Metamath Proof Explorer


Theorem sqrtcvallem4

Description: Equivalent to saying that the square of the real component of the square root of a complex number is a nonnegative real number. Lemma for sqrtcval . See resqrtval . (Contributed by RP, 11-May-2024)

Ref Expression
Assertion sqrtcvallem4 ( 𝐴 ∈ ℂ → 0 ≤ ( ( ( abs ‘ 𝐴 ) + ( ℜ ‘ 𝐴 ) ) / 2 ) )

Proof

Step Hyp Ref Expression
1 abscl ( 𝐴 ∈ ℂ → ( abs ‘ 𝐴 ) ∈ ℝ )
2 recl ( 𝐴 ∈ ℂ → ( ℜ ‘ 𝐴 ) ∈ ℝ )
3 1 2 readdcld ( 𝐴 ∈ ℂ → ( ( abs ‘ 𝐴 ) + ( ℜ ‘ 𝐴 ) ) ∈ ℝ )
4 2rp 2 ∈ ℝ+
5 4 a1i ( 𝐴 ∈ ℂ → 2 ∈ ℝ+ )
6 negcl ( 𝐴 ∈ ℂ → - 𝐴 ∈ ℂ )
7 6 releabsd ( 𝐴 ∈ ℂ → ( ℜ ‘ - 𝐴 ) ≤ ( abs ‘ - 𝐴 ) )
8 6 abscld ( 𝐴 ∈ ℂ → ( abs ‘ - 𝐴 ) ∈ ℝ )
9 6 recld ( 𝐴 ∈ ℂ → ( ℜ ‘ - 𝐴 ) ∈ ℝ )
10 8 9 subge0d ( 𝐴 ∈ ℂ → ( 0 ≤ ( ( abs ‘ - 𝐴 ) − ( ℜ ‘ - 𝐴 ) ) ↔ ( ℜ ‘ - 𝐴 ) ≤ ( abs ‘ - 𝐴 ) ) )
11 7 10 mpbird ( 𝐴 ∈ ℂ → 0 ≤ ( ( abs ‘ - 𝐴 ) − ( ℜ ‘ - 𝐴 ) ) )
12 absneg ( 𝐴 ∈ ℂ → ( abs ‘ - 𝐴 ) = ( abs ‘ 𝐴 ) )
13 reneg ( 𝐴 ∈ ℂ → ( ℜ ‘ - 𝐴 ) = - ( ℜ ‘ 𝐴 ) )
14 12 13 oveq12d ( 𝐴 ∈ ℂ → ( ( abs ‘ - 𝐴 ) − ( ℜ ‘ - 𝐴 ) ) = ( ( abs ‘ 𝐴 ) − - ( ℜ ‘ 𝐴 ) ) )
15 1 recnd ( 𝐴 ∈ ℂ → ( abs ‘ 𝐴 ) ∈ ℂ )
16 2 recnd ( 𝐴 ∈ ℂ → ( ℜ ‘ 𝐴 ) ∈ ℂ )
17 15 16 subnegd ( 𝐴 ∈ ℂ → ( ( abs ‘ 𝐴 ) − - ( ℜ ‘ 𝐴 ) ) = ( ( abs ‘ 𝐴 ) + ( ℜ ‘ 𝐴 ) ) )
18 14 17 eqtrd ( 𝐴 ∈ ℂ → ( ( abs ‘ - 𝐴 ) − ( ℜ ‘ - 𝐴 ) ) = ( ( abs ‘ 𝐴 ) + ( ℜ ‘ 𝐴 ) ) )
19 11 18 breqtrd ( 𝐴 ∈ ℂ → 0 ≤ ( ( abs ‘ 𝐴 ) + ( ℜ ‘ 𝐴 ) ) )
20 3 5 19 divge0d ( 𝐴 ∈ ℂ → 0 ≤ ( ( ( abs ‘ 𝐴 ) + ( ℜ ‘ 𝐴 ) ) / 2 ) )