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 y φ
Assertion mof * x φ y x φ x = y

Proof

Step Hyp Ref Expression
1 mof.1 y φ
2 df-mo * x φ z x φ x = z
3 nfv y x = z
4 1 3 nfim y φ x = z
5 4 nfal y x φ x = z
6 nfv z x φ x = y
7 equequ2 z = y x = z x = y
8 7 imbi2d z = y φ x = z φ x = y
9 8 albidv z = y x φ x = z x φ x = y
10 5 6 9 cbvexv1 z x φ x = z y x φ x = y
11 2 10 bitri * x φ y x φ x = y