Theorem preq12b 4820

Description: Equality relationship for two unordered pairs. (Contributed by NM, 17-Oct-1996.)

Hypotheses

Ref	Expression
preqr1.a	⊢ 𝐴 ∈ V
preqr1.b	⊢ 𝐵 ∈ V
preq12b.c	⊢ 𝐶 ∈ V
preq12b.d	⊢ 𝐷 ∈ V

Assertion

Ref	Expression
preq12b	⊢ ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))

Proof of Theorem preq12b

Step	Hyp	Ref	Expression
1		preqr1.a	. . . . . 6 ⊢ 𝐴 ∈ V
2	1	prid1 4733	. . . . 5 ⊢ 𝐴 ∈ {𝐴, 𝐵}
3		eleq2 2855	. . . . 5 ⊢ ({𝐴, 𝐵} = {𝐶, 𝐷} → (𝐴 ∈ {𝐴, 𝐵} ↔ 𝐴 ∈ {𝐶, 𝐷}))
4	2, 3	mpbii 236	. . . 4 ⊢ ({𝐴, 𝐵} = {𝐶, 𝐷} → 𝐴 ∈ {𝐶, 𝐷})
5	1	elpr 4619	. . . 4 ⊢ (𝐴 ∈ {𝐶, 𝐷} ↔ (𝐴 = 𝐶 ∨ 𝐴 = 𝐷))
6	4, 5	sylib 221	. . 3 ⊢ ({𝐴, 𝐵} = {𝐶, 𝐷} → (𝐴 = 𝐶 ∨ 𝐴 = 𝐷))
7		preq1 4704	. . . . . . . 8 ⊢ (𝐴 = 𝐶 → {𝐴, 𝐵} = {𝐶, 𝐵})
8	7	eqeq1d 2768	. . . . . . 7 ⊢ (𝐴 = 𝐶 → ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ {𝐶, 𝐵} = {𝐶, 𝐷}))
9		preqr1.b	. . . . . . . 8 ⊢ 𝐵 ∈ V
10		preq12b.d	. . . . . . . 8 ⊢ 𝐷 ∈ V
11	9, 10	preqr2 4819	. . . . . . 7 ⊢ ({𝐶, 𝐵} = {𝐶, 𝐷} → 𝐵 = 𝐷)
12	8, 11	biimtrdi 256	. . . . . 6 ⊢ (𝐴 = 𝐶 → ({𝐴, 𝐵} = {𝐶, 𝐷} → 𝐵 = 𝐷))
13	12	com12 33	. . . . 5 ⊢ ({𝐴, 𝐵} = {𝐶, 𝐷} → (𝐴 = 𝐶 → 𝐵 = 𝐷))
14	13	ancld 560	. . . 4 ⊢ ({𝐴, 𝐵} = {𝐶, 𝐷} → (𝐴 = 𝐶 → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
15		prcom 4703	. . . . . . 7 ⊢ {𝐶, 𝐷} = {𝐷, 𝐶}
16	15	eqeq2i 2779	. . . . . 6 ⊢ ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ {𝐴, 𝐵} = {𝐷, 𝐶})
17		preq1 4704	. . . . . . . . 9 ⊢ (𝐴 = 𝐷 → {𝐴, 𝐵} = {𝐷, 𝐵})
18	17	eqeq1d 2768	. . . . . . . 8 ⊢ (𝐴 = 𝐷 → ({𝐴, 𝐵} = {𝐷, 𝐶} ↔ {𝐷, 𝐵} = {𝐷, 𝐶}))
19		preq12b.c	. . . . . . . . 9 ⊢ 𝐶 ∈ V
20	9, 19	preqr2 4819	. . . . . . . 8 ⊢ ({𝐷, 𝐵} = {𝐷, 𝐶} → 𝐵 = 𝐶)
21	18, 20	biimtrdi 256	. . . . . . 7 ⊢ (𝐴 = 𝐷 → ({𝐴, 𝐵} = {𝐷, 𝐶} → 𝐵 = 𝐶))
22	21	com12 33	. . . . . 6 ⊢ ({𝐴, 𝐵} = {𝐷, 𝐶} → (𝐴 = 𝐷 → 𝐵 = 𝐶))
23	16, 22	sylbi 220	. . . . 5 ⊢ ({𝐴, 𝐵} = {𝐶, 𝐷} → (𝐴 = 𝐷 → 𝐵 = 𝐶))
24	23	ancld 560	. . . 4 ⊢ ({𝐴, 𝐵} = {𝐶, 𝐷} → (𝐴 = 𝐷 → (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
25	14, 24	orim12d 979	. . 3 ⊢ ({𝐴, 𝐵} = {𝐶, 𝐷} → ((𝐴 = 𝐶 ∨ 𝐴 = 𝐷) → ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶))))
26	6, 25	mpd 16	. 2 ⊢ ({𝐴, 𝐵} = {𝐶, 𝐷} → ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
27		preq12 4706	. . 3 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → {𝐴, 𝐵} = {𝐶, 𝐷})
28		preq12 4706	. . . 4 ⊢ ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → {𝐴, 𝐵} = {𝐷, 𝐶})
29		prcom 4703	. . . 4 ⊢ {𝐷, 𝐶} = {𝐶, 𝐷}
30	28, 29	eqtrdi 2817	. . 3 ⊢ ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → {𝐴, 𝐵} = {𝐶, 𝐷})
31	27, 30	jaoi 871	. 2 ⊢ (((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)) → {𝐴, 𝐵} = {𝐶, 𝐷})
32	26, 31	impbii 212	1 ⊢ ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   ∨ wo 861   = wceq 1570   ∈ wcel 2146  Vcvv 3458  {cpr 4596

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2738

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2745  df-cleq 2758  df-clel 2841  df-v 3460  df-un 3913  df-sn 4595  df-pr 4597

This theorem is used by:  opthpr  4821  preq12bg  4823  opeqpr  5493  opthhausdorff0  5506  axlowdimlem13  29341  upgrwlkdvdelem  30122  altopthsn  36473  sprsymrelfolem2  48282  prproropf1olem4  48295  reuopreuprim  48315  usgrexmpl2trifr  48842  rrx2xpref1o  49538