Theorem xpeq2 4626

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 2234	. . . 4 ⊢ (𝐴 = 𝐵 → (𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵))
2	1	anbi2d 461	. . 3 ⊢ (𝐴 = 𝐵 → ((𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐴) ↔ (𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐵)))
3	2	opabbidv 4055	. 2 ⊢ (𝐴 = 𝐵 → {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐴)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐵)})
4		df-xp 4617	. 2 ⊢ (𝐶 × 𝐴) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐴)}
5		df-xp 4617	. 2 ⊢ (𝐶 × 𝐵) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐵)}
6	3, 4, 5	3eqtr4g 2228	1 ⊢ (𝐴 = 𝐵 → (𝐶 × 𝐴) = (𝐶 × 𝐵))

Syntax hints:   → wi 4   ∧ wa 103   = wceq 1348   ∈ wcel 2141  {copab 4049   × cxp 4609

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-11 1499  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-opab 4051  df-xp 4617

This theorem is referenced by:  xpeq12  4630  xpeq2i  4632  xpeq2d  4635  xpeq0r  5033  xpdisj2  5036  pmvalg  6637  xpcomeng  6806  djueq12  7016  txuni2  13050  txbas  13052  txopn  13059  txrest  13070  txdis  13071  txdis1cn  13072  xmettxlem  13303  xmettx  13304