Theorem preleq 4556

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 4555	. . . . 5 ⊢ ¬ (𝐷 ∈ 𝐶 ∧ 𝐶 ∈ 𝐷)
2		eleq12 2242	. . . . . 6 ⊢ ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → (𝐴 ∈ 𝐵 ↔ 𝐷 ∈ 𝐶))
3	2	anbi1d 465	. . . . 5 ⊢ ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → ((𝐴 ∈ 𝐵 ∧ 𝐶 ∈ 𝐷) ↔ (𝐷 ∈ 𝐶 ∧ 𝐶 ∈ 𝐷)))
4	1, 3	mtbiri 675	. . . 4 ⊢ ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → ¬ (𝐴 ∈ 𝐵 ∧ 𝐶 ∈ 𝐷))
5	4	con2i 627	. . 3 ⊢ ((𝐴 ∈ 𝐵 ∧ 𝐶 ∈ 𝐷) → ¬ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶))
6	5	adantr 276	. 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 3772	. . . 4 ⊢ ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
12	11	biimpi 120	. . 3 ⊢ ({𝐴, 𝐵} = {𝐶, 𝐷} → ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
13	12	adantl 277	. 2 ⊢ (((𝐴 ∈ 𝐵 ∧ 𝐶 ∈ 𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
14	6, 13	ecased 1349	1 ⊢ (((𝐴 ∈ 𝐵 ∧ 𝐶 ∈ 𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 104   ∨ wo 708   = wceq 1353   ∈ wcel 2148  Vcvv 2739  {cpr 3595

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 614  ax-in2 615  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159  ax-setind 4538

This theorem depends on definitions:  df-bi 117  df-3an 980  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-ral 2460  df-v 2741  df-dif 3133  df-un 3135  df-sn 3600  df-pr 3601

This theorem is referenced by:  opthreg  4557