Theorem setindft 9425

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 1431	. . 3 ⊢ Ⅎx∀xℲyφ
2		nfv 1418	. . . . . 6 ⊢ Ⅎz∀xℲyφ
3		nfnf1 1433	. . . . . . 7 ⊢ ℲyℲyφ
4	3	nfal 1465	. . . . . 6 ⊢ Ⅎy∀xℲyφ
5		nfsbt 1847	. . . . . 6 ⊢ (∀xℲyφ → Ⅎy[z / x]φ)
6		nfv 1418	. . . . . . 7 ⊢ Ⅎz[y / x]φ
7	6	a1i 9	. . . . . 6 ⊢ (∀xℲyφ → Ⅎz[y / x]φ)
8		sbequ 1718	. . . . . . 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 9262	. . . . 5 ⊢ (∀xℲyφ → (∀z ∈ x [z / x]φ ↔ ∀y ∈ x [y / x]φ))
11	10	biimpd 132	. . . 4 ⊢ (∀xℲyφ → (∀z ∈ x [z / x]φ → ∀y ∈ x [y / x]φ))
12	11	imim1d 69	. . 3 ⊢ (∀xℲyφ → ((∀y ∈ x [y / x]φ → φ) → (∀z ∈ x [z / x]φ → φ)))
13	1, 12	alimd 1411	. 2 ⊢ (∀xℲyφ → (∀x(∀y ∈ x [y / x]φ → φ) → ∀x(∀z ∈ x [z / x]φ → φ)))
14		ax-setind 4220	. 2 ⊢ (∀x(∀z ∈ x [z / x]φ → φ) → ∀xφ)
15	13, 14	syl6 29	1 ⊢ (∀xℲyφ → (∀x(∀y ∈ x [y / x]φ → φ) → ∀xφ))

Syntax hints:   → wi 4   ↔ wb 98  ∀wal 1240  Ⅎwnf 1346  [wsb 1642  ∀wral 2300

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 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bndl 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019  ax-setind 4220

This theorem depends on definitions:  df-bi 110  df-nf 1347  df-sb 1643  df-cleq 2030  df-clel 2033  df-ral 2305

This theorem is referenced by:  setindf  9426