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	⊢ (∀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 1534	. . 3 ⊢ Ⅎ𝑥∀𝑥Ⅎ𝑦𝜑
2		nfv 1521	. . . . . 6 ⊢ Ⅎ𝑧∀𝑥Ⅎ𝑦𝜑
3		nfnf1 1537	. . . . . . 7 ⊢ Ⅎ𝑦Ⅎ𝑦𝜑
4	3	nfal 1569	. . . . . 6 ⊢ Ⅎ𝑦∀𝑥Ⅎ𝑦𝜑
5		nfsbt 1969	. . . . . 6 ⊢ (∀𝑥Ⅎ𝑦𝜑 → Ⅎ𝑦[𝑧 / 𝑥]𝜑)
6		nfv 1521	. . . . . . 7 ⊢ Ⅎ𝑧[𝑦 / 𝑥]𝜑
7	6	a1i 9	. . . . . 6 ⊢ (∀𝑥Ⅎ𝑦𝜑 → Ⅎ𝑧[𝑦 / 𝑥]𝜑)
8		sbequ 1833	. . . . . . 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 13823	. . . . 5 ⊢ (∀𝑥Ⅎ𝑦𝜑 → (∀𝑧 ∈ 𝑥 [𝑧 / 𝑥]𝜑 ↔ ∀𝑦 ∈ 𝑥 [𝑦 / 𝑥]𝜑))
11	10	biimpd 143	. . . 4 ⊢ (∀𝑥Ⅎ𝑦𝜑 → (∀𝑧 ∈ 𝑥 [𝑧 / 𝑥]𝜑 → ∀𝑦 ∈ 𝑥 [𝑦 / 𝑥]𝜑))
12	11	imim1d 75	. . 3 ⊢ (∀𝑥Ⅎ𝑦𝜑 → ((∀𝑦 ∈ 𝑥 [𝑦 / 𝑥]𝜑 → 𝜑) → (∀𝑧 ∈ 𝑥 [𝑧 / 𝑥]𝜑 → 𝜑)))
13	1, 12	alimd 1514	. 2 ⊢ (∀𝑥Ⅎ𝑦𝜑 → (∀𝑥(∀𝑦 ∈ 𝑥 [𝑦 / 𝑥]𝜑 → 𝜑) → ∀𝑥(∀𝑧 ∈ 𝑥 [𝑧 / 𝑥]𝜑 → 𝜑)))
14		ax-setind 4521	. 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 104  ∀wal 1346  Ⅎwnf 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