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Theorem axprlem4 5373
Description: Lemma for axpr 5374. If an existing set of empty sets corresponds to one element of the pair, then the element is included in any superset of the set whose existence is asserted by the axiom of replacement. (Contributed by Rohan Ridenour, 10-Aug-2023.) (Revised by BJ, 13-Aug-2023.) (Revised by Matthew House, 18-Sep-2025.)
Hypotheses
Ref Expression
axprlem4.1 𝑠𝑛𝜑
axprlem4.2 (𝜑 → (𝑛𝑠 → ∀𝑡 ¬ 𝑡𝑛))
axprlem4.3 (∀𝑛𝜑 → (if-(∃𝑛 𝑛𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦) ↔ 𝑤 = 𝑣))
Assertion
Ref Expression
axprlem4 (∀𝑠(∀𝑛𝑠𝑡 ¬ 𝑡𝑛𝑠𝑝) → (𝑤 = 𝑣 → ∃𝑠(𝑠𝑝 ∧ if-(∃𝑛 𝑛𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦))))
Distinct variable groups:   𝑤,𝑠   𝑣,𝑠
Allowed substitution hints:   𝜑(𝑥,𝑦,𝑤,𝑣,𝑡,𝑛,𝑠,𝑝)

Proof of Theorem axprlem4
StepHypRef Expression
1 axprlem4.1 . . 3 𝑠𝑛𝜑
2 axprlem4.2 . . . . . . . 8 (𝜑 → (𝑛𝑠 → ∀𝑡 ¬ 𝑡𝑛))
32alimi 1813 . . . . . . 7 (∀𝑛𝜑 → ∀𝑛(𝑛𝑠 → ∀𝑡 ¬ 𝑡𝑛))
43ralrid 3060 . . . . . 6 (∀𝑛𝜑 → ∀𝑛𝑠𝑡 ¬ 𝑡𝑛)
54imim1i 63 . . . . 5 ((∀𝑛𝑠𝑡 ¬ 𝑡𝑛𝑠𝑝) → (∀𝑛𝜑𝑠𝑝))
65ancrd 551 . . . 4 ((∀𝑛𝑠𝑡 ¬ 𝑡𝑛𝑠𝑝) → (∀𝑛𝜑 → (𝑠𝑝 ∧ ∀𝑛𝜑)))
76aleximi 1834 . . 3 (∀𝑠(∀𝑛𝑠𝑡 ¬ 𝑡𝑛𝑠𝑝) → (∃𝑠𝑛𝜑 → ∃𝑠(𝑠𝑝 ∧ ∀𝑛𝜑)))
81, 7mpi 20 . 2 (∀𝑠(∀𝑛𝑠𝑡 ¬ 𝑡𝑛𝑠𝑝) → ∃𝑠(𝑠𝑝 ∧ ∀𝑛𝜑))
9 axprlem4.3 . . . . 5 (∀𝑛𝜑 → (if-(∃𝑛 𝑛𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦) ↔ 𝑤 = 𝑣))
109biimprcd 250 . . . 4 (𝑤 = 𝑣 → (∀𝑛𝜑 → if-(∃𝑛 𝑛𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦)))
1110anim2d 613 . . 3 (𝑤 = 𝑣 → ((𝑠𝑝 ∧ ∀𝑛𝜑) → (𝑠𝑝 ∧ if-(∃𝑛 𝑛𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦))))
1211eximdv 1919 . 2 (𝑤 = 𝑣 → (∃𝑠(𝑠𝑝 ∧ ∀𝑛𝜑) → ∃𝑠(𝑠𝑝 ∧ if-(∃𝑛 𝑛𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦))))
138, 12syl5com 31 1 (∀𝑠(∀𝑛𝑠𝑡 ¬ 𝑡𝑛𝑠𝑝) → (𝑤 = 𝑣 → ∃𝑠(𝑠𝑝 ∧ if-(∃𝑛 𝑛𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦))))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 206  wa 395  if-wif 1063  wal 1540  wex 1781  wral 3052
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912
This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1782  df-ral 3053
This theorem is referenced by:  axpr  5374
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