Metamath Proof Explorer


Theorem nfraldw

Description: Deduction version of nfralw . Version of nfrald with a disjoint variable condition, which does not require ax-13 . (Contributed by NM, 15-Feb-2013) (Revised by Gino Giotto, 24-Sep-2024)

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

Proof

Step Hyp Ref Expression
1 nfraldw.1
 |-  F/ y ph
2 nfraldw.2
 |-  ( ph -> F/_ x A )
3 nfraldw.3
 |-  ( ph -> F/ x ps )
4 df-ral
 |-  ( A. y e. A ps <-> A. y ( y e. A -> ps ) )
5 2 nfcrd
 |-  ( ph -> F/ x y e. A )
6 5 3 nfimd
 |-  ( ph -> F/ x ( y e. A -> ps ) )
7 1 6 nfald
 |-  ( ph -> F/ x A. y ( y e. A -> ps ) )
8 4 7 nfxfrd
 |-  ( ph -> F/ x A. y e. A ps )