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Theorem axprlem4 5426
Description: Lemma for axpr 5427. 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 1811 . . . . . . 7 (∀𝑛𝜑 → ∀𝑛(𝑛𝑠 → ∀𝑡 ¬ 𝑡𝑛))
4 df-ral 3062 . . . . . . 7 (∀𝑛𝑠𝑡 ¬ 𝑡𝑛 ↔ ∀𝑛(𝑛𝑠 → ∀𝑡 ¬ 𝑡𝑛))
53, 4sylibr 234 . . . . . 6 (∀𝑛𝜑 → ∀𝑛𝑠𝑡 ¬ 𝑡𝑛)
65imim1i 63 . . . . 5 ((∀𝑛𝑠𝑡 ¬ 𝑡𝑛𝑠𝑝) → (∀𝑛𝜑𝑠𝑝))
76ancrd 551 . . . 4 ((∀𝑛𝑠𝑡 ¬ 𝑡𝑛𝑠𝑝) → (∀𝑛𝜑 → (𝑠𝑝 ∧ ∀𝑛𝜑)))
87aleximi 1832 . . 3 (∀𝑠(∀𝑛𝑠𝑡 ¬ 𝑡𝑛𝑠𝑝) → (∃𝑠𝑛𝜑 → ∃𝑠(𝑠𝑝 ∧ ∀𝑛𝜑)))
91, 8mpi 20 . 2 (∀𝑠(∀𝑛𝑠𝑡 ¬ 𝑡𝑛𝑠𝑝) → ∃𝑠(𝑠𝑝 ∧ ∀𝑛𝜑))
10 axprlem4.3 . . . . 5 (∀𝑛𝜑 → (if-(∃𝑛 𝑛𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦) ↔ 𝑤 = 𝑣))
1110biimprcd 250 . . . 4 (𝑤 = 𝑣 → (∀𝑛𝜑 → if-(∃𝑛 𝑛𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦)))
1211anim2d 612 . . 3 (𝑤 = 𝑣 → ((𝑠𝑝 ∧ ∀𝑛𝜑) → (𝑠𝑝 ∧ if-(∃𝑛 𝑛𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦))))
1312eximdv 1917 . 2 (𝑤 = 𝑣 → (∃𝑠(𝑠𝑝 ∧ ∀𝑛𝜑) → ∃𝑠(𝑠𝑝 ∧ if-(∃𝑛 𝑛𝑠, 𝑤 = 𝑥, 𝑤 = 𝑦))))
149, 13syl5com 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 1538  wex 1779  wral 3061
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910
This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1780  df-ral 3062
This theorem is referenced by:  axpr  5427
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