Theorem setindft 15975

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 1565	. . 3 ⊢ Ⅎ𝑥∀𝑥Ⅎ𝑦𝜑
2		nfv 1552	. . . . . 6 ⊢ Ⅎ𝑧∀𝑥Ⅎ𝑦𝜑
3		nfnf1 1568	. . . . . . 7 ⊢ Ⅎ𝑦Ⅎ𝑦𝜑
4	3	nfal 1600	. . . . . 6 ⊢ Ⅎ𝑦∀𝑥Ⅎ𝑦𝜑
5		nfsbt 2005	. . . . . 6 ⊢ (∀𝑥Ⅎ𝑦𝜑 → Ⅎ𝑦[𝑧 / 𝑥]𝜑)
6		nfv 1552	. . . . . . 7 ⊢ Ⅎ𝑧[𝑦 / 𝑥]𝜑
7	6	a1i 9	. . . . . 6 ⊢ (∀𝑥Ⅎ𝑦𝜑 → Ⅎ𝑧[𝑦 / 𝑥]𝜑)
8		sbequ 1864	. . . . . . 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 15798	. . . . 5 ⊢ (∀𝑥Ⅎ𝑦𝜑 → (∀𝑧 ∈ 𝑥 [𝑧 / 𝑥]𝜑 ↔ ∀𝑦 ∈ 𝑥 [𝑦 / 𝑥]𝜑))
11	10	biimpd 144	. . . 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 1545	. 2 ⊢ (∀𝑥Ⅎ𝑦𝜑 → (∀𝑥(∀𝑦 ∈ 𝑥 [𝑦 / 𝑥]𝜑 → 𝜑) → ∀𝑥(∀𝑧 ∈ 𝑥 [𝑧 / 𝑥]𝜑 → 𝜑)))
14		ax-setind 4589	. 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 105  ∀wal 1371  Ⅎwnf 1484  [wsb 1786  ∀wral 2485

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2188  ax-setind 4589

This theorem depends on definitions:  df-bi 117  df-nf 1485  df-sb 1787  df-cleq 2199  df-clel 2202  df-ral 2490

This theorem is referenced by:  setindf  15976