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Theorem poeq1 4036
 Description: Equality theorem for partial ordering predicate. (Contributed by NM, 27-Mar-1997.)
Assertion
Ref Expression
poeq1 (𝑅 = 𝑆 → (𝑅 Po 𝐴𝑆 Po 𝐴))

Proof of Theorem poeq1
Dummy variables 𝑥 𝑦 𝑧 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 breq 3766 . . . . . 6 (𝑅 = 𝑆 → (𝑥𝑅𝑥𝑥𝑆𝑥))
21notbid 592 . . . . 5 (𝑅 = 𝑆 → (¬ 𝑥𝑅𝑥 ↔ ¬ 𝑥𝑆𝑥))
3 breq 3766 . . . . . . 7 (𝑅 = 𝑆 → (𝑥𝑅𝑦𝑥𝑆𝑦))
4 breq 3766 . . . . . . 7 (𝑅 = 𝑆 → (𝑦𝑅𝑧𝑦𝑆𝑧))
53, 4anbi12d 442 . . . . . 6 (𝑅 = 𝑆 → ((𝑥𝑅𝑦𝑦𝑅𝑧) ↔ (𝑥𝑆𝑦𝑦𝑆𝑧)))
6 breq 3766 . . . . . 6 (𝑅 = 𝑆 → (𝑥𝑅𝑧𝑥𝑆𝑧))
75, 6imbi12d 223 . . . . 5 (𝑅 = 𝑆 → (((𝑥𝑅𝑦𝑦𝑅𝑧) → 𝑥𝑅𝑧) ↔ ((𝑥𝑆𝑦𝑦𝑆𝑧) → 𝑥𝑆𝑧)))
82, 7anbi12d 442 . . . 4 (𝑅 = 𝑆 → ((¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ↔ (¬ 𝑥𝑆𝑥 ∧ ((𝑥𝑆𝑦𝑦𝑆𝑧) → 𝑥𝑆𝑧))))
98ralbidv 2326 . . 3 (𝑅 = 𝑆 → (∀𝑧𝐴𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ↔ ∀𝑧𝐴𝑥𝑆𝑥 ∧ ((𝑥𝑆𝑦𝑦𝑆𝑧) → 𝑥𝑆𝑧))))
1092ralbidv 2348 . 2 (𝑅 = 𝑆 → (∀𝑥𝐴𝑦𝐴𝑧𝐴𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ↔ ∀𝑥𝐴𝑦𝐴𝑧𝐴𝑥𝑆𝑥 ∧ ((𝑥𝑆𝑦𝑦𝑆𝑧) → 𝑥𝑆𝑧))))
11 df-po 4033 . 2 (𝑅 Po 𝐴 ↔ ∀𝑥𝐴𝑦𝐴𝑧𝐴𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦𝑦𝑅𝑧) → 𝑥𝑅𝑧)))
12 df-po 4033 . 2 (𝑆 Po 𝐴 ↔ ∀𝑥𝐴𝑦𝐴𝑧𝐴𝑥𝑆𝑥 ∧ ((𝑥𝑆𝑦𝑦𝑆𝑧) → 𝑥𝑆𝑧)))
1310, 11, 123bitr4g 212 1 (𝑅 = 𝑆 → (𝑅 Po 𝐴𝑆 Po 𝐴))
 Colors of variables: wff set class Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 97   ↔ wb 98   = wceq 1243  ∀wral 2306   class class class wbr 3764   Po wpo 4031 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 544  ax-in2 545  ax-5 1336  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-4 1400  ax-17 1419  ax-ial 1427  ax-ext 2022 This theorem depends on definitions:  df-bi 110  df-nf 1350  df-cleq 2033  df-clel 2036  df-ral 2311  df-br 3765  df-po 4033 This theorem is referenced by:  soeq1  4052
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