Theorem setindf 14001

Description: Axiom of set-induction with a disjoint variable condition replaced with a nonfreeness 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 14000	. 2 ⊢ (∀𝑥Ⅎ𝑦𝜑 → (∀𝑥(∀𝑦 ∈ 𝑥 [𝑦 / 𝑥]𝜑 → 𝜑) → ∀𝑥𝜑))
2		setindf.nf	. 2 ⊢ Ⅎ𝑦𝜑
3	1, 2	mpg 1444	1 ⊢ (∀𝑥(∀𝑦 ∈ 𝑥 [𝑦 / 𝑥]𝜑 → 𝜑) → ∀𝑥𝜑)

Syntax hints:   → wi 4  ∀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: (None)