Theorem sbelx 1914

Description: Elimination of substitution. (Contributed by NM, 5-Aug-1993.)

Assertion

Ref	Expression
sbelx	|- ( ph <-> E. x ( x = y /\ [ x / y ] ph ) )

Distinct variable groups:   x, y   ph, x

Allowed substitution hint:   ph( y)

Proof of Theorem sbelx

Step	Hyp	Ref	Expression
1		ax-17 1459	. 2 |- ( ph -> A. x ph )
2	1	sb5rf 1773	1 |- ( ph <-> E. x ( x = y /\ [ x / y ] ph ) )

Syntax hints:   /\ wa 102   <-> wb 103   E.wex 1421   [wsb 1685

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 1376  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-11 1437  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467

This theorem depends on definitions:  df-bi 115  df-sb 1686

This theorem is referenced by:  sbel2x  1915