Metamath Proof Explorer

Theorem outsidene1

Description: Outsideness implies inequality. (Contributed by Scott Fenton, 18-Oct-2013) (Revised by Mario Carneiro, 19-Apr-2014)

Ref Expression
Assertion outsidene1 ⊢ N ∈ ℕ ∧ P ∈ 𝔼 ⁡ N ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N → P OutsideOf A B → A ≠ P

Proof

Step Hyp Ref Expression
1 broutsideof2 ⊢ N ∈ ℕ ∧ P ∈ 𝔼 ⁡ N ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N → P OutsideOf A B ↔ A ≠ P ∧ B ≠ P ∧ A Btwn P B ∨ B Btwn P A
2 simp1 ⊢ A ≠ P ∧ B ≠ P ∧ A Btwn P B ∨ B Btwn P A → A ≠ P
3 1 2 syl6bi ⊢ N ∈ ℕ ∧ P ∈ 𝔼 ⁡ N ∧ A ∈ 𝔼 ⁡ N ∧ B ∈ 𝔼 ⁡ N → P OutsideOf A B → A ≠ P