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Metamath Proof Explorer

Theorem ltrec1

Description: Reciprocal swap in a 'less than' relation. (Contributed by NM, 24-Feb-2005)

Ref Expression
Assertion ltrec1 ( ( ( 𝐴 ∈ ℝ ∧ 0 < 𝐴 ) ∧ ( 𝐵 ∈ ℝ ∧ 0 < 𝐵 ) ) → ( ( 1 / 𝐴 ) < 𝐵 ↔ ( 1 / 𝐵 ) < 𝐴 ) )

Proof

Step Hyp Ref Expression
1 gt0ne0 ⊢ ( ( 𝐴 ∈ ℝ ∧ 0 < 𝐴 ) → 𝐴 ≠ 0 )
2 rereccl ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐴 ≠ 0 ) → ( 1 / 𝐴 ) ∈ ℝ )
3 1 2 syldan ⊢ ( ( 𝐴 ∈ ℝ ∧ 0 < 𝐴 ) → ( 1 / 𝐴 ) ∈ ℝ )
4 recgt0 ⊢ ( ( 𝐴 ∈ ℝ ∧ 0 < 𝐴 ) → 0 < ( 1 / 𝐴 ) )
5 3 4 jca ⊢ ( ( 𝐴 ∈ ℝ ∧ 0 < 𝐴 ) → ( ( 1 / 𝐴 ) ∈ ℝ ∧ 0 < ( 1 / 𝐴 ) ) )
6 ltrec ⊢ ( ( ( ( 1 / 𝐴 ) ∈ ℝ ∧ 0 < ( 1 / 𝐴 ) ) ∧ ( 𝐵 ∈ ℝ ∧ 0 < 𝐵 ) ) → ( ( 1 / 𝐴 ) < 𝐵 ↔ ( 1 / 𝐵 ) < ( 1 / ( 1 / 𝐴 ) ) ) )
7 5 6 sylan ⊢ ( ( ( 𝐴 ∈ ℝ ∧ 0 < 𝐴 ) ∧ ( 𝐵 ∈ ℝ ∧ 0 < 𝐵 ) ) → ( ( 1 / 𝐴 ) < 𝐵 ↔ ( 1 / 𝐵 ) < ( 1 / ( 1 / 𝐴 ) ) ) )
8 recn ⊢ ( 𝐴 ∈ ℝ → 𝐴 ∈ ℂ )
9 recrec ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) → ( 1 / ( 1 / 𝐴 ) ) = 𝐴 )
10 8 9 sylan ⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐴 ≠ 0 ) → ( 1 / ( 1 / 𝐴 ) ) = 𝐴 )
11 1 10 syldan ⊢ ( ( 𝐴 ∈ ℝ ∧ 0 < 𝐴 ) → ( 1 / ( 1 / 𝐴 ) ) = 𝐴 )
12 11 adantr ⊢ ( ( ( 𝐴 ∈ ℝ ∧ 0 < 𝐴 ) ∧ ( 𝐵 ∈ ℝ ∧ 0 < 𝐵 ) ) → ( 1 / ( 1 / 𝐴 ) ) = 𝐴 )
13 12 breq2d ⊢ ( ( ( 𝐴 ∈ ℝ ∧ 0 < 𝐴 ) ∧ ( 𝐵 ∈ ℝ ∧ 0 < 𝐵 ) ) → ( ( 1 / 𝐵 ) < ( 1 / ( 1 / 𝐴 ) ) ↔ ( 1 / 𝐵 ) < 𝐴 ) )
14 7 13 bitrd ⊢ ( ( ( 𝐴 ∈ ℝ ∧ 0 < 𝐴 ) ∧ ( 𝐵 ∈ ℝ ∧ 0 < 𝐵 ) ) → ( ( 1 / 𝐴 ) < 𝐵 ↔ ( 1 / 𝐵 ) < 𝐴 ) )