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Theorem sbel2x 1916
Description: Elimination of double substitution. (Contributed by NM, 5-Aug-1993.)
Assertion
Ref Expression
sbel2x  |-  ( ph  <->  E. x E. y ( ( x  =  z  /\  y  =  w )  /\  [ y  /  w ] [
x  /  z ]
ph ) )
Distinct variable groups:    x, y, z   
y, w    ph, x, y
Allowed substitution hints:    ph( z, w)

Proof of Theorem sbel2x
StepHypRef Expression
1 sbelx 1915 . . . . 5  |-  ( [ x  /  z ]
ph 
<->  E. y ( y  =  w  /\  [
y  /  w ] [ x  /  z ] ph ) )
21anbi2i 445 . . . 4  |-  ( ( x  =  z  /\  [ x  /  z ]
ph )  <->  ( x  =  z  /\  E. y
( y  =  w  /\  [ y  /  w ] [ x  / 
z ] ph )
) )
32exbii 1537 . . 3  |-  ( E. x ( x  =  z  /\  [ x  /  z ] ph ) 
<->  E. x ( x  =  z  /\  E. y ( y  =  w  /\  [ y  /  w ] [
x  /  z ]
ph ) ) )
4 sbelx 1915 . . 3  |-  ( ph  <->  E. x ( x  =  z  /\  [ x  /  z ] ph ) )
5 exdistr 1829 . . 3  |-  ( E. x E. y ( x  =  z  /\  ( y  =  w  /\  [ y  /  w ] [ x  / 
z ] ph )
)  <->  E. x ( x  =  z  /\  E. y ( y  =  w  /\  [ y  /  w ] [
x  /  z ]
ph ) ) )
63, 4, 53bitr4i 210 . 2  |-  ( ph  <->  E. x E. y ( x  =  z  /\  ( y  =  w  /\  [ y  /  w ] [ x  / 
z ] ph )
) )
7 anass 393 . . 3  |-  ( ( ( x  =  z  /\  y  =  w )  /\  [ y  /  w ] [
x  /  z ]
ph )  <->  ( x  =  z  /\  (
y  =  w  /\  [ y  /  w ] [ x  /  z ] ph ) ) )
872exbii 1538 . 2  |-  ( E. x E. y ( ( x  =  z  /\  y  =  w )  /\  [ y  /  w ] [
x  /  z ]
ph )  <->  E. x E. y ( x  =  z  /\  ( y  =  w  /\  [
y  /  w ] [ x  /  z ] ph ) ) )
96, 8bitr4i 185 1  |-  ( ph  <->  E. x E. y ( ( x  =  z  /\  y  =  w )  /\  [ y  /  w ] [
x  /  z ]
ph ) )
Colors of variables: wff set class
Syntax hints:    /\ wa 102    <-> wb 103   E.wex 1422   [wsb 1686
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1377  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-11 1438  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468
This theorem depends on definitions:  df-bi 115  df-sb 1687
This theorem is referenced by: (None)
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