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 ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | sbequ | ⊢ ( 𝑦 = 𝑧 → ( [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝑧 / 𝑥 ] 𝜑 ) ) | |
| 2 | 1 | pm5.74i | ⊢ ( ( 𝑦 = 𝑧 → [ 𝑦 / 𝑥 ] 𝜑 ) ↔ ( 𝑦 = 𝑧 → [ 𝑧 / 𝑥 ] 𝜑 ) ) |
| 3 | 2 | albii | ⊢ ( ∀ 𝑦 ( 𝑦 = 𝑧 → [ 𝑦 / 𝑥 ] 𝜑 ) ↔ ∀ 𝑦 ( 𝑦 = 𝑧 → [ 𝑧 / 𝑥 ] 𝜑 ) ) |
| 4 | sb6 | ⊢ ( [ 𝑧 / 𝑦 ] [ 𝑦 / 𝑥 ] 𝜑 ↔ ∀ 𝑦 ( 𝑦 = 𝑧 → [ 𝑦 / 𝑥 ] 𝜑 ) ) | |
| 5 | sb6 | ⊢ ( [ 𝑧 / 𝑦 ] [ 𝑧 / 𝑥 ] 𝜑 ↔ ∀ 𝑦 ( 𝑦 = 𝑧 → [ 𝑧 / 𝑥 ] 𝜑 ) ) | |
| 6 | 3 4 5 | 3bitr4i | ⊢ ( [ 𝑧 / 𝑦 ] [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝑧 / 𝑦 ] [ 𝑧 / 𝑥 ] 𝜑 ) |