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 ) )