Theorem xpeq2 4679

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 2260	. . . 4 ⊢ (𝐴 = 𝐵 → (𝑦 ∈ 𝐴 ↔ 𝑦 ∈ 𝐵))
2	1	anbi2d 464	. . 3 ⊢ (𝐴 = 𝐵 → ((𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐴) ↔ (𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐵)))
3	2	opabbidv 4100	. 2 ⊢ (𝐴 = 𝐵 → {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐴)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐵)})
4		df-xp 4670	. 2 ⊢ (𝐶 × 𝐴) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐴)}
5		df-xp 4670	. 2 ⊢ (𝐶 × 𝐵) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐵)}
6	3, 4, 5	3eqtr4g 2254	1 ⊢ (𝐴 = 𝐵 → (𝐶 × 𝐴) = (𝐶 × 𝐵))

Syntax hints:   → wi 4   ∧ wa 104   = wceq 1364   ∈ wcel 2167  {copab 4094   × cxp 4662

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1461  ax-7 1462  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-11 1520  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-i5r 1549  ax-ext 2178

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1475  df-sb 1777  df-clab 2183  df-cleq 2189  df-clel 2192  df-opab 4096  df-xp 4670

This theorem is referenced by:  xpeq12  4683  xpeq2i  4685  xpeq2d  4688  xpeq0r  5093  xpdisj2  5096  pmvalg  6727  xpcomeng  6896  djueq12  7114  txuni2  14576  txbas  14578  txopn  14585  txrest  14596  txdis  14597  txdis1cn  14598  xmettxlem  14829  xmettx  14830