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Theorem sb6a 2068
Description: Equivalence for substitution. (Contributed by NM, 5-Aug-1993.)
Assertion
Ref Expression
sb6a  |-  ( [ y  /  x ] ph 
<-> 
A. x ( x  =  y  ->  [ x  /  y ] ph ) )
Distinct variable group:    x, y
Allowed substitution hints:    ph( x, y)

Proof of Theorem sb6a
StepHypRef Expression
1 sb6 2051 . 2  |-  ( [ y  /  x ] ph 
<-> 
A. x ( x  =  y  ->  ph )
)
2 sbequ12 1872 . . . . 5  |-  ( y  =  x  ->  ( ph 
<->  [ x  /  y ] ph ) )
32equcoms 1666 . . . 4  |-  ( x  =  y  ->  ( ph 
<->  [ x  /  y ] ph ) )
43pm5.74i 236 . . 3  |-  ( ( x  =  y  ->  ph )  <->  ( x  =  y  ->  [ x  /  y ] ph ) )
54albii 1556 . 2  |-  ( A. x ( x  =  y  ->  ph )  <->  A. x
( x  =  y  ->  [ x  / 
y ] ph )
)
61, 5bitri 240 1  |-  ( [ y  /  x ] ph 
<-> 
A. x ( x  =  y  ->  [ x  /  y ] ph ) )
Colors of variables: wff set class
Syntax hints:    -> wi 4    <-> wb 176   A.wal 1530   [wsb 1638
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1536  ax-5 1547  ax-17 1606  ax-9 1644  ax-8 1661  ax-6 1715  ax-7 1720  ax-11 1727  ax-12 1878
This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1310  df-ex 1532  df-nf 1535  df-sb 1639
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