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Theorem poeq1 5480
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 5071 . . . . . 6 (𝑅 = 𝑆 → (𝑥𝑅𝑥𝑥𝑆𝑥))
21notbid 320 . . . . 5 (𝑅 = 𝑆 → (¬ 𝑥𝑅𝑥 ↔ ¬ 𝑥𝑆𝑥))
3 breq 5071 . . . . . . 7 (𝑅 = 𝑆 → (𝑥𝑅𝑦𝑥𝑆𝑦))
4 breq 5071 . . . . . . 7 (𝑅 = 𝑆 → (𝑦𝑅𝑧𝑦𝑆𝑧))
53, 4anbi12d 632 . . . . . 6 (𝑅 = 𝑆 → ((𝑥𝑅𝑦𝑦𝑅𝑧) ↔ (𝑥𝑆𝑦𝑦𝑆𝑧)))
6 breq 5071 . . . . . 6 (𝑅 = 𝑆 → (𝑥𝑅𝑧𝑥𝑆𝑧))
75, 6imbi12d 347 . . . . 5 (𝑅 = 𝑆 → (((𝑥𝑅𝑦𝑦𝑅𝑧) → 𝑥𝑅𝑧) ↔ ((𝑥𝑆𝑦𝑦𝑆𝑧) → 𝑥𝑆𝑧)))
82, 7anbi12d 632 . . . 4 (𝑅 = 𝑆 → ((¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ↔ (¬ 𝑥𝑆𝑥 ∧ ((𝑥𝑆𝑦𝑦𝑆𝑧) → 𝑥𝑆𝑧))))
98ralbidv 3200 . . 3 (𝑅 = 𝑆 → (∀𝑧𝐴𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ↔ ∀𝑧𝐴𝑥𝑆𝑥 ∧ ((𝑥𝑆𝑦𝑦𝑆𝑧) → 𝑥𝑆𝑧))))
1092ralbidv 3202 . 2 (𝑅 = 𝑆 → (∀𝑥𝐴𝑦𝐴𝑧𝐴𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ↔ ∀𝑥𝐴𝑦𝐴𝑧𝐴𝑥𝑆𝑥 ∧ ((𝑥𝑆𝑦𝑦𝑆𝑧) → 𝑥𝑆𝑧))))
11 df-po 5477 . 2 (𝑅 Po 𝐴 ↔ ∀𝑥𝐴𝑦𝐴𝑧𝐴𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦𝑦𝑅𝑧) → 𝑥𝑅𝑧)))
12 df-po 5477 . 2 (𝑆 Po 𝐴 ↔ ∀𝑥𝐴𝑦𝐴𝑧𝐴𝑥𝑆𝑥 ∧ ((𝑥𝑆𝑦𝑦𝑆𝑧) → 𝑥𝑆𝑧)))
1310, 11, 123bitr4g 316 1 (𝑅 = 𝑆 → (𝑅 Po 𝐴𝑆 Po 𝐴))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 208  wa 398   = wceq 1536  wral 3141   class class class wbr 5069   Po wpo 5475
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1969  ax-7 2014  ax-8 2115  ax-9 2123  ax-ext 2796
This theorem depends on definitions:  df-bi 209  df-an 399  df-ex 1780  df-cleq 2817  df-clel 2896  df-ral 3146  df-br 5070  df-po 5477
This theorem is referenced by:  soeq1  5497
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