Theorem sbelx 1344

Description: Elimination of substitution.

Assertion

Ref	Expression
sbelx	|- (ph <-> E.x(x = y /\ [x / y]ph))

Distinct variable groups:   x,y   ph,x

Proof of Theorem sbelx

Step	Hyp	Ref	Expression
1		sbid2v 1343	. 2 |- ([y / x][x / y]ph <-> ph)
2		sb5 1268	. 2 |- ([y / x][x / y]ph <-> E.x(x = y /\ [x / y]ph))
3	1, 2	bitr3 175	1 |- (ph <-> E.x(x = y /\ [x / y]ph))

Syntax hints:   <-> wb 146   /\ wa 223   = wceq 956  E.wex 980  [wsbc 1170

This theorem is referenced by:  sbel2x 1345

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 962  ax-gen 963  ax-8 964  ax-10 966  ax-12 968  ax-17 971  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  ax-10o 1140  ax-16 1210  ax-11o 1218

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 981  df-sb 1172

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