Theorem nfcdeq 2821

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 1702	. 2 |- ( [ y / x ] ph <-> ph )
3		nfv 1462	. . 3 |- F/ x ps
4		nfcdeq.2	. . . 4 |- CondEq ( x = y -> ( ph <-> ps ) )
5	4	cdeqri 2810	. . 3 |- ( x = y -> ( ph <-> ps ) )
6	3, 5	sbie 1716	. 2 |- ( [ y / x ] ph <-> ps )
7	2, 6	bitr3i 184	1 |- ( ph <-> ps )

Syntax hints:   <-> wb 103   F/wnf 1390   [wsb 1687  CondEqwcdeq 2807

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1377  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468

This theorem depends on definitions:  df-bi 115  df-nf 1391  df-sb 1688  df-cdeq 2808

This theorem is referenced by:  nfccdeq  2822