 4, 5	3eqtr4g 2097	1 ⊢ (𝐴 = 𝐵 → (𝐶 × 𝐴) = (𝐶 × 𝐵))

Syntax hints:   → wi 4   ∧ wa 97   = wceq 1243   ∈ wcel 1393  {copab 3817   × cxp 4343

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-11 1397  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-opab 3819  df-xp 4351

This theorem is referenced by:  xpeq12  4364  xpeq2i  4366  xpeq2d  4369  xpeq0r  4746  xpdisj2  4748  xpcomeng  6302