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Theorem 2sb5 2274
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 2261 . 2  |-  ( [ z  /  x ] [ w  /  y ] ph  <->  E. x ( x  =  z  /\  [
w  /  y ]
ph ) )
2 19.42v 1836 . . . 4  |-  ( E. y ( x  =  z  /\  ( y  =  w  /\  ph ) )  <->  ( x  =  z  /\  E. y
( y  =  w  /\  ph ) ) )
3 anass 655 . . . . 5  |-  ( ( ( x  =  z  /\  y  =  w )  /\  ph )  <->  ( x  =  z  /\  ( y  =  w  /\  ph ) ) )
43exbii 1720 . . . 4  |-  ( E. y ( ( x  =  z  /\  y  =  w )  /\  ph ) 
<->  E. y ( x  =  z  /\  (
y  =  w  /\  ph ) ) )
5 sb5 2261 . . . . 5  |-  ( [ w  /  y ]
ph 
<->  E. y ( y  =  w  /\  ph ) )
65anbi2i 701 . . . 4  |-  ( ( x  =  z  /\  [ w  /  y ]
ph )  <->  ( x  =  z  /\  E. y
( y  =  w  /\  ph ) ) )
72, 4, 63bitr4ri 282 . . 3  |-  ( ( x  =  z  /\  [ w  /  y ]
ph )  <->  E. y
( ( x  =  z  /\  y  =  w )  /\  ph ) )
87exbii 1720 . 2  |-  ( E. x ( x  =  z  /\  [ w  /  y ] ph ) 
<->  E. x E. y
( ( x  =  z  /\  y  =  w )  /\  ph ) )
91, 8bitri 253 1  |-  ( [ z  /  x ] [ w  /  y ] ph  <->  E. x E. y
( ( x  =  z  /\  y  =  w )  /\  ph ) )
Colors of variables: wff setvar class
Syntax hints:    <-> wb 188    /\ wa 371   E.wex 1665   [wsb 1799
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1671  ax-4 1684  ax-5 1760  ax-6 1807  ax-7 1853  ax-10 1917  ax-12 1935  ax-13 2093
This theorem depends on definitions:  df-bi 189  df-an 373  df-ex 1666  df-nf 1670  df-sb 1800
This theorem is referenced by:  opelopabsbALT  4713
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