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Theorem 2sb5 2064
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 2052 . 2  |-  ( [ z  /  x ] [ w  /  y ] ph  <->  E. x ( x  =  z  /\  [
w  /  y ]
ph ) )
2 19.42v 1858 . . . 4  |-  ( E. y ( x  =  z  /\  ( y  =  w  /\  ph ) )  <->  ( x  =  z  /\  E. y
( y  =  w  /\  ph ) ) )
3 anass 630 . . . . 5  |-  ( ( ( x  =  z  /\  y  =  w )  /\  ph )  <->  ( x  =  z  /\  ( y  =  w  /\  ph ) ) )
43exbii 1572 . . . 4  |-  ( E. y ( ( x  =  z  /\  y  =  w )  /\  ph ) 
<->  E. y ( x  =  z  /\  (
y  =  w  /\  ph ) ) )
5 sb5 2052 . . . . 5  |-  ( [ w  /  y ]
ph 
<->  E. y ( y  =  w  /\  ph ) )
65anbi2i 675 . . . 4  |-  ( ( x  =  z  /\  [ w  /  y ]
ph )  <->  ( x  =  z  /\  E. y
( y  =  w  /\  ph ) ) )
72, 4, 63bitr4ri 269 . . 3  |-  ( ( x  =  z  /\  [ w  /  y ]
ph )  <->  E. y
( ( x  =  z  /\  y  =  w )  /\  ph ) )
87exbii 1572 . 2  |-  ( E. x ( x  =  z  /\  [ w  /  y ] ph ) 
<->  E. x E. y
( ( x  =  z  /\  y  =  w )  /\  ph ) )
91, 8bitri 240 1  |-  ( [ z  /  x ] [ w  /  y ] ph  <->  E. x E. y
( ( x  =  z  /\  y  =  w )  /\  ph ) )
Colors of variables: wff set class
Syntax hints:    <-> wb 176    /\ wa 358   E.wex 1531   [wsb 1638
This theorem is referenced by:  pm11.07  2067  opelopabsbOLD  4289
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1536  ax-5 1547  ax-17 1606  ax-9 1644  ax-8 1661  ax-6 1715  ax-7 1720  ax-11 1727  ax-12 1878
This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1310  df-ex 1532  df-nf 1535  df-sb 1639
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