Metamath Proof Explorer


Theorem mof

Description: Version of df-mo with disjoint variable condition replaced by nonfreeness hypothesis. (Contributed by NM, 8-Mar-1995) Extract dfmo from this proof, and prove mof from it (as of 30-Sep-2022, directly from df-mo ). (Revised by Wolf Lammen, 28-May-2019) Avoid ax-13 . (Revised by Wolf Lammen, 16-Oct-2022)

Ref Expression
Hypothesis mof.1 yφ
Assertion mof *xφyxφx=y

Proof

Step Hyp Ref Expression
1 mof.1 yφ
2 df-mo *xφzxφx=z
3 nfv yx=z
4 1 3 nfim yφx=z
5 4 nfal yxφx=z
6 nfv zxφx=y
7 equequ2 z=yx=zx=y
8 7 imbi2d z=yφx=zφx=y
9 8 albidv z=yxφx=zxφx=y
10 5 6 9 cbvexv1 zxφx=zyxφx=y
11 2 10 bitri *xφyxφx=y