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

Theorem sb6x 2123
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 2119 . 2  |-  ( [ y  /  x ] ph 
<-> 
ph )
3 biidd 230 . . 3  |-  ( x  =  y  ->  ( ph 
<-> 
ph ) )
41, 3equsal 2000 . 2  |-  ( A. x ( x  =  y  ->  ph )  <->  ph )
52, 4bitr4i 245 1  |-  ( [ y  /  x ] ph 
<-> 
A. x ( x  =  y  ->  ph )
)
Colors of variables: wff set class
Syntax hints:    -> wi 4    <-> wb 178   A.wal 1550   F/wnf 1554   [wsb 1659
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1556  ax-5 1567  ax-17 1627  ax-9 1667  ax-8 1688  ax-6 1745  ax-11 1762  ax-12 1951
This theorem depends on definitions:  df-bi 179  df-an 362  df-ex 1552  df-nf 1555  df-sb 1660
  Copyright terms: Public domain W3C validator