Theorem xpeq1 4745

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 2295	. . . 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 4160	. 2 ⊢ (𝐴 = 𝐵 → {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐶)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)})
4		df-xp 4737	. 2 ⊢ (𝐴 × 𝐶) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐶)}
5		df-xp 4737	. 2 ⊢ (𝐵 × 𝐶) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)}
6	3, 4, 5	3eqtr4g 2289	1 ⊢ (𝐴 = 𝐵 → (𝐴 × 𝐶) = (𝐵 × 𝐶))

Syntax hints:   → wi 4   ∧ wa 104   = wceq 1398   ∈ wcel 2202  {copab 4154   × cxp 4729

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 2213

This theorem depends on definitions:  df-bi 117  df-tru 1401  df-nf 1510  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-opab 4156  df-xp 4737

This theorem is referenced by:  xpeq12  4750  xpeq1i  4751  xpeq1d  4754  opthprc  4783  reseq2  5014  xpeq0r  5166  xpdisj1  5168  xpima1  5190  pmvalg  6871  xpsneng  7049  xpcomeng  7055  xpdom2g  7059  xpfi  7167  exmidomni  7384  exmidfodomrlemim  7455  hashxp  11136  txuni2  15050  txbas  15052  txopn  15059  txrest  15070  txdis  15071  txdis1cn  15072  xmettxlem  15303  xmettx  15304  dvmptid  15510  incistruhgr  16014