reserve f,f1,f2,g for PartFunc of REAL,REAL;
reserve A for non empty closed_interval Subset of REAL;
reserve p,r,x,x0 for Real;
reserve n for Element of NAT;
reserve Z for open Subset of REAL;

theorem
  A = [.x+2*n*PI,x+(2*n+1)*PI.] implies integral(sin-cos,A) = 2*cos x + 2*sin x
proof
  assume A = [.x+2*n*PI,x+(2*n+1)*PI.];
  then upper_bound A=x+(2*n+1)*PI & lower_bound A=x+2*n*PI by Th37;
  then
  integral(sin-cos,A) = (-cos).(x+(2*n+1)*PI) - (-cos).(x+2*n*PI) - (sin.(
  x+(2*n+1)*PI) - sin.(x+2*n*PI)) by Th78
    .= -(cos.(x+(2*n+1)*PI)) - (-cos).(x+2*n*PI) - (sin.(x+(2*n+1)*PI) - sin
  .(x+2*n*PI)) by VALUED_1:8
    .= -(cos(x+(2*n+1)*PI)) - -(cos(x+2*n*PI)) - (sin(x+(2*n+1)*PI) - sin(x+
  2*n*PI)) by VALUED_1:8
    .= -(-cos x) - -(cos(x+2*n*PI)) - (sin(x+(2*n+1)*PI) - sin(x+2*n*PI)) by
Th4
    .= -(-cos x) - -cos x - (sin(x+(2*n+1)*PI) - sin(x+2*n*PI)) by Th3
    .= -(-cos x) - -cos x - (-sin x - sin(x+2*n*PI)) by Th2
    .= cos x + cos x - (-sin x - sin x) by Th1
    .= 2* cos x + 2*sin x;
  hence thesis;
end;
