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Theorem 2sb5 2292
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 2279 . 2  |-  ( [ z  /  x ] [ w  /  y ] ph  <->  E. x ( x  =  z  /\  [
w  /  y ]
ph ) )
2 19.42v 1842 . . . 4  |-  ( E. y ( x  =  z  /\  ( y  =  w  /\  ph ) )  <->  ( x  =  z  /\  E. y
( y  =  w  /\  ph ) ) )
3 anass 661 . . . . 5  |-  ( ( ( x  =  z  /\  y  =  w )  /\  ph )  <->  ( x  =  z  /\  ( y  =  w  /\  ph ) ) )
43exbii 1726 . . . 4  |-  ( E. y ( ( x  =  z  /\  y  =  w )  /\  ph ) 
<->  E. y ( x  =  z  /\  (
y  =  w  /\  ph ) ) )
5 sb5 2279 . . . . 5  |-  ( [ w  /  y ]
ph 
<->  E. y ( y  =  w  /\  ph ) )
65anbi2i 708 . . . 4  |-  ( ( x  =  z  /\  [ w  /  y ]
ph )  <->  ( x  =  z  /\  E. y
( y  =  w  /\  ph ) ) )
72, 4, 63bitr4ri 286 . . 3  |-  ( ( x  =  z  /\  [ w  /  y ]
ph )  <->  E. y
( ( x  =  z  /\  y  =  w )  /\  ph ) )
87exbii 1726 . 2  |-  ( E. x ( x  =  z  /\  [ w  /  y ] ph ) 
<->  E. x E. y
( ( x  =  z  /\  y  =  w )  /\  ph ) )
91, 8bitri 257 1  |-  ( [ z  /  x ] [ w  /  y ] ph  <->  E. x E. y
( ( x  =  z  /\  y  =  w )  /\  ph ) )
Colors of variables: wff setvar class
Syntax hints:    <-> wb 189    /\ wa 376   E.wex 1671   [wsb 1805
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1677  ax-4 1690  ax-5 1766  ax-6 1813  ax-7 1859  ax-10 1932  ax-12 1950  ax-13 2104
This theorem depends on definitions:  df-bi 190  df-an 378  df-ex 1672  df-nf 1676  df-sb 1806
This theorem is referenced by:  opelopabsbALT  4710
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