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Theorem setindft 14000
Description: Axiom of set-induction with a disjoint variable condition replaced with a nonfreeness hypothesis. (Contributed by BJ, 22-Nov-2019.)
Assertion
Ref Expression
setindft (∀𝑥𝑦𝜑 → (∀𝑥(∀𝑦𝑥 [𝑦 / 𝑥]𝜑𝜑) → ∀𝑥𝜑))
Distinct variable group:   𝑥,𝑦
Allowed substitution hints:   𝜑(𝑥,𝑦)

Proof of Theorem setindft
Dummy variable 𝑧 is distinct from all other variables.
StepHypRef Expression
1 nfa1 1534 . . 3 𝑥𝑥𝑦𝜑
2 nfv 1521 . . . . . 6 𝑧𝑥𝑦𝜑
3 nfnf1 1537 . . . . . . 7 𝑦𝑦𝜑
43nfal 1569 . . . . . 6 𝑦𝑥𝑦𝜑
5 nfsbt 1969 . . . . . 6 (∀𝑥𝑦𝜑 → Ⅎ𝑦[𝑧 / 𝑥]𝜑)
6 nfv 1521 . . . . . . 7 𝑧[𝑦 / 𝑥]𝜑
76a1i 9 . . . . . 6 (∀𝑥𝑦𝜑 → Ⅎ𝑧[𝑦 / 𝑥]𝜑)
8 sbequ 1833 . . . . . . 7 (𝑧 = 𝑦 → ([𝑧 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑))
98a1i 9 . . . . . 6 (∀𝑥𝑦𝜑 → (𝑧 = 𝑦 → ([𝑧 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑)))
102, 4, 5, 7, 9cbvrald 13823 . . . . 5 (∀𝑥𝑦𝜑 → (∀𝑧𝑥 [𝑧 / 𝑥]𝜑 ↔ ∀𝑦𝑥 [𝑦 / 𝑥]𝜑))
1110biimpd 143 . . . 4 (∀𝑥𝑦𝜑 → (∀𝑧𝑥 [𝑧 / 𝑥]𝜑 → ∀𝑦𝑥 [𝑦 / 𝑥]𝜑))
1211imim1d 75 . . 3 (∀𝑥𝑦𝜑 → ((∀𝑦𝑥 [𝑦 / 𝑥]𝜑𝜑) → (∀𝑧𝑥 [𝑧 / 𝑥]𝜑𝜑)))
131, 12alimd 1514 . 2 (∀𝑥𝑦𝜑 → (∀𝑥(∀𝑦𝑥 [𝑦 / 𝑥]𝜑𝜑) → ∀𝑥(∀𝑧𝑥 [𝑧 / 𝑥]𝜑𝜑)))
14 ax-setind 4521 . 2 (∀𝑥(∀𝑧𝑥 [𝑧 / 𝑥]𝜑𝜑) → ∀𝑥𝜑)
1513, 14syl6 33 1 (∀𝑥𝑦𝜑 → (∀𝑥(∀𝑦𝑥 [𝑦 / 𝑥]𝜑𝜑) → ∀𝑥𝜑))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 104  wal 1346  wnf 1453  [wsb 1755  wral 2448
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152  ax-setind 4521
This theorem depends on definitions:  df-bi 116  df-nf 1454  df-sb 1756  df-cleq 2163  df-clel 2166  df-ral 2453
This theorem is referenced by:  setindf  14001
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