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Theorem poeq1 4027
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 3757 . . . . . 6 (𝑅 = 𝑆 → (x𝑅xx𝑆x))
21notbid 591 . . . . 5 (𝑅 = 𝑆 → (¬ x𝑅x ↔ ¬ x𝑆x))
3 breq 3757 . . . . . . 7 (𝑅 = 𝑆 → (x𝑅yx𝑆y))
4 breq 3757 . . . . . . 7 (𝑅 = 𝑆 → (y𝑅zy𝑆z))
53, 4anbi12d 442 . . . . . 6 (𝑅 = 𝑆 → ((x𝑅y y𝑅z) ↔ (x𝑆y y𝑆z)))
6 breq 3757 . . . . . 6 (𝑅 = 𝑆 → (x𝑅zx𝑆z))
75, 6imbi12d 223 . . . . 5 (𝑅 = 𝑆 → (((x𝑅y y𝑅z) → x𝑅z) ↔ ((x𝑆y y𝑆z) → x𝑆z)))
82, 7anbi12d 442 . . . 4 (𝑅 = 𝑆 → ((¬ x𝑅x ((x𝑅y y𝑅z) → x𝑅z)) ↔ (¬ x𝑆x ((x𝑆y y𝑆z) → x𝑆z))))
98ralbidv 2320 . . 3 (𝑅 = 𝑆 → (z Ax𝑅x ((x𝑅y y𝑅z) → x𝑅z)) ↔ z Ax𝑆x ((x𝑆y y𝑆z) → x𝑆z))))
1092ralbidv 2342 . 2 (𝑅 = 𝑆 → (x A y A z Ax𝑅x ((x𝑅y y𝑅z) → x𝑅z)) ↔ x A y A z Ax𝑆x ((x𝑆y y𝑆z) → x𝑆z))))
11 df-po 4024 . 2 (𝑅 Po Ax A y A z Ax𝑅x ((x𝑅y y𝑅z) → x𝑅z)))
12 df-po 4024 . 2 (𝑆 Po Ax A y A z Ax𝑆x ((x𝑆y y𝑆z) → 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 1242  wral 2300   class class class wbr 3755   Po wpo 4022
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 1333  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-4 1397  ax-17 1416  ax-ial 1424  ax-ext 2019
This theorem depends on definitions:  df-bi 110  df-nf 1347  df-cleq 2030  df-clel 2033  df-ral 2305  df-br 3756  df-po 4024
This theorem is referenced by:  soeq1  4043
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