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	|- F/ y ph

Assertion

Ref	Expression
setindf	|- ( A. x ( A. y e. x [ y / x ] ph -> ph ) -> A. x ph )

Distinct variable group:   x, y

Allowed substitution hints:   ph( x, y)

Proof of Theorem setindf

Step	Hyp	Ref	Expression
1		setindft 14000	. 2 |- ( A. x F/ y ph -> ( A. x ( A. y e. x [ y / x ] ph -> ph ) -> A. x ph ) )
2		setindf.nf	. 2 |- F/ y ph
3	1, 2	mpg 1444	1 |- ( A. x ( A. y e. x [ y / x ] ph -> ph ) -> A. x ph )

Syntax hints:   -> wi 4   A.wal 1346   F/wnf 1453   [wsb 1755   A.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)