Theorem preleq 4298

Description: Equality of two unordered pairs when one member of each pair contains the other member. (Contributed by NM, 16-Oct-1996.)

Hypotheses

Ref	Expression
preleq.1	⊢ 𝐴 ∈ V
preleq.2	⊢ 𝐵 ∈ V
preleq.3	⊢ 𝐶 ∈ V
preleq.4	⊢ 𝐷 ∈ V

Assertion

Ref	Expression
preleq	⊢ (((𝐴 ∈ 𝐵 ∧ 𝐶 ∈ 𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))

Proof of Theorem preleq

Step	Hyp	Ref	Expression
1		en2lp 4297	. . . . 5 ⊢ ¬ (𝐷 ∈ 𝐶 ∧ 𝐶 ∈ 𝐷)
2		eleq12 2143	. . . . . 6 ⊢ ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → (𝐴 ∈ 𝐵 ↔ 𝐷 ∈ 𝐶))
3	2	anbi1d 452	. . . . 5 ⊢ ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → ((𝐴 ∈ 𝐵 ∧ 𝐶 ∈ 𝐷) ↔ (𝐷 ∈ 𝐶 ∧ 𝐶 ∈ 𝐷)))
4	1, 3	mtbiri 632	. . . 4 ⊢ ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → ¬ (𝐴 ∈ 𝐵 ∧ 𝐶 ∈ 𝐷))
5	4	con2i 589	. . 3 ⊢ ((𝐴 ∈ 𝐵 ∧ 𝐶 ∈ 𝐷) → ¬ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶))
6	5	adantr 270	. 2 ⊢ (((𝐴 ∈ 𝐵 ∧ 𝐶 ∈ 𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → ¬ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶))
7		preleq.1	. . . . 5 ⊢ 𝐴 ∈ V
8		preleq.2	. . . . 5 ⊢ 𝐵 ∈ V
9		preleq.3	. . . . 5 ⊢ 𝐶 ∈ V
10		preleq.4	. . . . 5 ⊢ 𝐷 ∈ V
11	7, 8, 9, 10	preq12b 3562	. . . 4 ⊢ ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
12	11	biimpi 118	. . 3 ⊢ ({𝐴, 𝐵} = {𝐶, 𝐷} → ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
13	12	adantl 271	. 2 ⊢ (((𝐴 ∈ 𝐵 ∧ 𝐶 ∈ 𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
14	6, 13	ecased 1280	1 ⊢ (((𝐴 ∈ 𝐵 ∧ 𝐶 ∈ 𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 102   ∨ wo 661   = wceq 1284   ∈ wcel 1433  Vcvv 2601  {cpr 3399

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-in1 576  ax-in2 577  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-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063  ax-setind 4280

This theorem depends on definitions:  df-bi 115  df-3an 921  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-ral 2353  df-v 2603  df-dif 2975  df-un 2977  df-sn 3404  df-pr 3405

This theorem is referenced by:  opthreg  4299