Theorem preleq 4465

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 4464	. . . . 5 ⊢ ¬ (𝐷 ∈ 𝐶 ∧ 𝐶 ∈ 𝐷)
2		eleq12 2202	. . . . . 6 ⊢ ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → (𝐴 ∈ 𝐵 ↔ 𝐷 ∈ 𝐶))
3	2	anbi1d 460	. . . . 5 ⊢ ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → ((𝐴 ∈ 𝐵 ∧ 𝐶 ∈ 𝐷) ↔ (𝐷 ∈ 𝐶 ∧ 𝐶 ∈ 𝐷)))
4	1, 3	mtbiri 664	. . . 4 ⊢ ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → ¬ (𝐴 ∈ 𝐵 ∧ 𝐶 ∈ 𝐷))
5	4	con2i 616	. . 3 ⊢ ((𝐴 ∈ 𝐵 ∧ 𝐶 ∈ 𝐷) → ¬ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶))
6	5	adantr 274	. 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 3692	. . . 4 ⊢ ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
12	11	biimpi 119	. . 3 ⊢ ({𝐴, 𝐵} = {𝐶, 𝐷} → ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
13	12	adantl 275	. 2 ⊢ (((𝐴 ∈ 𝐵 ∧ 𝐶 ∈ 𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
14	6, 13	ecased 1327	1 ⊢ (((𝐴 ∈ 𝐵 ∧ 𝐶 ∈ 𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 103   ∨ wo 697   = wceq 1331   ∈ wcel 1480  Vcvv 2681  {cpr 3523

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 603  ax-in2 604  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2119  ax-setind 4447

This theorem depends on definitions:  df-bi 116  df-3an 964  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2124  df-cleq 2130  df-clel 2133  df-nfc 2268  df-ral 2419  df-v 2683  df-dif 3068  df-un 3070  df-sn 3528  df-pr 3529

This theorem is referenced by:  opthreg  4466