Theorem sbelx 2458

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		sbid2v 2457	. 2 |- ( [ y / x ] [ x / y ] ph <-> ph )
2		sb5 2430	. 2 |- ( [ y / x ] [ x / y ] ph <-> E. x ( x = y /\ [ x / y ] ph ) )
3	1, 2	bitr3i 266	1 |- ( ph <-> E. x ( x = y /\ [ x / y ] ph ) )

Syntax hints:   <-> wb 196   /\ wa 384   E.wex 1704   [wsb 1880

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-10 2019  ax-12 2047  ax-13 2246

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-ex 1705  df-nf 1710  df-sb 1881

This theorem is referenced by:  pm13.196a  38615