Theorem setindft 10760

Description: Axiom of set-induction with a DV condition replaced with a non-freeness hypothesis (Contributed by BJ, 22-Nov-2019.)

Assertion

Ref	Expression
setindft	⊢ (∀𝑥Ⅎ𝑦𝜑 → (∀𝑥(∀𝑦 ∈ 𝑥 [𝑦 / 𝑥]𝜑 → 𝜑) → ∀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.

Step	Hyp	Ref	Expression
1		nfa1 1474	. . 3 ⊢ Ⅎ𝑥∀𝑥Ⅎ𝑦𝜑
2		nfv 1461	. . . . . 6 ⊢ Ⅎ𝑧∀𝑥Ⅎ𝑦𝜑
3		nfnf1 1476	. . . . . . 7 ⊢ Ⅎ𝑦Ⅎ𝑦𝜑
4	3	nfal 1508	. . . . . 6 ⊢ Ⅎ𝑦∀𝑥Ⅎ𝑦𝜑
5		nfsbt 1891	. . . . . 6 ⊢ (∀𝑥Ⅎ𝑦𝜑 → Ⅎ𝑦[𝑧 / 𝑥]𝜑)
6		nfv 1461	. . . . . . 7 ⊢ Ⅎ𝑧[𝑦 / 𝑥]𝜑
7	6	a1i 9	. . . . . 6 ⊢ (∀𝑥Ⅎ𝑦𝜑 → Ⅎ𝑧[𝑦 / 𝑥]𝜑)
8		sbequ 1761	. . . . . . 7 ⊢ (𝑧 = 𝑦 → ([𝑧 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑))
9	8	a1i 9	. . . . . 6 ⊢ (∀𝑥Ⅎ𝑦𝜑 → (𝑧 = 𝑦 → ([𝑧 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑)))
10	2, 4, 5, 7, 9	cbvrald 10598	. . . . 5 ⊢ (∀𝑥Ⅎ𝑦𝜑 → (∀𝑧 ∈ 𝑥 [𝑧 / 𝑥]𝜑 ↔ ∀𝑦 ∈ 𝑥 [𝑦 / 𝑥]𝜑))
11	10	biimpd 142	. . . 4 ⊢ (∀𝑥Ⅎ𝑦𝜑 → (∀𝑧 ∈ 𝑥 [𝑧 / 𝑥]𝜑 → ∀𝑦 ∈ 𝑥 [𝑦 / 𝑥]𝜑))
12	11	imim1d 74	. . 3 ⊢ (∀𝑥Ⅎ𝑦𝜑 → ((∀𝑦 ∈ 𝑥 [𝑦 / 𝑥]𝜑 → 𝜑) → (∀𝑧 ∈ 𝑥 [𝑧 / 𝑥]𝜑 → 𝜑)))
13	1, 12	alimd 1454	. 2 ⊢ (∀𝑥Ⅎ𝑦𝜑 → (∀𝑥(∀𝑦 ∈ 𝑥 [𝑦 / 𝑥]𝜑 → 𝜑) → ∀𝑥(∀𝑧 ∈ 𝑥 [𝑧 / 𝑥]𝜑 → 𝜑)))
14		ax-setind 4280	. 2 ⊢ (∀𝑥(∀𝑧 ∈ 𝑥 [𝑧 / 𝑥]𝜑 → 𝜑) → ∀𝑥𝜑)
15	13, 14	syl6 33	1 ⊢ (∀𝑥Ⅎ𝑦𝜑 → (∀𝑥(∀𝑦 ∈ 𝑥 [𝑦 / 𝑥]𝜑 → 𝜑) → ∀𝑥𝜑))

Syntax hints:   → wi 4   ↔ wb 103  ∀wal 1282  Ⅎwnf 1389  [wsb 1685  ∀wral 2348

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063  ax-setind 4280

This theorem depends on definitions:  df-bi 115  df-nf 1390  df-sb 1686  df-cleq 2074  df-clel 2077  df-ral 2353

This theorem is referenced by:  setindf  10761