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Theorem 2sb5 1959
Description: Equivalence for double substitution. (Contributed by NM, 3-Feb-2005.)
Assertion
Ref Expression
2sb5  |-  ( [ z  /  x ] [ w  /  y ] ph  <->  E. x E. y
( ( x  =  z  /\  y  =  w )  /\  ph ) )
Distinct variable groups:    x, y, z   
y, w
Allowed substitution hints:    ph( x, y, z, w)

Proof of Theorem 2sb5
StepHypRef Expression
1 sb5 1860 . 2  |-  ( [ z  /  x ] [ w  /  y ] ph  <->  E. x ( x  =  z  /\  [
w  /  y ]
ph ) )
2 19.42v 1879 . . . 4  |-  ( E. y ( x  =  z  /\  ( y  =  w  /\  ph ) )  <->  ( x  =  z  /\  E. y
( y  =  w  /\  ph ) ) )
3 anass 399 . . . . 5  |-  ( ( ( x  =  z  /\  y  =  w )  /\  ph )  <->  ( x  =  z  /\  ( y  =  w  /\  ph ) ) )
43exbii 1585 . . . 4  |-  ( E. y ( ( x  =  z  /\  y  =  w )  /\  ph ) 
<->  E. y ( x  =  z  /\  (
y  =  w  /\  ph ) ) )
5 sb5 1860 . . . . 5  |-  ( [ w  /  y ]
ph 
<->  E. y ( y  =  w  /\  ph ) )
65anbi2i 453 . . . 4  |-  ( ( x  =  z  /\  [ w  /  y ]
ph )  <->  ( x  =  z  /\  E. y
( y  =  w  /\  ph ) ) )
72, 4, 63bitr4ri 212 . . 3  |-  ( ( x  =  z  /\  [ w  /  y ]
ph )  <->  E. y
( ( x  =  z  /\  y  =  w )  /\  ph ) )
87exbii 1585 . 2  |-  ( E. x ( x  =  z  /\  [ w  /  y ] ph ) 
<->  E. x E. y
( ( x  =  z  /\  y  =  w )  /\  ph ) )
91, 8bitri 183 1  |-  ( [ z  /  x ] [ w  /  y ] ph  <->  E. x E. y
( ( x  =  z  /\  y  =  w )  /\  ph ) )
Colors of variables: wff set class
Syntax hints:    /\ wa 103    <-> wb 104   E.wex 1469   [wsb 1736
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1424  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-11 1485  ax-4 1488  ax-17 1507  ax-i9 1511  ax-ial 1515
This theorem depends on definitions:  df-bi 116  df-sb 1737
This theorem is referenced by:  opelopabsbALT  4185
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