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Theorem supeq1 8507
 Description: Equality theorem for supremum. (Contributed by NM, 22-May-1999.)
Assertion
Ref Expression
supeq1 (𝐵 = 𝐶 → sup(𝐵, 𝐴, 𝑅) = sup(𝐶, 𝐴, 𝑅))

Proof of Theorem supeq1
Dummy variables 𝑥 𝑦 𝑧 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 raleq 3287 . . . . 5 (𝐵 = 𝐶 → (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ↔ ∀𝑦𝐶 ¬ 𝑥𝑅𝑦))
2 rexeq 3288 . . . . . . 7 (𝐵 = 𝐶 → (∃𝑧𝐵 𝑦𝑅𝑧 ↔ ∃𝑧𝐶 𝑦𝑅𝑧))
32imbi2d 329 . . . . . 6 (𝐵 = 𝐶 → ((𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧) ↔ (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧)))
43ralbidv 3135 . . . . 5 (𝐵 = 𝐶 → (∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧) ↔ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧)))
51, 4anbi12d 616 . . . 4 (𝐵 = 𝐶 → ((∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧)) ↔ (∀𝑦𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧))))
65rabbidv 3339 . . 3 (𝐵 = 𝐶 → {𝑥𝐴 ∣ (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧))} = {𝑥𝐴 ∣ (∀𝑦𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧))})
76unieqd 4584 . 2 (𝐵 = 𝐶 {𝑥𝐴 ∣ (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧))} = {𝑥𝐴 ∣ (∀𝑦𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧))})
8 df-sup 8504 . 2 sup(𝐵, 𝐴, 𝑅) = {𝑥𝐴 ∣ (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧))}
9 df-sup 8504 . 2 sup(𝐶, 𝐴, 𝑅) = {𝑥𝐴 ∣ (∀𝑦𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧))}
107, 8, 93eqtr4g 2830 1 (𝐵 = 𝐶 → sup(𝐵, 𝐴, 𝑅) = sup(𝐶, 𝐴, 𝑅))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 382   = wceq 1631  ∀wral 3061  ∃wrex 3062  {crab 3065  ∪ cuni 4574   class class class wbr 4786  supcsup 8502 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1870  ax-4 1885  ax-5 1991  ax-6 2057  ax-7 2093  ax-9 2154  ax-10 2174  ax-11 2190  ax-12 2203  ax-ext 2751 This theorem depends on definitions:  df-bi 197  df-an 383  df-or 837  df-tru 1634  df-ex 1853  df-nf 1858  df-sb 2050  df-clab 2758  df-cleq 2764  df-clel 2767  df-nfc 2902  df-ral 3066  df-rex 3067  df-rab 3070  df-uni 4575  df-sup 8504 This theorem is referenced by:  supeq1d  8508  supeq1i  8509  infeq1  8538  bndth  22977  ioorval  23562  uniioombllem6  23576  mdegcl  24049  suplesup  40071  supminfxr  40210
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