Theorem nfeqf 2014

Description: A variable is effectively not free in an equality if it is not either of the involved variables. F/ version of ax-12o 2192. (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

Step	Hyp	Ref	Expression
1		nfnae 2008	. . 3 |- F/ z -. A. z z = x
2		nfnae 2008	. . 3 |- F/ z -. A. z z = y
3	1, 2	nfan 1842	. 2 |- F/ z ( -. A. z z = x /\ -. A. z z = y )
4		ax12o 1976	. . 3 |- ( -. A. z z = x -> ( -. A. z z = y -> ( x = y -> A. z x = y ) ) )
5	4	imp 419	. 2 |- ( ( -. A. z z = x /\ -. A. z z = y ) -> ( x = y -> A. z x = y ) )
6	3, 5	nfd 1778	1 |- ( ( -. A. z z = x /\ -. A. z z = y ) -> F/ z x = y )

Syntax hints:   -. wn 3   -> wi 4   /\ wa 359   A.wal 1546   F/wnf 1550

This theorem is referenced by:  dvelimh  2015  equvini  2040  equviniOLD  2041  equveli  2042  equveliOLD  2043  nfsb4t  2129  sbcom  2138  nfeud2  2266

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1552  ax-5 1563  ax-17 1623  ax-9 1662  ax-8 1683  ax-6 1740  ax-7 1745  ax-11 1757  ax-12 1946

This theorem depends on definitions:  df-bi 178  df-an 361  df-tru 1325  df-ex 1548  df-nf 1551