Metamath Proof Explorer


Theorem le2subd

Description: Subtracting both sides of two 'less than or equal to' relations. (Contributed by Mario Carneiro, 27-May-2016)

Ref Expression
Hypotheses leidd.1
|- ( ph -> A e. RR )
ltnegd.2
|- ( ph -> B e. RR )
ltadd1d.3
|- ( ph -> C e. RR )
lt2addd.4
|- ( ph -> D e. RR )
le2addd.5
|- ( ph -> A <_ C )
le2addd.6
|- ( ph -> B <_ D )
Assertion le2subd
|- ( ph -> ( A - D ) <_ ( C - B ) )

Proof

Step Hyp Ref Expression
1 leidd.1
 |-  ( ph -> A e. RR )
2 ltnegd.2
 |-  ( ph -> B e. RR )
3 ltadd1d.3
 |-  ( ph -> C e. RR )
4 lt2addd.4
 |-  ( ph -> D e. RR )
5 le2addd.5
 |-  ( ph -> A <_ C )
6 le2addd.6
 |-  ( ph -> B <_ D )
7 le2sub
 |-  ( ( ( A e. RR /\ D e. RR ) /\ ( C e. RR /\ B e. RR ) ) -> ( ( A <_ C /\ B <_ D ) -> ( A - D ) <_ ( C - B ) ) )
8 1 4 3 2 7 syl22anc
 |-  ( ph -> ( ( A <_ C /\ B <_ D ) -> ( A - D ) <_ ( C - B ) ) )
9 5 6 8 mp2and
 |-  ( ph -> ( A - D ) <_ ( C - B ) )