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Theorem supeq3 9397
Description: Equality theorem for supremum. (Contributed by Scott Fenton, 13-Jun-2018.)
Assertion
Ref Expression
supeq3 (𝑅 = 𝑆 → sup(𝐴, 𝐵, 𝑅) = sup(𝐴, 𝐵, 𝑆))

Proof of Theorem supeq3
Dummy variables 𝑥 𝑦 𝑧 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 breq 5104 . . . . . . 7 (𝑅 = 𝑆 → (𝑥𝑅𝑦𝑥𝑆𝑦))
21notbid 320 . . . . . 6 (𝑅 = 𝑆 → (¬ 𝑥𝑅𝑦 ↔ ¬ 𝑥𝑆𝑦))
32ralbidv 3187 . . . . 5 (𝑅 = 𝑆 → (∀𝑦𝐴 ¬ 𝑥𝑅𝑦 ↔ ∀𝑦𝐴 ¬ 𝑥𝑆𝑦))
4 breq 5104 . . . . . . 7 (𝑅 = 𝑆 → (𝑦𝑅𝑥𝑦𝑆𝑥))
5 breq 5104 . . . . . . . 8 (𝑅 = 𝑆 → (𝑦𝑅𝑧𝑦𝑆𝑧))
65rexbidv 3188 . . . . . . 7 (𝑅 = 𝑆 → (∃𝑧𝐴 𝑦𝑅𝑧 ↔ ∃𝑧𝐴 𝑦𝑆𝑧))
74, 6imbi12d 346 . . . . . 6 (𝑅 = 𝑆 → ((𝑦𝑅𝑥 → ∃𝑧𝐴 𝑦𝑅𝑧) ↔ (𝑦𝑆𝑥 → ∃𝑧𝐴 𝑦𝑆𝑧)))
87ralbidv 3187 . . . . 5 (𝑅 = 𝑆 → (∀𝑦𝐵 (𝑦𝑅𝑥 → ∃𝑧𝐴 𝑦𝑅𝑧) ↔ ∀𝑦𝐵 (𝑦𝑆𝑥 → ∃𝑧𝐴 𝑦𝑆𝑧)))
93, 8anbi12d 641 . . . 4 (𝑅 = 𝑆 → ((∀𝑦𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐵 (𝑦𝑅𝑥 → ∃𝑧𝐴 𝑦𝑅𝑧)) ↔ (∀𝑦𝐴 ¬ 𝑥𝑆𝑦 ∧ ∀𝑦𝐵 (𝑦𝑆𝑥 → ∃𝑧𝐴 𝑦𝑆𝑧))))
109rabbidv 3423 . . 3 (𝑅 = 𝑆 → {𝑥𝐵 ∣ (∀𝑦𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐵 (𝑦𝑅𝑥 → ∃𝑧𝐴 𝑦𝑅𝑧))} = {𝑥𝐵 ∣ (∀𝑦𝐴 ¬ 𝑥𝑆𝑦 ∧ ∀𝑦𝐵 (𝑦𝑆𝑥 → ∃𝑧𝐴 𝑦𝑆𝑧))})
1110unieqd 4880 . 2 (𝑅 = 𝑆 {𝑥𝐵 ∣ (∀𝑦𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐵 (𝑦𝑅𝑥 → ∃𝑧𝐴 𝑦𝑅𝑧))} = {𝑥𝐵 ∣ (∀𝑦𝐴 ¬ 𝑥𝑆𝑦 ∧ ∀𝑦𝐵 (𝑦𝑆𝑥 → ∃𝑧𝐴 𝑦𝑆𝑧))})
12 df-sup 9390 . 2 sup(𝐴, 𝐵, 𝑅) = {𝑥𝐵 ∣ (∀𝑦𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐵 (𝑦𝑅𝑥 → ∃𝑧𝐴 𝑦𝑅𝑧))}
13 df-sup 9390 . 2 sup(𝐴, 𝐵, 𝑆) = {𝑥𝐵 ∣ (∀𝑦𝐴 ¬ 𝑥𝑆𝑦 ∧ ∀𝑦𝐵 (𝑦𝑆𝑥 → ∃𝑧𝐴 𝑦𝑆𝑧))}
1411, 12, 133eqtr4g 2824 1 (𝑅 = 𝑆 → sup(𝐴, 𝐵, 𝑅) = sup(𝐴, 𝐵, 𝑆))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 399   = wceq 1562  wral 3078  wrex 3088  {crab 3416   cuni 4867   class class class wbr 5102  supcsup 9388
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1817  ax-4 1831  ax-5 1932  ax-6 1989  ax-7 2030  ax-8 2146  ax-9 2154  ax-ext 2736
This theorem depends on definitions:  df-bi 209  df-an 400  df-tru 1565  df-ex 1802  df-sb 2093  df-clab 2743  df-cleq 2756  df-clel 2839  df-ral 3079  df-rex 3089  df-rab 3417  df-v 3458  df-ss 3923  df-uni 4868  df-br 5103  df-sup 9390
This theorem is referenced by:  infeq3  9429
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