Metamath Proof Explorer


Theorem ertr

Description: An equivalence relation is transitive. (Contributed by NM, 4-Jun-1995) (Revised by Mario Carneiro, 12-Aug-2015)

Ref Expression
Hypothesis ersymb.1 ⊢ φ → R Er X
Assertion ertr ⊢ φ → A R B ∧ B R C → A R C

Proof

Step Hyp Ref Expression
1 ersymb.1 ⊢ φ → R Er X
2 errel ⊢ R Er X → Rel ⁡ R
3 1 2 syl ⊢ φ → Rel ⁡ R
4 simpr ⊢ A R B ∧ B R C → B R C
5 brrelex1 ⊢ Rel ⁡ R ∧ B R C → B ∈ V
6 3 4 5 syl2an ⊢ φ ∧ A R B ∧ B R C → B ∈ V
7 simpr ⊢ φ ∧ A R B ∧ B R C → A R B ∧ B R C
8 breq2 ⊢ x = B → A R x ↔ A R B
9 breq1 ⊢ x = B → x R C ↔ B R C
10 8 9 anbi12d ⊢ x = B → A R x ∧ x R C ↔ A R B ∧ B R C
11 6 7 10 spcedv ⊢ φ ∧ A R B ∧ B R C → ∃ x A R x ∧ x R C
12 simpl ⊢ A R B ∧ B R C → A R B
13 brrelex1 ⊢ Rel ⁡ R ∧ A R B → A ∈ V
14 3 12 13 syl2an ⊢ φ ∧ A R B ∧ B R C → A ∈ V
15 brrelex2 ⊢ Rel ⁡ R ∧ B R C → C ∈ V
16 3 4 15 syl2an ⊢ φ ∧ A R B ∧ B R C → C ∈ V
17 brcog ⊢ A ∈ V ∧ C ∈ V → A R ∘ R C ↔ ∃ x A R x ∧ x R C
18 14 16 17 syl2anc ⊢ φ ∧ A R B ∧ B R C → A R ∘ R C ↔ ∃ x A R x ∧ x R C
19 11 18 mpbird ⊢ φ ∧ A R B ∧ B R C → A R ∘ R C
20 19 ex ⊢ φ → A R B ∧ B R C → A R ∘ R C
21 df-er ⊢ R Er X ↔ Rel ⁡ R ∧ dom ⁡ R = X ∧ R -1 ∪ R ∘ R ⊆ R
22 21 simp3bi ⊢ R Er X → R -1 ∪ R ∘ R ⊆ R
23 1 22 syl ⊢ φ → R -1 ∪ R ∘ R ⊆ R
24 23 unssbd ⊢ φ → R ∘ R ⊆ R
25 24 ssbrd ⊢ φ → A R ∘ R C → A R C
26 20 25 syld ⊢ φ → A R B ∧ B R C → A R C