Description: Transitivity of subclass relationship. Exercise 5 of TakeutiZaring p. 17. (Contributed by NM, 24-Jun-1993) (Proof shortened by Andrew Salmon, 14-Jun-2011) Avoid axioms. (Revised by GG, 19-May-2025)
| Ref | Expression | ||
|---|---|---|---|
| Assertion | sstr2 | ⊢ ( 𝐴 ⊆ 𝐵 → ( 𝐵 ⊆ 𝐶 → 𝐴 ⊆ 𝐶 ) ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | imim1 | ⊢ ( ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) → ( ( 𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐶 ) → ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶 ) ) ) | |
| 2 | 1 | al2imi | ⊢ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) → ( ∀ 𝑥 ( 𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐶 ) → ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶 ) ) ) |
| 3 | df-ss | ⊢ ( 𝐴 ⊆ 𝐵 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) ) | |
| 4 | df-ss | ⊢ ( 𝐵 ⊆ 𝐶 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐶 ) ) | |
| 5 | df-ss | ⊢ ( 𝐴 ⊆ 𝐶 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶 ) ) | |
| 6 | 4 5 | imbi12i | ⊢ ( ( 𝐵 ⊆ 𝐶 → 𝐴 ⊆ 𝐶 ) ↔ ( ∀ 𝑥 ( 𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐶 ) → ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶 ) ) ) |
| 7 | 2 3 6 | 3imtr4i | ⊢ ( 𝐴 ⊆ 𝐵 → ( 𝐵 ⊆ 𝐶 → 𝐴 ⊆ 𝐶 ) ) |