Metamath Proof Explorer


Theorem sbcom3vv

Description: Substituting y for x and then z for y is equivalent to substituting z for both x and y . Version of sbcom3 with a disjoint variable condition using fewer axioms. (Contributed by NM, 27-May-1997) (Revised by Giovanni Mascellani, 8-Apr-2018) (Revised by BJ, 30-Dec-2020) (Proof shortened by Wolf Lammen, 19-Jan-2023)

Ref Expression
Assertion sbcom3vv z y y x φ z y z x φ

Proof

Step Hyp Ref Expression
1 sbequ y = z y x φ z x φ
2 1 sbbiiev z y y x φ z y z x φ