Theorem sb6x 2384

Description: Equivalence involving substitution for a variable not free. (Contributed by NM, 2-Jun-1993.) (Revised by Mario Carneiro, 4-Oct-2016.)

Hypothesis

Ref	Expression
sb6x.1	|- F/ x ph

Assertion

Ref	Expression
sb6x	|- ( [ y / x ] ph <-> A. x ( x = y -> ph ) )

Proof of Theorem sb6x

Step	Hyp	Ref	Expression
1		sb6x.1	. . 3 |- F/ x ph
2	1	sbf 2380	. 2 |- ( [ y / x ] ph <-> ph )
3		biidd 252	. . 3 |- ( x = y -> ( ph <-> ph ) )
4	1, 3	equsal 2291	. 2 |- ( A. x ( x = y -> ph ) <-> ph )
5	2, 4	bitr4i 267	1 |- ( [ y / x ] ph <-> A. x ( x = y -> ph ) )

Syntax hints:   -> wi 4   <-> wb 196   A.wal 1481   F/wnf 1708   [wsb 1880

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-12 2047  ax-13 2246

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-ex 1705  df-nf 1710  df-sb 1881

This theorem is referenced by: (None)