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Theorem supeq1 8897
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 3403 . . . . 5 (𝐵 = 𝐶 → (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ↔ ∀𝑦𝐶 ¬ 𝑥𝑅𝑦))
2 rexeq 3404 . . . . . . 7 (𝐵 = 𝐶 → (∃𝑧𝐵 𝑦𝑅𝑧 ↔ ∃𝑧𝐶 𝑦𝑅𝑧))
32imbi2d 342 . . . . . 6 (𝐵 = 𝐶 → ((𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧) ↔ (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧)))
43ralbidv 3194 . . . . 5 (𝐵 = 𝐶 → (∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧) ↔ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧)))
51, 4anbi12d 630 . . . 4 (𝐵 = 𝐶 → ((∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧)) ↔ (∀𝑦𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧))))
65rabbidv 3478 . . 3 (𝐵 = 𝐶 → {𝑥𝐴 ∣ (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧))} = {𝑥𝐴 ∣ (∀𝑦𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧))})
76unieqd 4840 . 2 (𝐵 = 𝐶 {𝑥𝐴 ∣ (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧))} = {𝑥𝐴 ∣ (∀𝑦𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧))})
8 df-sup 8894 . 2 sup(𝐵, 𝐴, 𝑅) = {𝑥𝐴 ∣ (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧))}
9 df-sup 8894 . 2 sup(𝐶, 𝐴, 𝑅) = {𝑥𝐴 ∣ (∀𝑦𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧))}
107, 8, 93eqtr4g 2878 1 (𝐵 = 𝐶 → sup(𝐵, 𝐴, 𝑅) = sup(𝐶, 𝐴, 𝑅))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 396   = wceq 1528  wral 3135  wrex 3136  {crab 3139   cuni 4830   class class class wbr 5057  supcsup 8892
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1787  ax-4 1801  ax-5 1902  ax-6 1961  ax-7 2006  ax-8 2107  ax-9 2115  ax-ext 2790
This theorem depends on definitions:  df-bi 208  df-an 397  df-ex 1772  df-sb 2061  df-clab 2797  df-cleq 2811  df-clel 2890  df-ral 3140  df-rex 3141  df-rab 3144  df-uni 4831  df-sup 8894
This theorem is referenced by:  supeq1d  8898  supeq1i  8899  infeq1  8928  bndth  23489  ioorval  24102  uniioombllem6  24116  mdegcl  24590  suplesup  41483  supminfxr  41616  prproropf1olem2  43543  prproropf1olem3  43544  prproropf1olem4  43545  prproropf1o  43546  prproropreud  43548
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