Theorem xpid11m 4654

Description: The cross product of a class with itself is one-to-one. (Contributed by Jim Kingdon, 8-Dec-2018.)

Assertion

Ref	Expression
xpid11m	⊢ ((∃𝑥 𝑥 ∈ 𝐴 ∧ ∃𝑥 𝑥 ∈ 𝐵) → ((𝐴 × 𝐴) = (𝐵 × 𝐵) ↔ 𝐴 = 𝐵))

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Proof of Theorem xpid11m

Step	Hyp	Ref	Expression
1		dmxpm 4652	. . . . . 6 ⊢ (∃𝑥 𝑥 ∈ 𝐴 → dom (𝐴 × 𝐴) = 𝐴)
2	1	adantr 270	. . . . 5 ⊢ ((∃𝑥 𝑥 ∈ 𝐴 ∧ ∃𝑥 𝑥 ∈ 𝐵) → dom (𝐴 × 𝐴) = 𝐴)
3		dmeq 4632	. . . . 5 ⊢ ((𝐴 × 𝐴) = (𝐵 × 𝐵) → dom (𝐴 × 𝐴) = dom (𝐵 × 𝐵))
4	2, 3	sylan9req 2141	. . . 4 ⊢ (((∃𝑥 𝑥 ∈ 𝐴 ∧ ∃𝑥 𝑥 ∈ 𝐵) ∧ (𝐴 × 𝐴) = (𝐵 × 𝐵)) → 𝐴 = dom (𝐵 × 𝐵))
5		dmxpm 4652	. . . . 5 ⊢ (∃𝑥 𝑥 ∈ 𝐵 → dom (𝐵 × 𝐵) = 𝐵)
6	5	ad2antlr 473	. . . 4 ⊢ (((∃𝑥 𝑥 ∈ 𝐴 ∧ ∃𝑥 𝑥 ∈ 𝐵) ∧ (𝐴 × 𝐴) = (𝐵 × 𝐵)) → dom (𝐵 × 𝐵) = 𝐵)
7	4, 6	eqtrd 2120	. . 3 ⊢ (((∃𝑥 𝑥 ∈ 𝐴 ∧ ∃𝑥 𝑥 ∈ 𝐵) ∧ (𝐴 × 𝐴) = (𝐵 × 𝐵)) → 𝐴 = 𝐵)
8	7	ex 113	. 2 ⊢ ((∃𝑥 𝑥 ∈ 𝐴 ∧ ∃𝑥 𝑥 ∈ 𝐵) → ((𝐴 × 𝐴) = (𝐵 × 𝐵) → 𝐴 = 𝐵))
9		xpeq12 4455	. . 3 ⊢ ((𝐴 = 𝐵 ∧ 𝐴 = 𝐵) → (𝐴 × 𝐴) = (𝐵 × 𝐵))
10	9	anidms 389	. 2 ⊢ (𝐴 = 𝐵 → (𝐴 × 𝐴) = (𝐵 × 𝐵))
11	8, 10	impbid1 140	1 ⊢ ((∃𝑥 𝑥 ∈ 𝐴 ∧ ∃𝑥 𝑥 ∈ 𝐵) → ((𝐴 × 𝐴) = (𝐵 × 𝐵) ↔ 𝐴 = 𝐵))

Syntax hints:   → wi 4   ∧ wa 102   ↔ wb 103   = wceq 1289  ∃wex 1426   ∈ wcel 1438   × cxp 4434  dom cdm 4436

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-io 665  ax-5 1381  ax-7 1382  ax-gen 1383  ax-ie1 1427  ax-ie2 1428  ax-8 1440  ax-10 1441  ax-11 1442  ax-i12 1443  ax-bndl 1444  ax-4 1445  ax-14 1450  ax-17 1464  ax-i9 1468  ax-ial 1472  ax-i5r 1473  ax-ext 2070  ax-sep 3955  ax-pow 4007  ax-pr 4034

This theorem depends on definitions:  df-bi 115  df-3an 926  df-tru 1292  df-nf 1395  df-sb 1693  df-eu 1951  df-mo 1952  df-clab 2075  df-cleq 2081  df-clel 2084  df-nfc 2217  df-ral 2364  df-v 2621  df-un 3003  df-in 3005  df-ss 3012  df-pw 3429  df-sn 3450  df-pr 3451  df-op 3453  df-br 3844  df-opab 3898  df-xp 4442  df-dm 4446

This theorem is referenced by: (None)