Theorem sbelx 2124

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 2123	. 2 |- ([y / x][x / y]ph <-> ph)
2		sb5 2100	. 2 |- ([y / x][x / y]ph <-> E.x(x = y /\ [x / y]ph))
3	1, 2	bitr3i 242	1 |- (ph <-> E.x(x = y /\ [x / y]ph))

This proof depends on syntax axioms:   <-> wb 176   /\ wa 358  E.wex 1541  [wsb 1648

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925

This proof depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649

This theorem is used by:  sbel2x  2125