Theorem xpeq1 4768

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 2298	. . . 4 ⊢ (𝐴 = 𝐵 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
2	1	anbi1d 465	. . 3 ⊢ (𝐴 = 𝐵 → ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐶) ↔ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)))
3	2	opabbidv 4181	. 2 ⊢ (𝐴 = 𝐵 → {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐶)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)})
4		df-xp 4760	. 2 ⊢ (𝐴 × 𝐶) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐶)}
5		df-xp 4760	. 2 ⊢ (𝐵 × 𝐶) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)}
6	3, 4, 5	3eqtr4g 2292	1 ⊢ (𝐴 = 𝐵 → (𝐴 × 𝐶) = (𝐵 × 𝐶))

Syntax hints:   → wi 4   ∧ wa 104   = wceq 1398   ∈ wcel 2205  {copab 4175   × cxp 4752

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 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-11 1555  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2216

This theorem depends on definitions:  df-bi 117  df-tru 1401  df-nf 1510  df-sb 1812  df-clab 2221  df-cleq 2227  df-clel 2230  df-opab 4177  df-xp 4760

This theorem is referenced by:  xpeq12  4773  xpeq1i  4774  xpeq1d  4777  opthprc  4806  reseq2  5038  xpeq0r  5190  xpdisj1  5192  xpima1  5214  pmvalg  6906  xpsneng  7086  xpcomeng  7092  xpdom2g  7096  xpfi  7205  exmidomni  7446  exmidfodomrlemim  7517  hashxp  11216  txuni2  15247  txbas  15249  txopn  15256  txrest  15267  txdis  15268  txdis1cn  15269  xmettxlem  15500  xmettx  15501  dvmptid  15707  incistruhgr  16211