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Theorem papsym 7602
Description: An apartness is symmetric. (Contributed by Jim Kingdon, 27-May-2026.)
Hypotheses
Ref Expression
papsym.r (𝜑𝑅 Ap 𝐴)
papsym.x (𝜑𝑋𝐴)
papsym.y (𝜑𝑌𝐴)
papsym.ap (𝜑𝑋𝑅𝑌)
Assertion
Ref Expression
papsym (𝜑𝑌𝑅𝑋)

Proof of Theorem papsym
Dummy variables 𝑥 𝑦 𝑧 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 papsym.ap . 2 (𝜑𝑋𝑅𝑌)
2 breq2 4129 . . . 4 (𝑦 = 𝑌 → (𝑋𝑅𝑦𝑋𝑅𝑌))
3 breq1 4128 . . . 4 (𝑦 = 𝑌 → (𝑦𝑅𝑋𝑌𝑅𝑋))
42, 3imbi12d 234 . . 3 (𝑦 = 𝑌 → ((𝑋𝑅𝑦𝑦𝑅𝑋) ↔ (𝑋𝑅𝑌𝑌𝑅𝑋)))
5 breq1 4128 . . . . . 6 (𝑥 = 𝑋 → (𝑥𝑅𝑦𝑋𝑅𝑦))
6 breq2 4129 . . . . . 6 (𝑥 = 𝑋 → (𝑦𝑅𝑥𝑦𝑅𝑋))
75, 6imbi12d 234 . . . . 5 (𝑥 = 𝑋 → ((𝑥𝑅𝑦𝑦𝑅𝑥) ↔ (𝑋𝑅𝑦𝑦𝑅𝑋)))
87ralbidv 2550 . . . 4 (𝑥 = 𝑋 → (∀𝑦𝐴 (𝑥𝑅𝑦𝑦𝑅𝑥) ↔ ∀𝑦𝐴 (𝑋𝑅𝑦𝑦𝑅𝑋)))
9 papsym.r . . . . . 6 (𝜑𝑅 Ap 𝐴)
10 df-pap 7598 . . . . . 6 (𝑅 Ap 𝐴 ↔ ((𝑅 ⊆ (𝐴 × 𝐴) ∧ ∀𝑥𝐴 ¬ 𝑥𝑅𝑥) ∧ (∀𝑥𝐴𝑦𝐴 (𝑥𝑅𝑦𝑦𝑅𝑥) ∧ ∀𝑥𝐴𝑦𝐴𝑧𝐴 (𝑥𝑅𝑦 → (𝑥𝑅𝑧𝑦𝑅𝑧)))))
119, 10sylib 122 . . . . 5 (𝜑 → ((𝑅 ⊆ (𝐴 × 𝐴) ∧ ∀𝑥𝐴 ¬ 𝑥𝑅𝑥) ∧ (∀𝑥𝐴𝑦𝐴 (𝑥𝑅𝑦𝑦𝑅𝑥) ∧ ∀𝑥𝐴𝑦𝐴𝑧𝐴 (𝑥𝑅𝑦 → (𝑥𝑅𝑧𝑦𝑅𝑧)))))
1211simprld 536 . . . 4 (𝜑 → ∀𝑥𝐴𝑦𝐴 (𝑥𝑅𝑦𝑦𝑅𝑥))
13 papsym.x . . . 4 (𝜑𝑋𝐴)
148, 12, 13rspcdva 2934 . . 3 (𝜑 → ∀𝑦𝐴 (𝑋𝑅𝑦𝑦𝑅𝑋))
15 papsym.y . . 3 (𝜑𝑌𝐴)
164, 14, 15rspcdva 2934 . 2 (𝜑 → (𝑋𝑅𝑌𝑌𝑅𝑋))
171, 16mpd 13 1 (𝜑𝑌𝑅𝑋)
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 104  wo 720   = wceq 1402  wcel 2209  wral 2528  wss 3220   class class class wbr 4125   × cxp 4767   Ap wap 7597
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-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220
This theorem depends on definitions:  df-bi 117  df-3an 1011  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-ral 2533  df-v 2823  df-un 3224  df-sn 3711  df-pr 3712  df-op 3714  df-br 4126  df-pap 7598
This theorem is referenced by:  aprlring  14573
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