MPE Home Metamath Proof Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >  sb6x Unicode version

Theorem sb6x 1969
Description: Equivalence involving substitution for a variable not free. (Contributed by NM, 5-Aug-1993.) (Revised by Mario Carneiro, 4-Oct-2016.)
Hypothesis
Ref Expression
sb6x.1  |-  F/ x ph
Assertion
Ref Expression
sb6x  |-  ( [ y  /  x ] ph 
<-> 
A. x ( x  =  y  ->  ph )
)

Proof of Theorem sb6x
StepHypRef Expression
1 sb6x.1 . . 3  |-  F/ x ph
21sbf 1966 . 2  |-  ( [ y  /  x ] ph 
<-> 
ph )
3 biidd 228 . . 3  |-  ( x  =  y  ->  ( ph 
<-> 
ph ) )
41, 3equsal 1900 . 2  |-  ( A. x ( x  =  y  ->  ph )  <->  ph )
52, 4bitr4i 243 1  |-  ( [ y  /  x ] ph 
<-> 
A. x ( x  =  y  ->  ph )
)
Colors of variables: wff set class
Syntax hints:    -> wi 4    <-> wb 176   A.wal 1527   F/wnf 1531   [wsb 1629
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1533  ax-5 1544  ax-17 1603  ax-9 1635  ax-8 1643  ax-6 1703  ax-7 1708  ax-11 1715  ax-12 1866
This theorem depends on definitions:  df-bi 177  df-an 360  df-ex 1529  df-nf 1532  df-sb 1630
  Copyright terms: Public domain W3C validator