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Theorem supeq1 6984
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 2672 . . . . 5 (𝐵 = 𝐶 → (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ↔ ∀𝑦𝐶 ¬ 𝑥𝑅𝑦))
2 rexeq 2673 . . . . . . 7 (𝐵 = 𝐶 → (∃𝑧𝐵 𝑦𝑅𝑧 ↔ ∃𝑧𝐶 𝑦𝑅𝑧))
32imbi2d 230 . . . . . 6 (𝐵 = 𝐶 → ((𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧) ↔ (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧)))
43ralbidv 2477 . . . . 5 (𝐵 = 𝐶 → (∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧) ↔ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧)))
51, 4anbi12d 473 . . . 4 (𝐵 = 𝐶 → ((∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧)) ↔ (∀𝑦𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧))))
65rabbidv 2726 . . 3 (𝐵 = 𝐶 → {𝑥𝐴 ∣ (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧))} = {𝑥𝐴 ∣ (∀𝑦𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧))})
76unieqd 3820 . 2 (𝐵 = 𝐶 {𝑥𝐴 ∣ (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧))} = {𝑥𝐴 ∣ (∀𝑦𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧))})
8 df-sup 6982 . 2 sup(𝐵, 𝐴, 𝑅) = {𝑥𝐴 ∣ (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧))}
9 df-sup 6982 . 2 sup(𝐶, 𝐴, 𝑅) = {𝑥𝐴 ∣ (∀𝑦𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧))}
107, 8, 93eqtr4g 2235 1 (𝐵 = 𝐶 → sup(𝐵, 𝐴, 𝑅) = sup(𝐶, 𝐴, 𝑅))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 104   = wceq 1353  wral 2455  wrex 2456  {crab 2459   cuni 3809   class class class wbr 4003  supcsup 6980
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159
This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-ral 2460  df-rex 2461  df-rab 2464  df-uni 3810  df-sup 6982
This theorem is referenced by:  supeq1d  6985  supeq1i  6986  infeq1  7009
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