Theorem sbelx 1343

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 1342	. 2 |- ([y / x][x / y]ph <-> ph)
2		sb5 1267	. 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 955  E.wex 979  [wsbc 1169

This theorem is referenced by:  sbel2x 1344

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 961  ax-gen 962  ax-8 963  ax-10 965  ax-12 967  ax-17 970  ax-4 972  ax-5o 974  ax-6o 977  ax-9o 1122  ax-10o 1139  ax-16 1209  ax-11o 1217

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 980  df-sb 1171

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