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Theorem supeq1 6963
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 2665 . . . . 5 (𝐵 = 𝐶 → (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ↔ ∀𝑦𝐶 ¬ 𝑥𝑅𝑦))
2 rexeq 2666 . . . . . . 7 (𝐵 = 𝐶 → (∃𝑧𝐵 𝑦𝑅𝑧 ↔ ∃𝑧𝐶 𝑦𝑅𝑧))
32imbi2d 229 . . . . . 6 (𝐵 = 𝐶 → ((𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧) ↔ (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧)))
43ralbidv 2470 . . . . 5 (𝐵 = 𝐶 → (∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧) ↔ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧)))
51, 4anbi12d 470 . . . 4 (𝐵 = 𝐶 → ((∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧)) ↔ (∀𝑦𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧))))
65rabbidv 2719 . . 3 (𝐵 = 𝐶 → {𝑥𝐴 ∣ (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧))} = {𝑥𝐴 ∣ (∀𝑦𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧))})
76unieqd 3807 . 2 (𝐵 = 𝐶 {𝑥𝐴 ∣ (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧))} = {𝑥𝐴 ∣ (∀𝑦𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧))})
8 df-sup 6961 . 2 sup(𝐵, 𝐴, 𝑅) = {𝑥𝐴 ∣ (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧))}
9 df-sup 6961 . 2 sup(𝐶, 𝐴, 𝑅) = {𝑥𝐴 ∣ (∀𝑦𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧))}
107, 8, 93eqtr4g 2228 1 (𝐵 = 𝐶 → sup(𝐵, 𝐴, 𝑅) = sup(𝐶, 𝐴, 𝑅))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 103   = wceq 1348  wral 2448  wrex 2449  {crab 2452   cuni 3796   class class class wbr 3989  supcsup 6959
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152
This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-ral 2453  df-rex 2454  df-rab 2457  df-uni 3797  df-sup 6961
This theorem is referenced by:  supeq1d  6964  supeq1i  6965  infeq1  6988
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