Theorem sbelx 1985

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 1514	. 2 |- ( ph -> A. x ph )
2	1	sb5rf 1840	1 |- ( ph <-> E. x ( x = y /\ [ x / y ] ph ) )

Syntax hints:   /\ wa 103   <-> wb 104   E.wex 1480   [wsb 1750

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 1435  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-11 1494  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522

This theorem depends on definitions:  df-bi 116  df-sb 1751

This theorem is referenced by:  sbel2x  1986