Theorem eqab 2367

Description: One direction of eqabb 2368. (Contributed by Wolf Lammen, 13-Feb-2025.)

Assertion

Ref	Expression
eqab	|- ( A. x ( x e. A <-> ph ) -> A = { x | ph } )

Distinct variable group:   x, A

Allowed substitution hint:   ph( x)

Proof of Theorem eqab

Step	Hyp	Ref	Expression
1		abid1 2366	. 2 |- A = { x | x e. A }
2		abbi 2351	. 2 |- ( A. x ( x e. A <-> ph ) -> { x | x e. A } = { x | ph } )
3	1, 2	eqtrid 2277	1 |- ( A. x ( x e. A <-> ph ) -> A = { x | ph } )

Syntax hints:   -> wi 4   <-> wb 105   A.wal 1396   = wceq 1398   e. wcel 2203   {cab 2218

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-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2214

This theorem depends on definitions:  df-bi 117  df-tru 1401  df-nf 1510  df-sb 1812  df-clab 2219  df-cleq 2225  df-clel 2228

This theorem is referenced by: (None)