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Theorem supeq3 9381
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 5092 . . . . . . 7 (𝑅 = 𝑆 → (𝑥𝑅𝑦𝑥𝑆𝑦))
21notbid 320 . . . . . 6 (𝑅 = 𝑆 → (¬ 𝑥𝑅𝑦 ↔ ¬ 𝑥𝑆𝑦))
32ralbidv 3175 . . . . 5 (𝑅 = 𝑆 → (∀𝑦𝐴 ¬ 𝑥𝑅𝑦 ↔ ∀𝑦𝐴 ¬ 𝑥𝑆𝑦))
4 breq 5092 . . . . . . 7 (𝑅 = 𝑆 → (𝑦𝑅𝑥𝑦𝑆𝑥))
5 breq 5092 . . . . . . . 8 (𝑅 = 𝑆 → (𝑦𝑅𝑧𝑦𝑆𝑧))
65rexbidv 3176 . . . . . . 7 (𝑅 = 𝑆 → (∃𝑧𝐴 𝑦𝑅𝑧 ↔ ∃𝑧𝐴 𝑦𝑆𝑧))
74, 6imbi12d 346 . . . . . 6 (𝑅 = 𝑆 → ((𝑦𝑅𝑥 → ∃𝑧𝐴 𝑦𝑅𝑧) ↔ (𝑦𝑆𝑥 → ∃𝑧𝐴 𝑦𝑆𝑧)))
87ralbidv 3175 . . . . 5 (𝑅 = 𝑆 → (∀𝑦𝐵 (𝑦𝑅𝑥 → ∃𝑧𝐴 𝑦𝑅𝑧) ↔ ∀𝑦𝐵 (𝑦𝑆𝑥 → ∃𝑧𝐴 𝑦𝑆𝑧)))
93, 8anbi12d 640 . . . 4 (𝑅 = 𝑆 → ((∀𝑦𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐵 (𝑦𝑅𝑥 → ∃𝑧𝐴 𝑦𝑅𝑧)) ↔ (∀𝑦𝐴 ¬ 𝑥𝑆𝑦 ∧ ∀𝑦𝐵 (𝑦𝑆𝑥 → ∃𝑧𝐴 𝑦𝑆𝑧))))
109rabbidv 3411 . . 3 (𝑅 = 𝑆 → {𝑥𝐵 ∣ (∀𝑦𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐵 (𝑦𝑅𝑥 → ∃𝑧𝐴 𝑦𝑅𝑧))} = {𝑥𝐵 ∣ (∀𝑦𝐴 ¬ 𝑥𝑆𝑦 ∧ ∀𝑦𝐵 (𝑦𝑆𝑥 → ∃𝑧𝐴 𝑦𝑆𝑧))})
1110unieqd 4868 . 2 (𝑅 = 𝑆 {𝑥𝐵 ∣ (∀𝑦𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐵 (𝑦𝑅𝑥 → ∃𝑧𝐴 𝑦𝑅𝑧))} = {𝑥𝐵 ∣ (∀𝑦𝐴 ¬ 𝑥𝑆𝑦 ∧ ∀𝑦𝐵 (𝑦𝑆𝑥 → ∃𝑧𝐴 𝑦𝑆𝑧))})
12 df-sup 9374 . 2 sup(𝐴, 𝐵, 𝑅) = {𝑥𝐵 ∣ (∀𝑦𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐵 (𝑦𝑅𝑥 → ∃𝑧𝐴 𝑦𝑅𝑧))}
13 df-sup 9374 . 2 sup(𝐴, 𝐵, 𝑆) = {𝑥𝐵 ∣ (∀𝑦𝐴 ¬ 𝑥𝑆𝑦 ∧ ∀𝑦𝐵 (𝑦𝑆𝑥 → ∃𝑧𝐴 𝑦𝑆𝑧))}
1411, 12, 133eqtr4g 2812 1 (𝑅 = 𝑆 → sup(𝐴, 𝐵, 𝑅) = sup(𝐴, 𝐵, 𝑆))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 398   = wceq 1550  wral 3066  wrex 3076  {crab 3404   cuni 4855   class class class wbr 5090  supcsup 9372
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1805  ax-4 1819  ax-5 1920  ax-6 1977  ax-7 2018  ax-8 2134  ax-9 2142  ax-ext 2724
This theorem depends on definitions:  df-bi 209  df-an 399  df-tru 1553  df-ex 1790  df-sb 2081  df-clab 2731  df-cleq 2744  df-clel 2827  df-ral 3067  df-rex 3077  df-rab 3405  df-v 3446  df-ss 3912  df-uni 4856  df-br 5091  df-sup 9374
This theorem is referenced by:  infeq3  9413
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