Metamath Proof Explorer


Theorem seglerflx

Description: Segment comparison is reflexive. Theorem 5.7 of Schwabhauser p. 42. (Contributed by Scott Fenton, 11-Oct-2013)

Ref Expression
Assertion seglerflx ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N → A B Seg ≤ A B

Proof

Step Hyp Ref Expression
1 simp3 ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N → B ∈ 𝔼 ⁡ N
2 btwntriv2 ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N → B Btwn A B
3 cgrrflx ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N → A B Cgr A B
4 breq1 ⊢ y = B → y Btwn A B ↔ B Btwn A B
5 opeq2 ⊢ y = B → A y = A B
6 5 breq2d ⊢ y = B → A B Cgr A y ↔ A B Cgr A B
7 4 6 anbi12d ⊢ y = B → y Btwn A B ∧ A B Cgr A y ↔ B Btwn A B ∧ A B Cgr A B
8 7 rspcev ⊢ B ∈ 𝔼 ⁡ N ∧ B Btwn A B ∧ A B Cgr A B → ∃ y ∈ 𝔼 ⁡ N y Btwn A B ∧ A B Cgr A y
9 1 2 3 8 syl12anc ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N → ∃ y ∈ 𝔼 ⁡ N y Btwn A B ∧ A B Cgr A y
10 simp1 ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N → N ∈ ℕ
11 simp2 ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N → A ∈ 𝔼 ⁡ N
12 brsegle ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N → A B Seg ≤ A B ↔ ∃ y ∈ 𝔼 ⁡ N y Btwn A B ∧ A B Cgr A y
13 10 11 1 11 1 12 syl122anc ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N → A B Seg ≤ A B ↔ ∃ y ∈ 𝔼 ⁡ N y Btwn A B ∧ A B Cgr A y
14 9 13 mpbird ⊢ N ∈ ℕ ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N → A B Seg ≤ A B