Theorem xpid11m 4575

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 4573	. . . . . 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 4553	. . . . 5 ⊢ ((𝐴 × 𝐴) = (𝐵 × 𝐵) → dom (𝐴 × 𝐴) = dom (𝐵 × 𝐵))
4	2, 3	sylan9req 2134	. . . 4 ⊢ (((∃𝑥 𝑥 ∈ 𝐴 ∧ ∃𝑥 𝑥 ∈ 𝐵) ∧ (𝐴 × 𝐴) = (𝐵 × 𝐵)) → 𝐴 = dom (𝐵 × 𝐵))
5		dmxpm 4573	. . . . 5 ⊢ (∃𝑥 𝑥 ∈ 𝐵 → dom (𝐵 × 𝐵) = 𝐵)
6	5	ad2antlr 472	. . . 4 ⊢ (((∃𝑥 𝑥 ∈ 𝐴 ∧ ∃𝑥 𝑥 ∈ 𝐵) ∧ (𝐴 × 𝐴) = (𝐵 × 𝐵)) → dom (𝐵 × 𝐵) = 𝐵)
7	4, 6	eqtrd 2113	. . 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 4382	. . 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 1284  ∃wex 1421   ∈ wcel 1433   × cxp 4361  dom cdm 4363

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 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-14 1445  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063  ax-sep 3896  ax-pow 3948  ax-pr 3964

This theorem depends on definitions:  df-bi 115  df-3an 921  df-tru 1287  df-nf 1390  df-sb 1686  df-eu 1944  df-mo 1945  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-ral 2353  df-v 2603  df-un 2977  df-in 2979  df-ss 2986  df-pw 3384  df-sn 3404  df-pr 3405  df-op 3407  df-br 3786  df-opab 3840  df-xp 4369  df-dm 4373

This theorem is referenced by: (None)