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Theorem axprlem4 5387
Description: Lemma for axpr 5388. 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 ∃𝑠∀𝑛𝜑
axprlem4.2 (𝜑 → (𝑛 ∈ 𝑠 → ∀𝑡 ¬ 𝑡 ∈ 𝑛))
axprlem4.3 (∀𝑛𝜑 → (if-(∃𝑛 𝑛 ∈ 𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦) ↔ 𝑤 = 𝑣))
Assertion
Ref Expression
axprlem4 (∀𝑠(∀𝑛 ∈ 𝑠 ∀𝑡 ¬ 𝑡 ∈ 𝑛 → 𝑠 ∈ 𝑝) → (𝑤 = 𝑣 → ∃𝑠(𝑠 ∈ 𝑝 ∧ if-(∃𝑛 𝑛 ∈ 𝑠, 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 ∃𝑠∀𝑛𝜑
2 axprlem4.2 . . . . . . . 8 (𝜑 → (𝑛 ∈ 𝑠 → ∀𝑡 ¬ 𝑡 ∈ 𝑛))
32alimi 1843 . . . . . . 7 (∀𝑛𝜑 → ∀𝑛(𝑛 ∈ 𝑠 → ∀𝑡 ¬ 𝑡 ∈ 𝑛))
43ralrid 3084 . . . . . 6 (∀𝑛𝜑 → ∀𝑛 ∈ 𝑠 ∀𝑡 ¬ 𝑡 ∈ 𝑛)
54imim1i 64 . . . . 5 ((∀𝑛 ∈ 𝑠 ∀𝑡 ¬ 𝑡 ∈ 𝑛 → 𝑠 ∈ 𝑝) → (∀𝑛𝜑 → 𝑠 ∈ 𝑝))
65ancrd 561 . . . 4 ((∀𝑛 ∈ 𝑠 ∀𝑡 ¬ 𝑡 ∈ 𝑛 → 𝑠 ∈ 𝑝) → (∀𝑛𝜑 → (𝑠 ∈ 𝑝 ∧ ∀𝑛𝜑)))
76aleximi 1864 . . 3 (∀𝑠(∀𝑛 ∈ 𝑠 ∀𝑡 ¬ 𝑡 ∈ 𝑛 → 𝑠 ∈ 𝑝) → (∃𝑠∀𝑛𝜑 → ∃𝑠(𝑠 ∈ 𝑝 ∧ ∀𝑛𝜑)))
81, 7mpi 21 . 2 (∀𝑠(∀𝑛 ∈ 𝑠 ∀𝑡 ¬ 𝑡 ∈ 𝑛 → 𝑠 ∈ 𝑝) → ∃𝑠(𝑠 ∈ 𝑝 ∧ ∀𝑛𝜑))
9 axprlem4.3 . . . . 5 (∀𝑛𝜑 → (if-(∃𝑛 𝑛 ∈ 𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦) ↔ 𝑤 = 𝑣))
109biimprcd 253 . . . 4 (𝑤 = 𝑣 → (∀𝑛𝜑 → if-(∃𝑛 𝑛 ∈ 𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦)))
1110anim2d 624 . . 3 (𝑤 = 𝑣 → ((𝑠 ∈ 𝑝 ∧ ∀𝑛𝜑) → (𝑠 ∈ 𝑝 ∧ if-(∃𝑛 𝑛 ∈ 𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦))))
1211eximdv 1949 . 2 (𝑤 = 𝑣 → (∃𝑠(𝑠 ∈ 𝑝 ∧ ∀𝑛𝜑) → ∃𝑠(𝑠 ∈ 𝑝 ∧ if-(∃𝑛 𝑛 ∈ 𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦))))
138, 12syl5com 32 1 (∀𝑠(∀𝑛 ∈ 𝑠 ∀𝑡 ¬ 𝑡 ∈ 𝑛 → 𝑠 ∈ 𝑝) → (𝑤 = 𝑣 → ∃𝑠(𝑠 ∈ 𝑝 ∧ if-(∃𝑛 𝑛 ∈ 𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦))))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401  if-wif 1078  ∀wal 1567  ∃wex 1811  ∀wral 3076
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942
This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-ral 3077
This theorem is used by:  axpr  5388
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