Metamath Proof Explorer


Theorem absltd

Description: Absolute value and 'less than' relation. (Contributed by Mario Carneiro, 29-May-2016)

Ref Expression
Hypotheses absltd.1
|- ( ph -> A e. RR )
absltd.2
|- ( ph -> B e. RR )
Assertion absltd
|- ( ph -> ( ( abs ` A ) < B <-> ( -u B < A /\ A < B ) ) )

Proof

Step Hyp Ref Expression
1 absltd.1
 |-  ( ph -> A e. RR )
2 absltd.2
 |-  ( ph -> B e. RR )
3 abslt
 |-  ( ( A e. RR /\ B e. RR ) -> ( ( abs ` A ) < B <-> ( -u B < A /\ A < B ) ) )
4 1 2 3 syl2anc
 |-  ( ph -> ( ( abs ` A ) < B <-> ( -u B < A /\ A < B ) ) )