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Theorem poeq1 4006
 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 3736 . . . . . 6 (𝑅 = 𝑆 → (x𝑅xx𝑆x))
21notbid 579 . . . . 5 (𝑅 = 𝑆 → (¬ x𝑅x ↔ ¬ x𝑆x))
3 breq 3736 . . . . . . 7 (𝑅 = 𝑆 → (x𝑅yx𝑆y))
4 breq 3736 . . . . . . 7 (𝑅 = 𝑆 → (y𝑅zy𝑆z))
53, 4anbi12d 445 . . . . . 6 (𝑅 = 𝑆 → ((x𝑅y y𝑅z) ↔ (x𝑆y y𝑆z)))
6 breq 3736 . . . . . 6 (𝑅 = 𝑆 → (x𝑅zx𝑆z))
75, 6imbi12d 223 . . . . 5 (𝑅 = 𝑆 → (((x𝑅y y𝑅z) → x𝑅z) ↔ ((x𝑆y y𝑆z) → x𝑆z)))
82, 7anbi12d 445 . . . 4 (𝑅 = 𝑆 → ((¬ x𝑅x ((x𝑅y y𝑅z) → x𝑅z)) ↔ (¬ x𝑆x ((x𝑆y y𝑆z) → x𝑆z))))
98ralbidv 2300 . . 3 (𝑅 = 𝑆 → (z Ax𝑅x ((x𝑅y y𝑅z) → x𝑅z)) ↔ z Ax𝑆x ((x𝑆y y𝑆z) → x𝑆z))))
1092ralbidv 2322 . 2 (𝑅 = 𝑆 → (x A y A z Ax𝑅x ((x𝑅y y𝑅z) → x𝑅z)) ↔ x A y A z Ax𝑆x ((x𝑆y y𝑆z) → x𝑆z))))
11 df-po 4003 . 2 (𝑅 Po Ax A y A z Ax𝑅x ((x𝑅y y𝑅z) → x𝑅z)))
12 df-po 4003 . 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 1226  ∀wral 2280   class class class wbr 3734   Po wpo 4001 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 532  ax-in2 533  ax-5 1312  ax-gen 1314  ax-ie1 1359  ax-ie2 1360  ax-4 1377  ax-17 1396  ax-ial 1405  ax-ext 2000 This theorem depends on definitions:  df-bi 110  df-nf 1326  df-cleq 2011  df-clel 2014  df-ral 2285  df-br 3735  df-po 4003 This theorem is referenced by:  soeq1  4022
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