Theorem setindf 13153

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

Hypothesis

Ref	Expression
setindf.nf	⊢ Ⅎ𝑦𝜑

Assertion

Ref	Expression
setindf	⊢ (∀𝑥(∀𝑦 ∈ 𝑥 [𝑦 / 𝑥]𝜑 → 𝜑) → ∀𝑥𝜑)

Distinct variable group:   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥,𝑦)

Proof of Theorem setindf

Step	Hyp	Ref	Expression
1		setindft 13152	. 2 ⊢ (∀𝑥Ⅎ𝑦𝜑 → (∀𝑥(∀𝑦 ∈ 𝑥 [𝑦 / 𝑥]𝜑 → 𝜑) → ∀𝑥𝜑))
2		setindf.nf	. 2 ⊢ Ⅎ𝑦𝜑
3	1, 2	mpg 1427	1 ⊢ (∀𝑥(∀𝑦 ∈ 𝑥 [𝑦 / 𝑥]𝜑 → 𝜑) → ∀𝑥𝜑)

Syntax hints:   → wi 4  ∀wal 1329  Ⅎwnf 1436  [wsb 1735  ∀wral 2414

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 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2119  ax-setind 4447

This theorem depends on definitions:  df-bi 116  df-nf 1437  df-sb 1736  df-cleq 2130  df-clel 2133  df-ral 2419

This theorem is referenced by: (None)