Metamath Proof Explorer


Theorem nfreud

Description: Deduction version of nfreu . Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by NM, 15-Feb-2013) (Revised by Mario Carneiro, 8-Oct-2016) (New usage is discouraged.)

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

Proof

Step Hyp Ref Expression
1 nfreud.1
 |-  F/ y ph
2 nfreud.2
 |-  ( ph -> F/_ x A )
3 nfreud.3
 |-  ( ph -> F/ x ps )
4 df-reu
 |-  ( E! y e. A ps <-> E! y ( y e. A /\ ps ) )
5 nfcvf
 |-  ( -. A. x x = y -> F/_ x y )
6 5 adantl
 |-  ( ( ph /\ -. A. x x = y ) -> F/_ x y )
7 2 adantr
 |-  ( ( ph /\ -. A. x x = y ) -> F/_ x A )
8 6 7 nfeld
 |-  ( ( ph /\ -. A. x x = y ) -> F/ x y e. A )
9 3 adantr
 |-  ( ( ph /\ -. A. x x = y ) -> F/ x ps )
10 8 9 nfand
 |-  ( ( ph /\ -. A. x x = y ) -> F/ x ( y e. A /\ ps ) )
11 1 10 nfeud2
 |-  ( ph -> F/ x E! y ( y e. A /\ ps ) )
12 4 11 nfxfrd
 |-  ( ph -> F/ x E! y e. A ps )