Metamath Proof Explorer


Theorem absnpncan2d

Description: Triangular inequality, combined with cancellation law for subtraction (applied twice). (Contributed by Glauco Siliprandi, 11-Dec-2019)

Ref Expression
Hypotheses absnpncan2d.a ( 𝜑𝐴 ∈ ℂ )
absnpncan2d.b ( 𝜑𝐵 ∈ ℂ )
absnpncan2d.c ( 𝜑𝐶 ∈ ℂ )
absnpncan2d.d ( 𝜑𝐷 ∈ ℂ )
Assertion absnpncan2d ( 𝜑 → ( abs ‘ ( 𝐴𝐷 ) ) ≤ ( ( ( abs ‘ ( 𝐴𝐵 ) ) + ( abs ‘ ( 𝐵𝐶 ) ) ) + ( abs ‘ ( 𝐶𝐷 ) ) ) )

Proof

Step Hyp Ref Expression
1 absnpncan2d.a ( 𝜑𝐴 ∈ ℂ )
2 absnpncan2d.b ( 𝜑𝐵 ∈ ℂ )
3 absnpncan2d.c ( 𝜑𝐶 ∈ ℂ )
4 absnpncan2d.d ( 𝜑𝐷 ∈ ℂ )
5 1 4 subcld ( 𝜑 → ( 𝐴𝐷 ) ∈ ℂ )
6 5 abscld ( 𝜑 → ( abs ‘ ( 𝐴𝐷 ) ) ∈ ℝ )
7 1 3 subcld ( 𝜑 → ( 𝐴𝐶 ) ∈ ℂ )
8 7 abscld ( 𝜑 → ( abs ‘ ( 𝐴𝐶 ) ) ∈ ℝ )
9 3 4 subcld ( 𝜑 → ( 𝐶𝐷 ) ∈ ℂ )
10 9 abscld ( 𝜑 → ( abs ‘ ( 𝐶𝐷 ) ) ∈ ℝ )
11 8 10 readdcld ( 𝜑 → ( ( abs ‘ ( 𝐴𝐶 ) ) + ( abs ‘ ( 𝐶𝐷 ) ) ) ∈ ℝ )
12 1 2 subcld ( 𝜑 → ( 𝐴𝐵 ) ∈ ℂ )
13 12 abscld ( 𝜑 → ( abs ‘ ( 𝐴𝐵 ) ) ∈ ℝ )
14 2 3 subcld ( 𝜑 → ( 𝐵𝐶 ) ∈ ℂ )
15 14 abscld ( 𝜑 → ( abs ‘ ( 𝐵𝐶 ) ) ∈ ℝ )
16 13 15 readdcld ( 𝜑 → ( ( abs ‘ ( 𝐴𝐵 ) ) + ( abs ‘ ( 𝐵𝐶 ) ) ) ∈ ℝ )
17 16 10 readdcld ( 𝜑 → ( ( ( abs ‘ ( 𝐴𝐵 ) ) + ( abs ‘ ( 𝐵𝐶 ) ) ) + ( abs ‘ ( 𝐶𝐷 ) ) ) ∈ ℝ )
18 1 4 3 abs3difd ( 𝜑 → ( abs ‘ ( 𝐴𝐷 ) ) ≤ ( ( abs ‘ ( 𝐴𝐶 ) ) + ( abs ‘ ( 𝐶𝐷 ) ) ) )
19 1 3 2 abs3difd ( 𝜑 → ( abs ‘ ( 𝐴𝐶 ) ) ≤ ( ( abs ‘ ( 𝐴𝐵 ) ) + ( abs ‘ ( 𝐵𝐶 ) ) ) )
20 8 16 10 19 leadd1dd ( 𝜑 → ( ( abs ‘ ( 𝐴𝐶 ) ) + ( abs ‘ ( 𝐶𝐷 ) ) ) ≤ ( ( ( abs ‘ ( 𝐴𝐵 ) ) + ( abs ‘ ( 𝐵𝐶 ) ) ) + ( abs ‘ ( 𝐶𝐷 ) ) ) )
21 6 11 17 18 20 letrd ( 𝜑 → ( abs ‘ ( 𝐴𝐷 ) ) ≤ ( ( ( abs ‘ ( 𝐴𝐵 ) ) + ( abs ‘ ( 𝐵𝐶 ) ) ) + ( abs ‘ ( 𝐶𝐷 ) ) ) )