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Theorem sbel2x 2202
Description: Elimination of double substitution. (Contributed by NM, 5-Aug-1993.)
Assertion
Ref Expression
sbel2x  |-  ( ph  <->  E. x E. y ( ( x  =  z  /\  y  =  w )  /\  [ y  /  w ] [
x  /  z ]
ph ) )
Distinct variable groups:    x, y,
z    y, w    ph, x, y
Allowed substitution hints:    ph( z, w)

Proof of Theorem sbel2x
StepHypRef Expression
1 sbelx 2201 . . . . 5  |-  ( [ x  /  z ]
ph 
<->  E. y ( y  =  w  /\  [
y  /  w ] [ x  /  z ] ph ) )
21anbi2i 676 . . . 4  |-  ( ( x  =  z  /\  [ x  /  z ]
ph )  <->  ( x  =  z  /\  E. y
( y  =  w  /\  [ y  /  w ] [ x  / 
z ] ph )
) )
32exbii 1592 . . 3  |-  ( E. x ( x  =  z  /\  [ x  /  z ] ph ) 
<->  E. x ( x  =  z  /\  E. y ( y  =  w  /\  [ y  /  w ] [
x  /  z ]
ph ) ) )
4 sbelx 2201 . . 3  |-  ( ph  <->  E. x ( x  =  z  /\  [ x  /  z ] ph ) )
5 exdistr 1929 . . 3  |-  ( E. x E. y ( x  =  z  /\  ( y  =  w  /\  [ y  /  w ] [ x  / 
z ] ph )
)  <->  E. x ( x  =  z  /\  E. y ( y  =  w  /\  [ y  /  w ] [
x  /  z ]
ph ) ) )
63, 4, 53bitr4i 269 . 2  |-  ( ph  <->  E. x E. y ( x  =  z  /\  ( y  =  w  /\  [ y  /  w ] [ x  / 
z ] ph )
) )
7 anass 631 . . 3  |-  ( ( ( x  =  z  /\  y  =  w )  /\  [ y  /  w ] [
x  /  z ]
ph )  <->  ( x  =  z  /\  (
y  =  w  /\  [ y  /  w ] [ x  /  z ] ph ) ) )
872exbii 1593 . 2  |-  ( E. x E. y ( ( x  =  z  /\  y  =  w )  /\  [ y  /  w ] [
x  /  z ]
ph )  <->  E. x E. y ( x  =  z  /\  ( y  =  w  /\  [
y  /  w ] [ x  /  z ] ph ) ) )
96, 8bitr4i 244 1  |-  ( ph  <->  E. x E. y ( ( x  =  z  /\  y  =  w )  /\  [ y  /  w ] [
x  /  z ]
ph ) )
Colors of variables: wff set class
Syntax hints:    <-> wb 177    /\ wa 359   E.wex 1550   [wsb 1658
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1555  ax-5 1566  ax-17 1626  ax-9 1666  ax-8 1687  ax-6 1744  ax-7 1749  ax-11 1761  ax-12 1950
This theorem depends on definitions:  df-bi 178  df-or 360  df-an 361  df-tru 1328  df-ex 1551  df-nf 1554  df-sb 1659
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