Theorem xpeq1 4415

Description: Equality theorem for cross product. (Contributed by NM, 4-Jul-1994.)

Assertion

Ref	Expression
xpeq1	⊢ (𝐴 = 𝐵 → (𝐴 × 𝐶) = (𝐵 × 𝐶))

Proof of Theorem xpeq1

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

Step	Hyp	Ref	Expression
1		eleq2 2146	. . . 4 ⊢ (𝐴 = 𝐵 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
2	1	anbi1d 453	. . 3 ⊢ (𝐴 = 𝐵 → ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐶) ↔ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)))
3	2	opabbidv 3870	. 2 ⊢ (𝐴 = 𝐵 → {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐶)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)})
4		df-xp 4407	. 2 ⊢ (𝐴 × 𝐶) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐶)}
5		df-xp 4407	. 2 ⊢ (𝐵 × 𝐶) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)}
6	3, 4, 5	3eqtr4g 2140	1 ⊢ (𝐴 = 𝐵 → (𝐴 × 𝐶) = (𝐵 × 𝐶))

Syntax hints:   → wi 4   ∧ wa 102   = wceq 1285   ∈ wcel 1434  {copab 3864   × cxp 4399

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 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-11 1438  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469  ax-ext 2065

This theorem depends on definitions:  df-bi 115  df-tru 1288  df-nf 1391  df-sb 1688  df-clab 2070  df-cleq 2076  df-clel 2079  df-opab 3866  df-xp 4407

This theorem is referenced by:  xpeq12  4420  xpeq1i  4421  xpeq1d  4424  opthprc  4447  reseq2  4666  xpeq0r  4808  xpdisj1  4809  xpima1  4831  pmvalg  6346  xpsneng  6468  xpcomeng  6474  xpdom2g  6478  xpfi  6565  exmidomni  6703  exmidfodomrlemim  6730  hashxp  10069