Metamath Proof Explorer


Theorem le2halvesd

Description: A sum is less than the whole if each term is less than half. (Contributed by Thierry Arnoux, 29-Nov-2017)

Ref Expression
Hypotheses le2halvesd.1
|- ( ph -> A e. RR )
le2halvesd.2
|- ( ph -> B e. RR )
le2halvesd.3
|- ( ph -> C e. RR )
le2halvesd.4
|- ( ph -> A <_ ( C / 2 ) )
le2halvesd.5
|- ( ph -> B <_ ( C / 2 ) )
Assertion le2halvesd
|- ( ph -> ( A + B ) <_ C )

Proof

Step Hyp Ref Expression
1 le2halvesd.1
 |-  ( ph -> A e. RR )
2 le2halvesd.2
 |-  ( ph -> B e. RR )
3 le2halvesd.3
 |-  ( ph -> C e. RR )
4 le2halvesd.4
 |-  ( ph -> A <_ ( C / 2 ) )
5 le2halvesd.5
 |-  ( ph -> B <_ ( C / 2 ) )
6 3 rehalfcld
 |-  ( ph -> ( C / 2 ) e. RR )
7 1 2 6 6 4 5 le2addd
 |-  ( ph -> ( A + B ) <_ ( ( C / 2 ) + ( C / 2 ) ) )
8 3 recnd
 |-  ( ph -> C e. CC )
9 8 2halvesd
 |-  ( ph -> ( ( C / 2 ) + ( C / 2 ) ) = C )
10 7 9 breqtrd
 |-  ( ph -> ( A + B ) <_ C )