Theorem abbib 2356

Description: Equal class abstractions require equivalent formulas, and conversely. (Contributed by NM, 25-Nov-2013.) (Revised by Mario Carneiro, 11-Aug-2016.) Remove dependency on ax-8 1557 and df-clel 2234 (by avoiding use of cleqh 2338). (Revised by BJ, 23-Jun-2019.) Definitial form. (Revised by Wolf Lammen, 23-Feb-2025.)

Assertion

Ref	Expression
abbib	|- ( { x | ph } = { x | ps } <-> A. x ( ph <-> ps ) )

Proof of Theorem abbib

Dummy variable y is distinct from all other variables.

Step	Hyp	Ref	Expression
1		dfcleq 2232	. 2 |- ( { x | ph } = { x | ps } <-> A. y ( y e. { x | ph } <-> y e. { x | ps } ) )
2		nfsab1 2228	. . . 4 |- F/ x y e. { x | ph }
3		nfsab1 2228	. . . 4 |- F/ x y e. { x | ps }
4	2, 3	nfbi 1642	. . 3 |- F/ x ( y e. { x | ph } <-> y e. { x | ps } )
5		nfv 1581	. . 3 |- F/ y ( ph <-> ps )
6		df-clab 2225	. . . . 5 |- ( y e. { x | ph } <-> [ y / x ] ph )
7		sbequ12r 1825	. . . . 5 |- ( y = x -> ( [ y / x ] ph <-> ph ) )
8	6, 7	bitrid 192	. . . 4 |- ( y = x -> ( y e. { x | ph } <-> ph ) )
9		df-clab 2225	. . . . 5 |- ( y e. { x | ps } <-> [ y / x ] ps )
10		sbequ12r 1825	. . . . 5 |- ( y = x -> ( [ y / x ] ps <-> ps ) )
11	9, 10	bitrid 192	. . . 4 |- ( y = x -> ( y e. { x | ps } <-> ps ) )
12	8, 11	bibi12d 235	. . 3 |- ( y = x -> ( ( y e. { x | ph } <-> y e. { x | ps } ) <-> ( ph <-> ps ) ) )
13	4, 5, 12	cbvalv1 1804	. 2 |- ( A. y ( y e. { x | ph } <-> y e. { x | ps } ) <-> A. x ( ph <-> ps ) )
14	1, 13	bitri 184	1 |- ( { x | ph } = { x | ps } <-> A. x ( ph <-> ps ) )

Syntax hints:   <-> wb 105   A.wal 1400   = wceq 1402   [wsb 1815   e. wcel 2209   {cab 2224

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-11 1559  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This theorem depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231

This theorem is referenced by:  eqabb  2374