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Metamath Proof Explorer

Theorem avglt1

Description: Ordering property for average. (Contributed by Mario Carneiro, 28-May-2014)

Ref Expression
Assertion avglt1 ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( 𝐴 < 𝐵 ↔ 𝐴 < ( ( 𝐴 + 𝐵 ) / 2 ) ) )

Proof

Step Hyp Ref Expression
1 ltadd2 ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐴 ∈ ℝ ) → ( 𝐴 < 𝐵 ↔ ( 𝐴 + 𝐴 ) < ( 𝐴 + 𝐵 ) ) )
2 1 3anidm13 ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( 𝐴 < 𝐵 ↔ ( 𝐴 + 𝐴 ) < ( 𝐴 + 𝐵 ) ) )
3 simpl ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → 𝐴 ∈ ℝ )
4 3 recnd ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → 𝐴 ∈ ℂ )
5 times2 ⊢ ( 𝐴 ∈ ℂ → ( 𝐴 · 2 ) = ( 𝐴 + 𝐴 ) )
6 4 5 syl ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( 𝐴 · 2 ) = ( 𝐴 + 𝐴 ) )
7 6 breq1d ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( ( 𝐴 · 2 ) < ( 𝐴 + 𝐵 ) ↔ ( 𝐴 + 𝐴 ) < ( 𝐴 + 𝐵 ) ) )
8 readdcl ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( 𝐴 + 𝐵 ) ∈ ℝ )
9 2re ⊢ 2 ∈ ℝ
10 2pos ⊢ 0 < 2
11 9 10 pm3.2i ⊢ ( 2 ∈ ℝ ∧ 0 < 2 )
12 11 a1i ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( 2 ∈ ℝ ∧ 0 < 2 ) )
13 ltmuldiv ⊢ ( ( 𝐴 ∈ ℝ ∧ ( 𝐴 + 𝐵 ) ∈ ℝ ∧ ( 2 ∈ ℝ ∧ 0 < 2 ) ) → ( ( 𝐴 · 2 ) < ( 𝐴 + 𝐵 ) ↔ 𝐴 < ( ( 𝐴 + 𝐵 ) / 2 ) ) )
14 3 8 12 13 syl3anc ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( ( 𝐴 · 2 ) < ( 𝐴 + 𝐵 ) ↔ 𝐴 < ( ( 𝐴 + 𝐵 ) / 2 ) ) )
15 2 7 14 3bitr2d ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ) → ( 𝐴 < 𝐵 ↔ 𝐴 < ( ( 𝐴 + 𝐵 ) / 2 ) ) )