Theorem xpid11m 4500

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 4498	. . . . . 6 ⊢ (∃x x ∈ A → dom (A × A) = A)
2	1	adantr 261	. . . . 5 ⊢ ((∃x x ∈ A ∧ ∃x x ∈ B) → dom (A × A) = A)
3		dmeq 4478	. . . . 5 ⊢ ((A × A) = (B × B) → dom (A × A) = dom (B × B))
4	2, 3	sylan9req 2090	. . . 4 ⊢ (((∃x x ∈ A ∧ ∃x x ∈ B) ∧ (A × A) = (B × B)) → A = dom (B × B))
5		dmxpm 4498	. . . . 5 ⊢ (∃x x ∈ B → dom (B × B) = B)
6	5	ad2antlr 458	. . . 4 ⊢ (((∃x x ∈ A ∧ ∃x x ∈ B) ∧ (A × A) = (B × B)) → dom (B × B) = B)
7	4, 6	eqtrd 2069	. . 3 ⊢ (((∃x x ∈ A ∧ ∃x x ∈ B) ∧ (A × A) = (B × B)) → A = B)
8	7	ex 108	. 2 ⊢ ((∃x x ∈ A ∧ ∃x x ∈ B) → ((A × A) = (B × B) → A = B))
9		xpeq12 4307	. . 3 ⊢ ((A = B ∧ A = B) → (A × A) = (B × B))
10	9	anidms 377	. 2 ⊢ (A = B → (A × A) = (B × B))
11	8, 10	impbid1 130	1 ⊢ ((∃x x ∈ A ∧ ∃x x ∈ B) → ((A × A) = (B × B) ↔ A = B))

Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98   = wceq 1242  ∃wex 1378   ∈ wcel 1390   × cxp 4286  dom cdm 4288

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bndl 1396  ax-4 1397  ax-14 1402  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019  ax-sep 3866  ax-pow 3918  ax-pr 3935

This theorem depends on definitions:  df-bi 110  df-3an 886  df-tru 1245  df-nf 1347  df-sb 1643  df-eu 1900  df-mo 1901  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-ral 2305  df-v 2553  df-un 2916  df-in 2918  df-ss 2925  df-pw 3353  df-sn 3373  df-pr 3374  df-op 3376  df-br 3756  df-opab 3810  df-xp 4294  df-dm 4298

This theorem is referenced by: (None)