Metamath Proof Explorer


Theorem sgnnbi

Description: Negative signum. (Contributed by Thierry Arnoux, 2-Oct-2018)

Ref Expression
Assertion sgnnbi ( 𝐴 ∈ ℝ* → ( ( sgn ‘ 𝐴 ) = - 1 ↔ 𝐴 < 0 ) )

Proof

Step Hyp Ref Expression
1 id ( 𝐴 ∈ ℝ*𝐴 ∈ ℝ* )
2 eqeq1 ( ( sgn ‘ 𝐴 ) = 0 → ( ( sgn ‘ 𝐴 ) = - 1 ↔ 0 = - 1 ) )
3 2 imbi1d ( ( sgn ‘ 𝐴 ) = 0 → ( ( ( sgn ‘ 𝐴 ) = - 1 → 𝐴 < 0 ) ↔ ( 0 = - 1 → 𝐴 < 0 ) ) )
4 eqeq1 ( ( sgn ‘ 𝐴 ) = 1 → ( ( sgn ‘ 𝐴 ) = - 1 ↔ 1 = - 1 ) )
5 4 imbi1d ( ( sgn ‘ 𝐴 ) = 1 → ( ( ( sgn ‘ 𝐴 ) = - 1 → 𝐴 < 0 ) ↔ ( 1 = - 1 → 𝐴 < 0 ) ) )
6 eqeq1 ( ( sgn ‘ 𝐴 ) = - 1 → ( ( sgn ‘ 𝐴 ) = - 1 ↔ - 1 = - 1 ) )
7 6 imbi1d ( ( sgn ‘ 𝐴 ) = - 1 → ( ( ( sgn ‘ 𝐴 ) = - 1 → 𝐴 < 0 ) ↔ ( - 1 = - 1 → 𝐴 < 0 ) ) )
8 neg1ne0 - 1 ≠ 0
9 8 nesymi ¬ 0 = - 1
10 9 pm2.21i ( 0 = - 1 → 𝐴 < 0 )
11 10 a1i ( ( 𝐴 ∈ ℝ*𝐴 = 0 ) → ( 0 = - 1 → 𝐴 < 0 ) )
12 neg1rr - 1 ∈ ℝ
13 neg1lt0 - 1 < 0
14 0lt1 0 < 1
15 0re 0 ∈ ℝ
16 1re 1 ∈ ℝ
17 12 15 16 lttri ( ( - 1 < 0 ∧ 0 < 1 ) → - 1 < 1 )
18 13 14 17 mp2an - 1 < 1
19 12 18 gtneii 1 ≠ - 1
20 19 neii ¬ 1 = - 1
21 20 pm2.21i ( 1 = - 1 → 𝐴 < 0 )
22 21 a1i ( ( 𝐴 ∈ ℝ* ∧ 0 < 𝐴 ) → ( 1 = - 1 → 𝐴 < 0 ) )
23 simp2 ( ( 𝐴 ∈ ℝ*𝐴 < 0 ∧ - 1 = - 1 ) → 𝐴 < 0 )
24 23 3expia ( ( 𝐴 ∈ ℝ*𝐴 < 0 ) → ( - 1 = - 1 → 𝐴 < 0 ) )
25 1 3 5 7 11 22 24 sgn3da ( 𝐴 ∈ ℝ* → ( ( sgn ‘ 𝐴 ) = - 1 → 𝐴 < 0 ) )
26 25 imp ( ( 𝐴 ∈ ℝ* ∧ ( sgn ‘ 𝐴 ) = - 1 ) → 𝐴 < 0 )
27 sgnn ( ( 𝐴 ∈ ℝ*𝐴 < 0 ) → ( sgn ‘ 𝐴 ) = - 1 )
28 26 27 impbida ( 𝐴 ∈ ℝ* → ( ( sgn ‘ 𝐴 ) = - 1 ↔ 𝐴 < 0 ) )