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Metamath Proof Explorer

Theorem abslt2sqd

Description: Comparison of the square of two numbers. (Contributed by Glauco Siliprandi, 24-Dec-2020)

Ref Expression
Hypotheses abslt2sqd.a ⊢ ( 𝜑 → 𝐴 ∈ ℝ )
abslt2sqd.b ⊢ ( 𝜑 → 𝐵 ∈ ℝ )
abslt2sqd.l ⊢ ( 𝜑 → ( abs ‘ 𝐴 ) < ( abs ‘ 𝐵 ) )
Assertion abslt2sqd ( 𝜑 → ( 𝐴 ↑ 2 ) < ( 𝐵 ↑ 2 ) )

Proof

Step Hyp Ref Expression
1 abslt2sqd.a ⊢ ( 𝜑 → 𝐴 ∈ ℝ )
2 abslt2sqd.b ⊢ ( 𝜑 → 𝐵 ∈ ℝ )
3 abslt2sqd.l ⊢ ( 𝜑 → ( abs ‘ 𝐴 ) < ( abs ‘ 𝐵 ) )
4 1 recnd ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
5 4 abscld ⊢ ( 𝜑 → ( abs ‘ 𝐴 ) ∈ ℝ )
6 4 absge0d ⊢ ( 𝜑 → 0 ≤ ( abs ‘ 𝐴 ) )
7 2 recnd ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
8 7 abscld ⊢ ( 𝜑 → ( abs ‘ 𝐵 ) ∈ ℝ )
9 7 absge0d ⊢ ( 𝜑 → 0 ≤ ( abs ‘ 𝐵 ) )
10 lt2sq ⊢ ( ( ( ( abs ‘ 𝐴 ) ∈ ℝ ∧ 0 ≤ ( abs ‘ 𝐴 ) ) ∧ ( ( abs ‘ 𝐵 ) ∈ ℝ ∧ 0 ≤ ( abs ‘ 𝐵 ) ) ) → ( ( abs ‘ 𝐴 ) < ( abs ‘ 𝐵 ) ↔ ( ( abs ‘ 𝐴 ) ↑ 2 ) < ( ( abs ‘ 𝐵 ) ↑ 2 ) ) )
11 5 6 8 9 10 syl22anc ⊢ ( 𝜑 → ( ( abs ‘ 𝐴 ) < ( abs ‘ 𝐵 ) ↔ ( ( abs ‘ 𝐴 ) ↑ 2 ) < ( ( abs ‘ 𝐵 ) ↑ 2 ) ) )
12 3 11 mpbid ⊢ ( 𝜑 → ( ( abs ‘ 𝐴 ) ↑ 2 ) < ( ( abs ‘ 𝐵 ) ↑ 2 ) )
13 absresq ⊢ ( 𝐴 ∈ ℝ → ( ( abs ‘ 𝐴 ) ↑ 2 ) = ( 𝐴 ↑ 2 ) )
14 1 13 syl ⊢ ( 𝜑 → ( ( abs ‘ 𝐴 ) ↑ 2 ) = ( 𝐴 ↑ 2 ) )
15 absresq ⊢ ( 𝐵 ∈ ℝ → ( ( abs ‘ 𝐵 ) ↑ 2 ) = ( 𝐵 ↑ 2 ) )
16 2 15 syl ⊢ ( 𝜑 → ( ( abs ‘ 𝐵 ) ↑ 2 ) = ( 𝐵 ↑ 2 ) )
17 14 16 breq12d ⊢ ( 𝜑 → ( ( ( abs ‘ 𝐴 ) ↑ 2 ) < ( ( abs ‘ 𝐵 ) ↑ 2 ) ↔ ( 𝐴 ↑ 2 ) < ( 𝐵 ↑ 2 ) ) )
18 12 17 mpbid ⊢ ( 𝜑 → ( 𝐴 ↑ 2 ) < ( 𝐵 ↑ 2 ) )