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Theorem preleq 8514
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 (((𝐴𝐵𝐶𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → (𝐴 = 𝐶𝐵 = 𝐷))

Proof of Theorem preleq
StepHypRef Expression
1 preleq.1 . . . . . . 7 𝐴 ∈ V
2 preleq.2 . . . . . . 7 𝐵 ∈ V
3 preleq.3 . . . . . . 7 𝐶 ∈ V
4 preleq.4 . . . . . . 7 𝐷 ∈ V
51, 2, 3, 4preq12b 4382 . . . . . 6 ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ ((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)))
65biimpi 206 . . . . 5 ({𝐴, 𝐵} = {𝐶, 𝐷} → ((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)))
76ord 392 . . . 4 ({𝐴, 𝐵} = {𝐶, 𝐷} → (¬ (𝐴 = 𝐶𝐵 = 𝐷) → (𝐴 = 𝐷𝐵 = 𝐶)))
8 en2lp 8510 . . . . 5 ¬ (𝐷𝐶𝐶𝐷)
9 eleq12 2691 . . . . . 6 ((𝐴 = 𝐷𝐵 = 𝐶) → (𝐴𝐵𝐷𝐶))
109anbi1d 741 . . . . 5 ((𝐴 = 𝐷𝐵 = 𝐶) → ((𝐴𝐵𝐶𝐷) ↔ (𝐷𝐶𝐶𝐷)))
118, 10mtbiri 317 . . . 4 ((𝐴 = 𝐷𝐵 = 𝐶) → ¬ (𝐴𝐵𝐶𝐷))
127, 11syl6 35 . . 3 ({𝐴, 𝐵} = {𝐶, 𝐷} → (¬ (𝐴 = 𝐶𝐵 = 𝐷) → ¬ (𝐴𝐵𝐶𝐷)))
1312con4d 114 . 2 ({𝐴, 𝐵} = {𝐶, 𝐷} → ((𝐴𝐵𝐶𝐷) → (𝐴 = 𝐶𝐵 = 𝐷)))
1413impcom 446 1 (((𝐴𝐵𝐶𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → (𝐴 = 𝐶𝐵 = 𝐷))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wo 383  wa 384   = wceq 1483  wcel 1990  Vcvv 3200  {cpr 4179
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602  ax-sep 4781  ax-nul 4789  ax-pr 4906  ax-reg 8497
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3an 1039  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-eu 2474  df-mo 2475  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-ne 2795  df-ral 2917  df-rex 2918  df-rab 2921  df-v 3202  df-sbc 3436  df-dif 3577  df-un 3579  df-in 3581  df-ss 3588  df-nul 3916  df-if 4087  df-sn 4178  df-pr 4180  df-op 4184  df-br 4654  df-opab 4713  df-eprel 5029  df-fr 5073
This theorem is referenced by:  opthreg  8515  dfac2  8953
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