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Theorem setindis 10092
Description: Axiom of set induction using implicit substitutions. (Contributed by BJ, 22-Nov-2019.)
Hypotheses
Ref Expression
setindis.nf0 𝑥𝜓
setindis.nf1 𝑥𝜒
setindis.nf2 𝑦𝜑
setindis.nf3 𝑦𝜓
setindis.1 (𝑥 = 𝑧 → (𝜑𝜓))
setindis.2 (𝑥 = 𝑦 → (𝜒𝜑))
Assertion
Ref Expression
setindis (∀𝑦(∀𝑧𝑦 𝜓𝜒) → ∀𝑥𝜑)
Distinct variable groups:   𝑥,𝑦,𝑧   𝜑,𝑧
Allowed substitution hints:   𝜑(𝑥,𝑦)   𝜓(𝑥,𝑦,𝑧)   𝜒(𝑥,𝑦,𝑧)

Proof of Theorem setindis
StepHypRef Expression
1 nfcv 2178 . . . . 5 𝑥𝑦
2 setindis.nf0 . . . . 5 𝑥𝜓
31, 2nfralxy 2360 . . . 4 𝑥𝑧𝑦 𝜓
4 setindis.nf1 . . . 4 𝑥𝜒
53, 4nfim 1464 . . 3 𝑥(∀𝑧𝑦 𝜓𝜒)
6 nfcv 2178 . . . . 5 𝑦𝑥
7 setindis.nf3 . . . . 5 𝑦𝜓
86, 7nfralxy 2360 . . . 4 𝑦𝑧𝑥 𝜓
9 setindis.nf2 . . . 4 𝑦𝜑
108, 9nfim 1464 . . 3 𝑦(∀𝑧𝑥 𝜓𝜑)
11 raleq 2505 . . . . 5 (𝑦 = 𝑥 → (∀𝑧𝑦 𝜓 ↔ ∀𝑧𝑥 𝜓))
1211biimprd 147 . . . 4 (𝑦 = 𝑥 → (∀𝑧𝑥 𝜓 → ∀𝑧𝑦 𝜓))
13 setindis.2 . . . . 5 (𝑥 = 𝑦 → (𝜒𝜑))
1413equcoms 1594 . . . 4 (𝑦 = 𝑥 → (𝜒𝜑))
1512, 14imim12d 68 . . 3 (𝑦 = 𝑥 → ((∀𝑧𝑦 𝜓𝜒) → (∀𝑧𝑥 𝜓𝜑)))
165, 10, 15cbv3 1630 . 2 (∀𝑦(∀𝑧𝑦 𝜓𝜒) → ∀𝑥(∀𝑧𝑥 𝜓𝜑))
17 setindis.1 . . . . . 6 (𝑥 = 𝑧 → (𝜑𝜓))
182, 17bj-sbime 9913 . . . . 5 ([𝑧 / 𝑥]𝜑𝜓)
1918ralimi 2384 . . . 4 (∀𝑧𝑥 [𝑧 / 𝑥]𝜑 → ∀𝑧𝑥 𝜓)
2019imim1i 54 . . 3 ((∀𝑧𝑥 𝜓𝜑) → (∀𝑧𝑥 [𝑧 / 𝑥]𝜑𝜑))
2120alimi 1344 . 2 (∀𝑥(∀𝑧𝑥 𝜓𝜑) → ∀𝑥(∀𝑧𝑥 [𝑧 / 𝑥]𝜑𝜑))
22 ax-setind 4262 . 2 (∀𝑥(∀𝑧𝑥 [𝑧 / 𝑥]𝜑𝜑) → ∀𝑥𝜑)
2316, 21, 223syl 17 1 (∀𝑦(∀𝑧𝑦 𝜓𝜒) → ∀𝑥𝜑)
Colors of variables: wff set class
Syntax hints:  wi 4  wal 1241  wnf 1349  [wsb 1645  wral 2306
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022  ax-setind 4262
This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-cleq 2033  df-clel 2036  df-nfc 2167  df-ral 2311
This theorem is referenced by:  bj-inf2vnlem4  10098  bj-findis  10104
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