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Theorem axprlem4 5362
Description: Lemma for axpr 5363. 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 1818 . . . . . . 7 (∀𝑛𝜑 → ∀𝑛(𝑛𝑠 → ∀𝑡 ¬ 𝑡𝑛))
43ralrid 3062 . . . . . 6 (∀𝑛𝜑 → ∀𝑛𝑠𝑡 ¬ 𝑡𝑛)
54imim1i 63 . . . . 5 ((∀𝑛𝑠𝑡 ¬ 𝑡𝑛𝑠𝑝) → (∀𝑛𝜑𝑠𝑝))
65ancrd 556 . . . 4 ((∀𝑛𝑠𝑡 ¬ 𝑡𝑛𝑠𝑝) → (∀𝑛𝜑 → (𝑠𝑝 ∧ ∀𝑛𝜑)))
76aleximi 1839 . . 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 251 . . . 4 (𝑤 = 𝑣 → (∀𝑛𝜑 → if-(∃𝑛 𝑛𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦)))
1110anim2d 618 . . 3 (𝑤 = 𝑣 → ((𝑠𝑝 ∧ ∀𝑛𝜑) → (𝑠𝑝 ∧ if-(∃𝑛 𝑛𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦))))
1211eximdv 1924 . 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 207  wa 396  if-wif 1068  wal 1545  wex 1786  wral 3054
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917
This theorem depends on definitions:  df-bi 208  df-an 397  df-ex 1787  df-ral 3055
This theorem is referenced by:  axpr  5363
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