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

Theorem nfeqf 1898
Description: A variable is effectively not free in an equality if it is not either of the involved variables.  F/ version of ax-12o 2081. (Contributed by Mario Carneiro, 6-Oct-2016.)
Assertion
Ref Expression
nfeqf  |-  ( ( -.  A. z  z  =  x  /\  -.  A. z  z  =  y )  ->  F/ z  x  =  y )

Proof of Theorem nfeqf
StepHypRef Expression
1 nfnae 1896 . . 3  |-  F/ z  -.  A. z  z  =  x
2 nfnae 1896 . . 3  |-  F/ z  -.  A. z  z  =  y
31, 2nfan 1771 . 2  |-  F/ z ( -.  A. z 
z  =  x  /\  -.  A. z  z  =  y )
4 ax12o 1875 . . 3  |-  ( -. 
A. z  z  =  x  ->  ( -.  A. z  z  =  y  ->  ( x  =  y  ->  A. z  x  =  y )
) )
54imp 418 . 2  |-  ( ( -.  A. z  z  =  x  /\  -.  A. z  z  =  y )  ->  ( x  =  y  ->  A. z  x  =  y )
)
63, 5nfd 1746 1  |-  ( ( -.  A. z  z  =  x  /\  -.  A. z  z  =  y )  ->  F/ z  x  =  y )
Colors of variables: wff set class
Syntax hints:   -. wn 3    -> wi 4    /\ wa 358   A.wal 1527   F/wnf 1531
This theorem is referenced by:  equvini  1927  equveli  1928  nfsb4t  2020  sbcom  2029  nfeud2  2155
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-tru 1310  df-ex 1529  df-nf 1532
  Copyright terms: Public domain W3C validator