Theorem nfcdeq 3432

Description: If we have a conditional equality proof, where ph is ph ( x ) and ps is ph ( y ), and ph ( x ) in fact does not have x free in it according to F/, then ph ( x ) <-> ph ( y ) unconditionally. This proves that F/ x ph is actually a not-free predicate. (Contributed by Mario Carneiro, 11-Aug-2016.)

Hypotheses

Ref	Expression
nfcdeq.1	|- F/ x ph
nfcdeq.2	|- CondEq ( x = y -> ( ph <-> ps ) )

Assertion

Ref	Expression
nfcdeq	|- ( ph <-> ps )

Distinct variable groups:   ps, x   ph, y

Allowed substitution hints:   ph( x)   ps( y)

Proof of Theorem nfcdeq

Step	Hyp	Ref	Expression
1		nfcdeq.1	. . 3 |- F/ x ph
2	1	sbf 2380	. 2 |- ( [ y / x ] ph <-> ph )
3		nfv 1843	. . 3 |- F/ x ps
4		nfcdeq.2	. . . 4 |- CondEq ( x = y -> ( ph <-> ps ) )
5	4	cdeqri 3421	. . 3 |- ( x = y -> ( ph <-> ps ) )
6	3, 5	sbie 2408	. 2 |- ( [ y / x ] ph <-> ps )
7	2, 6	bitr3i 266	1 |- ( ph <-> ps )

Syntax hints:   <-> wb 196   F/wnf 1708   [wsb 1880  CondEqwcdeq 3418

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-10 2019  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  df-cdeq 3419

This theorem is referenced by:  nfccdeq  3433