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Theorem supeq3 9409
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 5113 . . . . . . 7 (𝑅 = 𝑆 → (𝑥𝑅𝑦𝑥𝑆𝑦))
21notbid 321 . . . . . 6 (𝑅 = 𝑆 → (¬ 𝑥𝑅𝑦 ↔ ¬ 𝑥𝑆𝑦))
32ralbidv 3194 . . . . 5 (𝑅 = 𝑆 → (∀𝑦𝐴 ¬ 𝑥𝑅𝑦 ↔ ∀𝑦𝐴 ¬ 𝑥𝑆𝑦))
4 breq 5113 . . . . . . 7 (𝑅 = 𝑆 → (𝑦𝑅𝑥𝑦𝑆𝑥))
5 breq 5113 . . . . . . . 8 (𝑅 = 𝑆 → (𝑦𝑅𝑧𝑦𝑆𝑧))
65rexbidv 3195 . . . . . . 7 (𝑅 = 𝑆 → (∃𝑧𝐴 𝑦𝑅𝑧 ↔ ∃𝑧𝐴 𝑦𝑆𝑧))
74, 6imbi12d 347 . . . . . 6 (𝑅 = 𝑆 → ((𝑦𝑅𝑥 → ∃𝑧𝐴 𝑦𝑅𝑧) ↔ (𝑦𝑆𝑥 → ∃𝑧𝐴 𝑦𝑆𝑧)))
87ralbidv 3194 . . . . 5 (𝑅 = 𝑆 → (∀𝑦𝐵 (𝑦𝑅𝑥 → ∃𝑧𝐴 𝑦𝑅𝑧) ↔ ∀𝑦𝐵 (𝑦𝑆𝑥 → ∃𝑧𝐴 𝑦𝑆𝑧)))
93, 8anbi12d 643 . . . 4 (𝑅 = 𝑆 → ((∀𝑦𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐵 (𝑦𝑅𝑥 → ∃𝑧𝐴 𝑦𝑅𝑧)) ↔ (∀𝑦𝐴 ¬ 𝑥𝑆𝑦 ∧ ∀𝑦𝐵 (𝑦𝑆𝑥 → ∃𝑧𝐴 𝑦𝑆𝑧))))
109rabbidv 3429 . . 3 (𝑅 = 𝑆 → {𝑥𝐵 ∣ (∀𝑦𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐵 (𝑦𝑅𝑥 → ∃𝑧𝐴 𝑦𝑅𝑧))} = {𝑥𝐵 ∣ (∀𝑦𝐴 ¬ 𝑥𝑆𝑦 ∧ ∀𝑦𝐵 (𝑦𝑆𝑥 → ∃𝑧𝐴 𝑦𝑆𝑧))})
1110unieqd 4887 . 2 (𝑅 = 𝑆 {𝑥𝐵 ∣ (∀𝑦𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐵 (𝑦𝑅𝑥 → ∃𝑧𝐴 𝑦𝑅𝑧))} = {𝑥𝐵 ∣ (∀𝑦𝐴 ¬ 𝑥𝑆𝑦 ∧ ∀𝑦𝐵 (𝑦𝑆𝑥 → ∃𝑧𝐴 𝑦𝑆𝑧))})
12 df-sup 9402 . 2 sup(𝐴, 𝐵, 𝑅) = {𝑥𝐵 ∣ (∀𝑦𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐵 (𝑦𝑅𝑥 → ∃𝑧𝐴 𝑦𝑅𝑧))}
13 df-sup 9402 . 2 sup(𝐴, 𝐵, 𝑆) = {𝑥𝐵 ∣ (∀𝑦𝐴 ¬ 𝑥𝑆𝑦 ∧ ∀𝑦𝐵 (𝑦𝑆𝑥 → ∃𝑧𝐴 𝑦𝑆𝑧))}
1411, 12, 133eqtr4g 2829 1 (𝑅 = 𝑆 → sup(𝐴, 𝐵, 𝑅) = sup(𝐴, 𝐵, 𝑆))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 400   = wceq 1567  wral 3085  wrex 3095  {crab 3422   cuni 4874   class class class wbr 5111  supcsup 9400
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-ext 2741
This theorem depends on definitions:  df-bi 210  df-an 401  df-tru 1570  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-ral 3086  df-rex 3096  df-rab 3423  df-v 3463  df-ss 3928  df-uni 4875  df-br 5112  df-sup 9402
This theorem is referenced by:  infeq3  9441
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