Metamath Proof Explorer


Theorem lt2add

Description: Adding both sides of two 'less than' relations. Theorem I.25 of Apostol p. 20. (Contributed by NM, 15-Aug-1999) (Proof shortened by Mario Carneiro, 27-May-2016)

Ref Expression
Assertion lt2add
|- ( ( ( A e. RR /\ B e. RR ) /\ ( C e. RR /\ D e. RR ) ) -> ( ( A < C /\ B < D ) -> ( A + B ) < ( C + D ) ) )

Proof

Step Hyp Ref Expression
1 ltle
 |-  ( ( A e. RR /\ C e. RR ) -> ( A < C -> A <_ C ) )
2 1 ad2ant2r
 |-  ( ( ( A e. RR /\ B e. RR ) /\ ( C e. RR /\ D e. RR ) ) -> ( A < C -> A <_ C ) )
3 leltadd
 |-  ( ( ( A e. RR /\ B e. RR ) /\ ( C e. RR /\ D e. RR ) ) -> ( ( A <_ C /\ B < D ) -> ( A + B ) < ( C + D ) ) )
4 2 3 syland
 |-  ( ( ( A e. RR /\ B e. RR ) /\ ( C e. RR /\ D e. RR ) ) -> ( ( A < C /\ B < D ) -> ( A + B ) < ( C + D ) ) )