Theorem xpeq2 4378

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 2142	. . . 4 ⊢ (𝐴 = 𝐵 → (𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵))
2	1	anbi2d 451	. . 3 ⊢ (𝐴 = 𝐵 → ((𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐴) ↔ (𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐵)))
3	2	opabbidv 3844	. 2 ⊢ (𝐴 = 𝐵 → {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐴)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐵)})
4		df-xp 4369	. 2 ⊢ (𝐶 × 𝐴) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐴)}
5		df-xp 4369	. 2 ⊢ (𝐶 × 𝐵) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐵)}
6	3, 4, 5	3eqtr4g 2138	1 ⊢ (𝐴 = 𝐵 → (𝐶 × 𝐴) = (𝐶 × 𝐵))

Syntax hints:   → wi 4   ∧ wa 102   = wceq 1284   ∈ wcel 1433  {copab 3838   × cxp 4361

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-11 1437  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-opab 3840  df-xp 4369

This theorem is referenced by:  xpeq12  4382  xpeq2i  4384  xpeq2d  4387  xpeq0r  4766  xpdisj2  4768  xpcomeng  6325