Theorem sbelx 994

Description: Elimination of substitution.

Assertion

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

Distinct variable group(s):   x,y   ph,x

Proof of Theorem sbelx

Step	Hyp	Ref	Expression
1		sbid2v 993	. 2 |- ([y / x][x / y]ph <-> ph)
2		sb5 988	. 2 |- ([y / x][x / y]ph <-> E.x(x = y /\ [x / y]ph))
3	1, 2	bitr3 153	1 |- (ph <-> E.x(x = y /\ [x / y]ph))

Syntax hints:   <-> wb 127   /\ wa 196  E.wex 678   = weq 797  [wsb 852

This theorem is referenced by:  sbel2x 995

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853

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