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Theorem preleq 4306
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 en2lp 4305 . . . . 5 ¬ (𝐷𝐶𝐶𝐷)
2 eleq12 2118 . . . . . 6 ((𝐴 = 𝐷𝐵 = 𝐶) → (𝐴𝐵𝐷𝐶))
32anbi1d 446 . . . . 5 ((𝐴 = 𝐷𝐵 = 𝐶) → ((𝐴𝐵𝐶𝐷) ↔ (𝐷𝐶𝐶𝐷)))
41, 3mtbiri 610 . . . 4 ((𝐴 = 𝐷𝐵 = 𝐶) → ¬ (𝐴𝐵𝐶𝐷))
54con2i 567 . . 3 ((𝐴𝐵𝐶𝐷) → ¬ (𝐴 = 𝐷𝐵 = 𝐶))
65adantr 265 . 2 (((𝐴𝐵𝐶𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → ¬ (𝐴 = 𝐷𝐵 = 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
117, 8, 9, 10preq12b 3568 . . . 4 ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ ((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)))
1211biimpi 117 . . 3 ({𝐴, 𝐵} = {𝐶, 𝐷} → ((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)))
1312adantl 266 . 2 (((𝐴𝐵𝐶𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → ((𝐴 = 𝐶𝐵 = 𝐷) ∨ (𝐴 = 𝐷𝐵 = 𝐶)))
146, 13ecased 1255 1 (((𝐴𝐵𝐶𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → (𝐴 = 𝐶𝐵 = 𝐷))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 101  wo 639   = wceq 1259  wcel 1409  Vcvv 2574  {cpr 3403
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-in1 554  ax-in2 555  ax-io 640  ax-5 1352  ax-7 1353  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-8 1411  ax-10 1412  ax-11 1413  ax-i12 1414  ax-bndl 1415  ax-4 1416  ax-17 1435  ax-i9 1439  ax-ial 1443  ax-i5r 1444  ax-ext 2038  ax-setind 4289
This theorem depends on definitions:  df-bi 114  df-3an 898  df-tru 1262  df-nf 1366  df-sb 1662  df-clab 2043  df-cleq 2049  df-clel 2052  df-nfc 2183  df-ral 2328  df-v 2576  df-dif 2947  df-un 2949  df-sn 3408  df-pr 3409
This theorem is referenced by:  opthreg  4307
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