Theorem sbelx 2025

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 1549	. 2 |- ( ph -> A. x ph )
2	1	sb5rf 1875	1 |- ( ph <-> E. x ( x = y /\ [ x / y ] ph ) )

Syntax hints:   /\ wa 104   <-> wb 105   E.wex 1515   [wsb 1785

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 1470  ax-gen 1472  ax-ie1 1516  ax-ie2 1517  ax-8 1527  ax-11 1529  ax-4 1533  ax-17 1549  ax-i9 1553  ax-ial 1557

This theorem depends on definitions:  df-bi 117  df-sb 1786

This theorem is referenced by:  sbel2x  2026