Metamath Proof Explorer


Theorem abs2difd

Description: Difference of absolute values. (Contributed by Mario Carneiro, 29-May-2016)

Ref Expression
Hypotheses abscld.1
|- ( ph -> A e. CC )
abssubd.2
|- ( ph -> B e. CC )
Assertion abs2difd
|- ( ph -> ( ( abs ` A ) - ( abs ` B ) ) <_ ( abs ` ( A - B ) ) )

Proof

Step Hyp Ref Expression
1 abscld.1
 |-  ( ph -> A e. CC )
2 abssubd.2
 |-  ( ph -> B e. CC )
3 abs2dif
 |-  ( ( A e. CC /\ B e. CC ) -> ( ( abs ` A ) - ( abs ` B ) ) <_ ( abs ` ( A - B ) ) )
4 1 2 3 syl2anc
 |-  ( ph -> ( ( abs ` A ) - ( abs ` B ) ) <_ ( abs ` ( A - B ) ) )