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 ] ph <-> [ z / y ] [ z / x ] ph ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | sbequ | |- ( y = z -> ( [ y / x ] ph <-> [ z / x ] ph ) ) |
|
| 2 | 1 | pm5.74i | |- ( ( y = z -> [ y / x ] ph ) <-> ( y = z -> [ z / x ] ph ) ) |
| 3 | 2 | albii | |- ( A. y ( y = z -> [ y / x ] ph ) <-> A. y ( y = z -> [ z / x ] ph ) ) |
| 4 | sb6 | |- ( [ z / y ] [ y / x ] ph <-> A. y ( y = z -> [ y / x ] ph ) ) |
|
| 5 | sb6 | |- ( [ z / y ] [ z / x ] ph <-> A. y ( y = z -> [ z / x ] ph ) ) |
|
| 6 | 3 4 5 | 3bitr4i | |- ( [ z / y ] [ y / x ] ph <-> [ z / y ] [ z / x ] ph ) |