Metamath Proof Explorer


Theorem absnpncan3d

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

Ref Expression
Hypotheses absnpncan3d.a ( 𝜑𝐴 ∈ ℂ )
absnpncan3d.b ( 𝜑𝐵 ∈ ℂ )
absnpncan3d.c ( 𝜑𝐶 ∈ ℂ )
absnpncan3d.d ( 𝜑𝐷 ∈ ℂ )
absnpncan3d.e ( 𝜑𝐸 ∈ ℂ )
Assertion absnpncan3d ( 𝜑 → ( abs ‘ ( 𝐴𝐸 ) ) ≤ ( ( ( ( abs ‘ ( 𝐴𝐵 ) ) + ( abs ‘ ( 𝐵𝐶 ) ) ) + ( abs ‘ ( 𝐶𝐷 ) ) ) + ( abs ‘ ( 𝐷𝐸 ) ) ) )

Proof

Step Hyp Ref Expression
1 absnpncan3d.a ( 𝜑𝐴 ∈ ℂ )
2 absnpncan3d.b ( 𝜑𝐵 ∈ ℂ )
3 absnpncan3d.c ( 𝜑𝐶 ∈ ℂ )
4 absnpncan3d.d ( 𝜑𝐷 ∈ ℂ )
5 absnpncan3d.e ( 𝜑𝐸 ∈ ℂ )
6 1 5 subcld ( 𝜑 → ( 𝐴𝐸 ) ∈ ℂ )
7 6 abscld ( 𝜑 → ( abs ‘ ( 𝐴𝐸 ) ) ∈ ℝ )
8 1 4 subcld ( 𝜑 → ( 𝐴𝐷 ) ∈ ℂ )
9 8 abscld ( 𝜑 → ( abs ‘ ( 𝐴𝐷 ) ) ∈ ℝ )
10 4 5 subcld ( 𝜑 → ( 𝐷𝐸 ) ∈ ℂ )
11 10 abscld ( 𝜑 → ( abs ‘ ( 𝐷𝐸 ) ) ∈ ℝ )
12 9 11 readdcld ( 𝜑 → ( ( abs ‘ ( 𝐴𝐷 ) ) + ( abs ‘ ( 𝐷𝐸 ) ) ) ∈ ℝ )
13 1 2 subcld ( 𝜑 → ( 𝐴𝐵 ) ∈ ℂ )
14 13 abscld ( 𝜑 → ( abs ‘ ( 𝐴𝐵 ) ) ∈ ℝ )
15 2 3 subcld ( 𝜑 → ( 𝐵𝐶 ) ∈ ℂ )
16 15 abscld ( 𝜑 → ( abs ‘ ( 𝐵𝐶 ) ) ∈ ℝ )
17 14 16 readdcld ( 𝜑 → ( ( abs ‘ ( 𝐴𝐵 ) ) + ( abs ‘ ( 𝐵𝐶 ) ) ) ∈ ℝ )
18 3 4 subcld ( 𝜑 → ( 𝐶𝐷 ) ∈ ℂ )
19 18 abscld ( 𝜑 → ( abs ‘ ( 𝐶𝐷 ) ) ∈ ℝ )
20 17 19 readdcld ( 𝜑 → ( ( ( abs ‘ ( 𝐴𝐵 ) ) + ( abs ‘ ( 𝐵𝐶 ) ) ) + ( abs ‘ ( 𝐶𝐷 ) ) ) ∈ ℝ )
21 20 11 readdcld ( 𝜑 → ( ( ( ( abs ‘ ( 𝐴𝐵 ) ) + ( abs ‘ ( 𝐵𝐶 ) ) ) + ( abs ‘ ( 𝐶𝐷 ) ) ) + ( abs ‘ ( 𝐷𝐸 ) ) ) ∈ ℝ )
22 1 5 4 abs3difd ( 𝜑 → ( abs ‘ ( 𝐴𝐸 ) ) ≤ ( ( abs ‘ ( 𝐴𝐷 ) ) + ( abs ‘ ( 𝐷𝐸 ) ) ) )
23 1 2 3 4 absnpncan2d ( 𝜑 → ( abs ‘ ( 𝐴𝐷 ) ) ≤ ( ( ( abs ‘ ( 𝐴𝐵 ) ) + ( abs ‘ ( 𝐵𝐶 ) ) ) + ( abs ‘ ( 𝐶𝐷 ) ) ) )
24 9 20 11 23 leadd1dd ( 𝜑 → ( ( abs ‘ ( 𝐴𝐷 ) ) + ( abs ‘ ( 𝐷𝐸 ) ) ) ≤ ( ( ( ( abs ‘ ( 𝐴𝐵 ) ) + ( abs ‘ ( 𝐵𝐶 ) ) ) + ( abs ‘ ( 𝐶𝐷 ) ) ) + ( abs ‘ ( 𝐷𝐸 ) ) ) )
25 7 12 21 22 24 letrd ( 𝜑 → ( abs ‘ ( 𝐴𝐸 ) ) ≤ ( ( ( ( abs ‘ ( 𝐴𝐵 ) ) + ( abs ‘ ( 𝐵𝐶 ) ) ) + ( abs ‘ ( 𝐶𝐷 ) ) ) + ( abs ‘ ( 𝐷𝐸 ) ) ) )