Metamath Proof Explorer


Theorem nfrexdg

Description: Deduction version of nfrexg . Usage of this theorem is discouraged because it depends on ax-13 . See nfrexd for a version with a disjoint variable condition, but not requiring ax-13 . (Contributed by Mario Carneiro, 14-Oct-2016) (New usage is discouraged.)

Ref Expression
Hypotheses nfrexdg.1
|- F/ y ph
nfrexdg.2
|- ( ph -> F/_ x A )
nfrexdg.3
|- ( ph -> F/ x ps )
Assertion nfrexdg
|- ( ph -> F/ x E. y e. A ps )

Proof

Step Hyp Ref Expression
1 nfrexdg.1
 |-  F/ y ph
2 nfrexdg.2
 |-  ( ph -> F/_ x A )
3 nfrexdg.3
 |-  ( ph -> F/ x ps )
4 dfrex2
 |-  ( E. y e. A ps <-> -. A. y e. A -. ps )
5 3 nfnd
 |-  ( ph -> F/ x -. ps )
6 1 2 5 nfrald
 |-  ( ph -> F/ x A. y e. A -. ps )
7 6 nfnd
 |-  ( ph -> F/ x -. A. y e. A -. ps )
8 4 7 nfxfrd
 |-  ( ph -> F/ x E. y e. A ps )