Theorem sbelx 1970

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 1506	. 2 |- ( ph -> A. x ph )
2	1	sb5rf 1824	1 |- ( ph <-> E. x ( x = y /\ [ x / y ] ph ) )

Syntax hints:   /\ wa 103   <-> wb 104   E.wex 1468   [wsb 1735

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1423  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-11 1484  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514

This theorem depends on definitions:  df-bi 116  df-sb 1736

This theorem is referenced by:  sbel2x  1971