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
|- ( ph -> A e. CC )
absnpncan2d.b
|- ( ph -> B e. CC )
absnpncan2d.c
|- ( ph -> C e. CC )
absnpncan2d.d
|- ( ph -> D e. CC )
Assertion absnpncan2d
|- ( ph -> ( abs ` ( A - D ) ) <_ ( ( ( abs ` ( A - B ) ) + ( abs ` ( B - C ) ) ) + ( abs ` ( C - D ) ) ) )

Proof

Step Hyp Ref Expression
1 absnpncan2d.a
 |-  ( ph -> A e. CC )
2 absnpncan2d.b
 |-  ( ph -> B e. CC )
3 absnpncan2d.c
 |-  ( ph -> C e. CC )
4 absnpncan2d.d
 |-  ( ph -> D e. CC )
5 1 4 subcld
 |-  ( ph -> ( A - D ) e. CC )
6 5 abscld
 |-  ( ph -> ( abs ` ( A - D ) ) e. RR )
7 1 3 subcld
 |-  ( ph -> ( A - C ) e. CC )
8 7 abscld
 |-  ( ph -> ( abs ` ( A - C ) ) e. RR )
9 3 4 subcld
 |-  ( ph -> ( C - D ) e. CC )
10 9 abscld
 |-  ( ph -> ( abs ` ( C - D ) ) e. RR )
11 8 10 readdcld
 |-  ( ph -> ( ( abs ` ( A - C ) ) + ( abs ` ( C - D ) ) ) e. RR )
12 1 2 subcld
 |-  ( ph -> ( A - B ) e. CC )
13 12 abscld
 |-  ( ph -> ( abs ` ( A - B ) ) e. RR )
14 2 3 subcld
 |-  ( ph -> ( B - C ) e. CC )
15 14 abscld
 |-  ( ph -> ( abs ` ( B - C ) ) e. RR )
16 13 15 readdcld
 |-  ( ph -> ( ( abs ` ( A - B ) ) + ( abs ` ( B - C ) ) ) e. RR )
17 16 10 readdcld
 |-  ( ph -> ( ( ( abs ` ( A - B ) ) + ( abs ` ( B - C ) ) ) + ( abs ` ( C - D ) ) ) e. RR )
18 1 4 3 abs3difd
 |-  ( ph -> ( abs ` ( A - D ) ) <_ ( ( abs ` ( A - C ) ) + ( abs ` ( C - D ) ) ) )
19 1 3 2 abs3difd
 |-  ( ph -> ( abs ` ( A - C ) ) <_ ( ( abs ` ( A - B ) ) + ( abs ` ( B - C ) ) ) )
20 8 16 10 19 leadd1dd
 |-  ( ph -> ( ( abs ` ( A - C ) ) + ( abs ` ( C - D ) ) ) <_ ( ( ( abs ` ( A - B ) ) + ( abs ` ( B - C ) ) ) + ( abs ` ( C - D ) ) ) )
21 6 11 17 18 20 letrd
 |-  ( ph -> ( abs ` ( A - D ) ) <_ ( ( ( abs ` ( A - B ) ) + ( abs ` ( B - C ) ) ) + ( abs ` ( C - D ) ) ) )