Metamath Proof Explorer


Theorem sstr2

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

Proof

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