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Theorem 2sb6 1902
Description: Equivalence for double substitution. (Contributed by NM, 3-Feb-2005.)
Assertion
Ref Expression
2sb6  |-  ( [ z  /  x ] [ w  /  y ] ph  <->  A. x A. 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 2sb6
StepHypRef Expression
1 sb6 1808 . 2  |-  ( [ z  /  x ] [ w  /  y ] ph  <->  A. x ( x  =  z  ->  [ w  /  y ] ph ) )
2 19.21v 1795 . . . 4  |-  ( A. y ( x  =  z  ->  ( y  =  w  ->  ph )
)  <->  ( x  =  z  ->  A. y
( y  =  w  ->  ph ) ) )
3 impexp 259 . . . . 5  |-  ( ( ( x  =  z  /\  y  =  w )  ->  ph )  <->  ( x  =  z  ->  ( y  =  w  ->  ph )
) )
43albii 1400 . . . 4  |-  ( A. y ( ( x  =  z  /\  y  =  w )  ->  ph )  <->  A. y ( x  =  z  ->  ( y  =  w  ->  ph )
) )
5 sb6 1808 . . . . 5  |-  ( [ w  /  y ]
ph 
<-> 
A. y ( y  =  w  ->  ph )
)
65imbi2i 224 . . . 4  |-  ( ( x  =  z  ->  [ w  /  y ] ph )  <->  ( x  =  z  ->  A. y
( y  =  w  ->  ph ) ) )
72, 4, 63bitr4ri 211 . . 3  |-  ( ( x  =  z  ->  [ w  /  y ] ph )  <->  A. y
( ( x  =  z  /\  y  =  w )  ->  ph )
)
87albii 1400 . 2  |-  ( A. x ( x  =  z  ->  [ w  /  y ] ph ) 
<-> 
A. x A. y
( ( x  =  z  /\  y  =  w )  ->  ph )
)
91, 8bitri 182 1  |-  ( [ z  /  x ] [ w  /  y ] ph  <->  A. x A. y
( ( x  =  z  /\  y  =  w )  ->  ph )
)
Colors of variables: wff set class
Syntax hints:    -> wi 4    /\ wa 102    <-> wb 103   A.wal 1283   [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  ax-i5r 1469
This theorem depends on definitions:  df-bi 115  df-sb 1687
This theorem is referenced by: (None)
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