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Theorem preq12b 4794
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 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)))

Proof of Theorem preq12b
StepHypRef Expression
1 preqr1.a . . . . . 6 𝐴 ∈ V
21prid1 4707 . . . . 5 𝐴 ∈ {𝐴, 𝐵}
3 eleq2 2826 . . . . 5 ({𝐴, 𝐵} = {𝐶, 𝐷} → (𝐴 ∈ {𝐴, 𝐵} ↔ 𝐴 ∈ {𝐶, 𝐷}))
42, 3mpbii 233 . . . 4 ({𝐴, 𝐵} = {𝐶, 𝐷} → 𝐴 ∈ {𝐶, 𝐷})
51elpr 4593 . . . 4 (𝐴 ∈ {𝐶, 𝐷} ↔ (𝐴 = 𝐶𝐴 = 𝐷))
64, 5sylib 218 . . 3 ({𝐴, 𝐵} = {𝐶, 𝐷} → (𝐴 = 𝐶𝐴 = 𝐷))
7 preq1 4678 . . . . . . . 8 (𝐴 = 𝐶 → {𝐴, 𝐵} = {𝐶, 𝐵})
87eqeq1d 2739 . . . . . . 7 (𝐴 = 𝐶 → ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ {𝐶, 𝐵} = {𝐶, 𝐷}))
9 preqr1.b . . . . . . . 8 𝐵 ∈ V
10 preq12b.d . . . . . . . 8 𝐷 ∈ V
119, 10preqr2 4793 . . . . . . 7 ({𝐶, 𝐵} = {𝐶, 𝐷} → 𝐵 = 𝐷)
128, 11biimtrdi 253 . . . . . 6 (𝐴 = 𝐶 → ({𝐴, 𝐵} = {𝐶, 𝐷} → 𝐵 = 𝐷))
1312com12 32 . . . . 5 ({𝐴, 𝐵} = {𝐶, 𝐷} → (𝐴 = 𝐶𝐵 = 𝐷))
1413ancld 550 . . . 4 ({𝐴, 𝐵} = {𝐶, 𝐷} → (𝐴 = 𝐶 → (𝐴 = 𝐶𝐵 = 𝐷)))
15 prcom 4677 . . . . . . 7 {𝐶, 𝐷} = {𝐷, 𝐶}
1615eqeq2i 2750 . . . . . 6 ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ {𝐴, 𝐵} = {𝐷, 𝐶})
17 preq1 4678 . . . . . . . . 9 (𝐴 = 𝐷 → {𝐴, 𝐵} = {𝐷, 𝐵})
1817eqeq1d 2739 . . . . . . . 8 (𝐴 = 𝐷 → ({𝐴, 𝐵} = {𝐷, 𝐶} ↔ {𝐷, 𝐵} = {𝐷, 𝐶}))
19 preq12b.c . . . . . . . . 9 𝐶 ∈ V
209, 19preqr2 4793 . . . . . . . 8 ({𝐷, 𝐵} = {𝐷, 𝐶} → 𝐵 = 𝐶)
2118, 20biimtrdi 253 . . . . . . 7 (𝐴 = 𝐷 → ({𝐴, 𝐵} = {𝐷, 𝐶} → 𝐵 = 𝐶))
2221com12 32 . . . . . 6 ({𝐴, 𝐵} = {𝐷, 𝐶} → (𝐴 = 𝐷𝐵 = 𝐶))
2316, 22sylbi 217 . . . . 5 ({𝐴, 𝐵} = {𝐶, 𝐷} → (𝐴 = 𝐷𝐵 = 𝐶))
2423ancld 550 . . . 4 ({𝐴, 𝐵} = {𝐶, 𝐷} → (𝐴 = 𝐷 → (𝐴 = 𝐷𝐵 = 𝐶)))
2514, 24orim12d 967 . . 3 ({𝐴, 𝐵} = {𝐶, 𝐷} → ((𝐴 = 𝐶𝐴 = 𝐷) → ((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶))))
266, 25mpd 15 . 2 ({𝐴, 𝐵} = {𝐶, 𝐷} → ((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)))
27 preq12 4680 . . 3 ((𝐴 = 𝐶𝐵 = 𝐷) → {𝐴, 𝐵} = {𝐶, 𝐷})
28 preq12 4680 . . . 4 ((𝐴 = 𝐷𝐵 = 𝐶) → {𝐴, 𝐵} = {𝐷, 𝐶})
29 prcom 4677 . . . 4 {𝐷, 𝐶} = {𝐶, 𝐷}
3028, 29eqtrdi 2788 . . 3 ((𝐴 = 𝐷𝐵 = 𝐶) → {𝐴, 𝐵} = {𝐶, 𝐷})
3127, 30jaoi 858 . 2 (((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)) → {𝐴, 𝐵} = {𝐶, 𝐷})
3226, 31impbii 209 1 ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ ((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wa 395  wo 848   = wceq 1542  wcel 2114  Vcvv 3430  {cpr 4570
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1545  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-v 3432  df-un 3895  df-sn 4569  df-pr 4571
This theorem is referenced by:  opthpr  4795  preq12bg  4797  opeqpr  5453  opthhausdorff0  5466  axlowdimlem13  29037  upgrwlkdvdelem  29819  altopthsn  36159  sprsymrelfolem2  47965  prproropf1olem4  47978  reuopreuprim  47998  usgrexmpl2trifr  48525  rrx2xpref1o  49206
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