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Theorem supeq1 8558
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 3286 . . . . 5 (𝐵 = 𝐶 → (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ↔ ∀𝑦𝐶 ¬ 𝑥𝑅𝑦))
2 rexeq 3287 . . . . . . 7 (𝐵 = 𝐶 → (∃𝑧𝐵 𝑦𝑅𝑧 ↔ ∃𝑧𝐶 𝑦𝑅𝑧))
32imbi2d 331 . . . . . 6 (𝐵 = 𝐶 → ((𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧) ↔ (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧)))
43ralbidv 3133 . . . . 5 (𝐵 = 𝐶 → (∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧) ↔ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧)))
51, 4anbi12d 624 . . . 4 (𝐵 = 𝐶 → ((∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧)) ↔ (∀𝑦𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧))))
65rabbidv 3338 . . 3 (𝐵 = 𝐶 → {𝑥𝐴 ∣ (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧))} = {𝑥𝐴 ∣ (∀𝑦𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧))})
76unieqd 4604 . 2 (𝐵 = 𝐶 {𝑥𝐴 ∣ (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧))} = {𝑥𝐴 ∣ (∀𝑦𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧))})
8 df-sup 8555 . 2 sup(𝐵, 𝐴, 𝑅) = {𝑥𝐴 ∣ (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧))}
9 df-sup 8555 . 2 sup(𝐶, 𝐴, 𝑅) = {𝑥𝐴 ∣ (∀𝑦𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧))}
107, 8, 93eqtr4g 2824 1 (𝐵 = 𝐶 → sup(𝐵, 𝐴, 𝑅) = sup(𝐶, 𝐴, 𝑅))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 384   = wceq 1652  wral 3055  wrex 3056  {crab 3059   cuni 4594   class class class wbr 4809  supcsup 8553
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1890  ax-4 1904  ax-5 2005  ax-6 2070  ax-7 2105  ax-9 2164  ax-10 2183  ax-11 2198  ax-12 2211  ax-ext 2743
This theorem depends on definitions:  df-bi 198  df-an 385  df-or 874  df-tru 1656  df-ex 1875  df-nf 1879  df-sb 2063  df-clab 2752  df-cleq 2758  df-clel 2761  df-nfc 2896  df-ral 3060  df-rex 3061  df-rab 3064  df-uni 4595  df-sup 8555
This theorem is referenced by:  supeq1d  8559  supeq1i  8560  infeq1  8589  bndth  23036  ioorval  23632  uniioombllem6  23646  mdegcl  24120  suplesup  40193  supminfxr  40331
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