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Theorem sbcom3 2154
Description: Substituting  y for  x and then  z for  y is equivalent to substituting  z for both  x and  y. (Contributed by Giovanni Mascellani, 8-Apr-2018.) Remove dependency on ax-11 1843. (Revised by Wolf Lammen, 16-Sep-2018.) (Proof shortened by Wolf Lammen, 16-Sep-2018.)
Assertion
Ref Expression
sbcom3  |-  ( [ z  /  y ] [ y  /  x ] ph  <->  [ z  /  y ] [ z  /  x ] ph )

Proof of Theorem sbcom3
StepHypRef Expression
1 nfa1 1898 . . 3  |-  F/ y A. y  y  =  z
2 drsb2 2120 . . 3  |-  ( A. y  y  =  z  ->  ( [ y  /  x ] ph  <->  [ z  /  x ] ph )
)
31, 2sbbid 2145 . 2  |-  ( A. y  y  =  z  ->  ( [ z  / 
y ] [ y  /  x ] ph  <->  [ z  /  y ] [ z  /  x ] ph ) )
4 sb4b 2099 . . . 4  |-  ( -. 
A. y  y  =  z  ->  ( [
z  /  y ] [ y  /  x ] ph  <->  A. y ( y  =  z  ->  [ y  /  x ] ph ) ) )
5 sbequ 2118 . . . . . 6  |-  ( y  =  z  ->  ( [ y  /  x ] ph  <->  [ z  /  x ] ph ) )
65pm5.74i 245 . . . . 5  |-  ( ( y  =  z  ->  [ y  /  x ] ph )  <->  ( y  =  z  ->  [ z  /  x ] ph ) )
76albii 1641 . . . 4  |-  ( A. y ( y  =  z  ->  [ y  /  x ] ph )  <->  A. y ( y  =  z  ->  [ z  /  x ] ph )
)
84, 7syl6bb 261 . . 3  |-  ( -. 
A. y  y  =  z  ->  ( [
z  /  y ] [ y  /  x ] ph  <->  A. y ( y  =  z  ->  [ z  /  x ] ph ) ) )
9 sb4b 2099 . . 3  |-  ( -. 
A. y  y  =  z  ->  ( [
z  /  y ] [ z  /  x ] ph  <->  A. y ( y  =  z  ->  [ z  /  x ] ph ) ) )
108, 9bitr4d 256 . 2  |-  ( -. 
A. y  y  =  z  ->  ( [
z  /  y ] [ y  /  x ] ph  <->  [ z  /  y ] [ z  /  x ] ph ) )
113, 10pm2.61i 164 1  |-  ( [ z  /  y ] [ y  /  x ] ph  <->  [ z  /  y ] [ z  /  x ] ph )
Colors of variables: wff setvar class
Syntax hints:   -. wn 3    -> wi 4    <-> wb 184   A.wal 1393   [wsb 1740
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1619  ax-4 1632  ax-5 1705  ax-6 1748  ax-7 1791  ax-10 1838  ax-12 1855  ax-13 2000
This theorem depends on definitions:  df-bi 185  df-or 370  df-an 371  df-ex 1614  df-nf 1618  df-sb 1741
This theorem is referenced by:  sbco  2155  sbidm  2157  sbcom  2162  equsb3  2177  wl-equsb3  30209
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