let A be closed-interval Subset of REAL ; :: thesis: for Z being open Subset of REAL st A c= Z & Z = dom (cos - sin ) & (cos - sin ) | A is continuous holds
integral (cos - sin ),A = ((sin + cos ) . (upper_bound A)) - ((sin + cos ) . (lower_bound A))
let Z be open Subset of REAL ; :: thesis: ( A c= Z & Z = dom (cos - sin ) & (cos - sin ) | A is continuous implies integral (cos - sin ),A = ((sin + cos ) . (upper_bound A)) - ((sin + cos ) . (lower_bound A)) )
assume A1: ( A c= Z & Z = dom (cos - sin ) & (cos - sin ) | A is continuous ) ; :: thesis: integral (cos - sin ),A = ((sin + cos ) . (upper_bound A)) - ((sin + cos ) . (lower_bound A))
then A2: ( cos - sin is_integrable_on A & (cos - sin ) | A is bounded ) by INTEGRA5:10, INTEGRA5:11;
Z = (dom cos ) /\ (dom sin ) by A1, VALUED_1:12;
then AB: Z c= dom (sin + cos ) by VALUED_1:def 1;
then A3: sin + cos is_differentiable_on Z by FDIFF_7:38;
A4: for x being Real st x in dom ((sin + cos ) `| Z) holds
((sin + cos ) `| Z) . x = (cos - sin ) . x

proof

let x be Real; :: thesis: ( x in dom ((sin + cos ) `| Z) implies ((sin + cos ) `| Z) . x = (cos - sin ) . x )
assume x in dom ((sin + cos ) `| Z) ; :: thesis: ((sin + cos ) `| Z) . x = (cos - sin ) . x
then A5: x in Z by A3, FDIFF_1:def 8;
then ((sin + cos ) `| Z) . x = (cos . x) - (sin . x) by AB, FDIFF_7:38
.= (cos - sin ) . x by A1, A5, VALUED_1:13 ;
hence ((sin + cos ) `| Z) . x = (cos - sin ) . x ; :: thesis: verum

end;

dom ((sin + cos ) `| Z) = dom (cos - sin ) by A1, A3, FDIFF_1:def 8;
then (sin + cos ) `| Z = cos - sin by A4, PARTFUN1:34;
hence integral (cos - sin ),A = ((sin + cos ) . (upper_bound A)) - ((sin + cos ) . (lower_bound A)) by A1, A2, AB, FDIFF_7:38, INTEGRA5:13; :: thesis: verum