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Theorem axprlem4 5388
Description: Lemma for axpr 5389. 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 1844 . . . . . . 7 (∀𝑛𝜑 → ∀𝑛(𝑛 ∈ 𝑠 → ∀𝑡 ¬ 𝑡 ∈ 𝑛))
43ralrid 3085 . . . . . 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 1865 . . 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 1950 . 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 1568  ∃wex 1812  ∀wral 3077
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943
This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-ral 3078
This theorem is used by:  axpr  5389
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