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Theorem setindis 13859
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 2308 . . . . 5 𝑥𝑦
2 setindis.nf0 . . . . 5 𝑥𝜓
31, 2nfralxy 2504 . . . 4 𝑥𝑧𝑦 𝜓
4 setindis.nf1 . . . 4 𝑥𝜒
53, 4nfim 1560 . . 3 𝑥(∀𝑧𝑦 𝜓𝜒)
6 nfcv 2308 . . . . 5 𝑦𝑥
7 setindis.nf3 . . . . 5 𝑦𝜓
86, 7nfralxy 2504 . . . 4 𝑦𝑧𝑥 𝜓
9 setindis.nf2 . . . 4 𝑦𝜑
108, 9nfim 1560 . . 3 𝑦(∀𝑧𝑥 𝜓𝜑)
11 raleq 2661 . . . . 5 (𝑦 = 𝑥 → (∀𝑧𝑦 𝜓 ↔ ∀𝑧𝑥 𝜓))
1211biimprd 157 . . . 4 (𝑦 = 𝑥 → (∀𝑧𝑥 𝜓 → ∀𝑧𝑦 𝜓))
13 setindis.2 . . . . 5 (𝑥 = 𝑦 → (𝜒𝜑))
1413equcoms 1696 . . . 4 (𝑦 = 𝑥 → (𝜒𝜑))
1512, 14imim12d 74 . . 3 (𝑦 = 𝑥 → ((∀𝑧𝑦 𝜓𝜒) → (∀𝑧𝑥 𝜓𝜑)))
165, 10, 15cbv3 1730 . 2 (∀𝑦(∀𝑧𝑦 𝜓𝜒) → ∀𝑥(∀𝑧𝑥 𝜓𝜑))
17 setindis.1 . . . . . 6 (𝑥 = 𝑧 → (𝜑𝜓))
182, 17bj-sbime 13664 . . . . 5 ([𝑧 / 𝑥]𝜑𝜓)
1918ralimi 2529 . . . 4 (∀𝑧𝑥 [𝑧 / 𝑥]𝜑 → ∀𝑧𝑥 𝜓)
2019imim1i 60 . . 3 ((∀𝑧𝑥 𝜓𝜑) → (∀𝑧𝑥 [𝑧 / 𝑥]𝜑𝜑))
2120alimi 1443 . 2 (∀𝑥(∀𝑧𝑥 𝜓𝜑) → ∀𝑥(∀𝑧𝑥 [𝑧 / 𝑥]𝜑𝜑))
22 ax-setind 4514 . 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 1341  wnf 1448  [wsb 1750  wral 2444
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 699  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147  ax-setind 4514
This theorem depends on definitions:  df-bi 116  df-tru 1346  df-nf 1449  df-sb 1751  df-cleq 2158  df-clel 2161  df-nfc 2297  df-ral 2449
This theorem is referenced by:  bj-inf2vnlem4  13865  bj-findis  13871
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