Theorem xpeq1 4427

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 2148	. . . 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 3881	. 2 ⊢ (𝐴 = 𝐵 → {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐶)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)})
4		df-xp 4419	. 2 ⊢ (𝐴 × 𝐶) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐶)}
5		df-xp 4419	. 2 ⊢ (𝐵 × 𝐶) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)}
6	3, 4, 5	3eqtr4g 2142	1 ⊢ (𝐴 = 𝐵 → (𝐴 × 𝐶) = (𝐵 × 𝐶))

Syntax hints:   → wi 4   ∧ wa 102   = wceq 1287   ∈ wcel 1436  {copab 3875   × cxp 4411

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 1379  ax-7 1380  ax-gen 1381  ax-ie1 1425  ax-ie2 1426  ax-8 1438  ax-11 1440  ax-4 1443  ax-17 1462  ax-i9 1466  ax-ial 1470  ax-i5r 1471  ax-ext 2067

This theorem depends on definitions:  df-bi 115  df-tru 1290  df-nf 1393  df-sb 1690  df-clab 2072  df-cleq 2078  df-clel 2081  df-opab 3877  df-xp 4419

This theorem is referenced by:  xpeq12  4432  xpeq1i  4433  xpeq1d  4436  opthprc  4459  reseq2  4678  xpeq0r  4822  xpdisj1  4823  xpima1  4845  pmvalg  6370  xpsneng  6492  xpcomeng  6498  xpdom2g  6502  xpfi  6593  exmidomni  6745  exmidfodomrlemim  6774  hashxp  10152