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Theorem supeq3 9138
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 5072 . . . . . . 7 (𝑅 = 𝑆 → (𝑥𝑅𝑦𝑥𝑆𝑦))
21notbid 317 . . . . . 6 (𝑅 = 𝑆 → (¬ 𝑥𝑅𝑦 ↔ ¬ 𝑥𝑆𝑦))
32ralbidv 3120 . . . . 5 (𝑅 = 𝑆 → (∀𝑦𝐴 ¬ 𝑥𝑅𝑦 ↔ ∀𝑦𝐴 ¬ 𝑥𝑆𝑦))
4 breq 5072 . . . . . . 7 (𝑅 = 𝑆 → (𝑦𝑅𝑥𝑦𝑆𝑥))
5 breq 5072 . . . . . . . 8 (𝑅 = 𝑆 → (𝑦𝑅𝑧𝑦𝑆𝑧))
65rexbidv 3225 . . . . . . 7 (𝑅 = 𝑆 → (∃𝑧𝐴 𝑦𝑅𝑧 ↔ ∃𝑧𝐴 𝑦𝑆𝑧))
74, 6imbi12d 344 . . . . . 6 (𝑅 = 𝑆 → ((𝑦𝑅𝑥 → ∃𝑧𝐴 𝑦𝑅𝑧) ↔ (𝑦𝑆𝑥 → ∃𝑧𝐴 𝑦𝑆𝑧)))
87ralbidv 3120 . . . . 5 (𝑅 = 𝑆 → (∀𝑦𝐵 (𝑦𝑅𝑥 → ∃𝑧𝐴 𝑦𝑅𝑧) ↔ ∀𝑦𝐵 (𝑦𝑆𝑥 → ∃𝑧𝐴 𝑦𝑆𝑧)))
93, 8anbi12d 630 . . . 4 (𝑅 = 𝑆 → ((∀𝑦𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐵 (𝑦𝑅𝑥 → ∃𝑧𝐴 𝑦𝑅𝑧)) ↔ (∀𝑦𝐴 ¬ 𝑥𝑆𝑦 ∧ ∀𝑦𝐵 (𝑦𝑆𝑥 → ∃𝑧𝐴 𝑦𝑆𝑧))))
109rabbidv 3404 . . 3 (𝑅 = 𝑆 → {𝑥𝐵 ∣ (∀𝑦𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐵 (𝑦𝑅𝑥 → ∃𝑧𝐴 𝑦𝑅𝑧))} = {𝑥𝐵 ∣ (∀𝑦𝐴 ¬ 𝑥𝑆𝑦 ∧ ∀𝑦𝐵 (𝑦𝑆𝑥 → ∃𝑧𝐴 𝑦𝑆𝑧))})
1110unieqd 4850 . 2 (𝑅 = 𝑆 {𝑥𝐵 ∣ (∀𝑦𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐵 (𝑦𝑅𝑥 → ∃𝑧𝐴 𝑦𝑅𝑧))} = {𝑥𝐵 ∣ (∀𝑦𝐴 ¬ 𝑥𝑆𝑦 ∧ ∀𝑦𝐵 (𝑦𝑆𝑥 → ∃𝑧𝐴 𝑦𝑆𝑧))})
12 df-sup 9131 . 2 sup(𝐴, 𝐵, 𝑅) = {𝑥𝐵 ∣ (∀𝑦𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐵 (𝑦𝑅𝑥 → ∃𝑧𝐴 𝑦𝑅𝑧))}
13 df-sup 9131 . 2 sup(𝐴, 𝐵, 𝑆) = {𝑥𝐵 ∣ (∀𝑦𝐴 ¬ 𝑥𝑆𝑦 ∧ ∀𝑦𝐵 (𝑦𝑆𝑥 → ∃𝑧𝐴 𝑦𝑆𝑧))}
1411, 12, 133eqtr4g 2804 1 (𝑅 = 𝑆 → sup(𝐴, 𝐵, 𝑅) = sup(𝐴, 𝐵, 𝑆))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 395   = wceq 1539  wral 3063  wrex 3064  {crab 3067   cuni 4836   class class class wbr 5070  supcsup 9129
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709
This theorem depends on definitions:  df-bi 206  df-an 396  df-tru 1542  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-ral 3068  df-rex 3069  df-rab 3072  df-v 3424  df-in 3890  df-ss 3900  df-uni 4837  df-br 5071  df-sup 9131
This theorem is referenced by:  infeq3  9169
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