Theorem sbelx 2009

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 1537	. 2 |- ( ph -> A. x ph )
2	1	sb5rf 1863	1 |- ( ph <-> E. x ( x = y /\ [ x / y ] ph ) )

Syntax hints:   /\ wa 104   <-> wb 105   E.wex 1503   [wsb 1773

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 1458  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-11 1517  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545

This theorem depends on definitions:  df-bi 117  df-sb 1774

This theorem is referenced by:  sbel2x  2010