Metamath Proof Explorer


Theorem nfexd

Description: If x is not free in ps , then it is not free in E. y ps . (Contributed by Mario Carneiro, 24-Sep-2016)

Ref Expression
Hypotheses nfald.1
|- F/ y ph
nfald.2
|- ( ph -> F/ x ps )
Assertion nfexd
|- ( ph -> F/ x E. y ps )

Proof

Step Hyp Ref Expression
1 nfald.1
 |-  F/ y ph
2 nfald.2
 |-  ( ph -> F/ x ps )
3 df-ex
 |-  ( E. y ps <-> -. A. y -. ps )
4 2 nfnd
 |-  ( ph -> F/ x -. ps )
5 1 4 nfald
 |-  ( ph -> F/ x A. y -. ps )
6 5 nfnd
 |-  ( ph -> F/ x -. A. y -. ps )
7 3 6 nfxfrd
 |-  ( ph -> F/ x E. y ps )