 reserve A for non empty Subset of REAL;
 reserve A for non empty closed_interval Subset of REAL;

theorem Lm20B:
  for a,b,c be Real, f be Function of REAL,REAL st
  b > 0 & c > 0 &
  (for x be Real holds f.x = max(0,b-|. b*(x-a)/c .|))
    holds
  f.(lower_bound ['a-c,a+c']) = 0 & f.(a-c) = 0 &
  f.(upper_bound ['a-c,a+c']) = 0 & f.(a+c) = 0
proof
 let a,b,c be Real, f be Function of REAL,REAL;
 assume that
 A1: b > 0 and
 A2: c > 0 and
 A3: for x be Real holds f.x = max(0, b - |. b*(x-a)/c .|);
 A5: a < a+c & a-c < a by XREAL_1:44, XREAL_1:29,A2;
 A8: [.(lower_bound ['a-c,a+c']),(upper_bound ['a-c,a+c']).] = ['a-c,a+c']
  & ['a-c,a+c'] = [.a-c,a+c.] by XXREAL_0:2,A5,INTEGRA5:def 3,INTEGRA1:4;
 A6: f.(a-c) = max(0, b - |. b*((a-c)-a)/c .|) by A3
 .= max(0, b - |. (-1)*(c*b)/c .|)
 .= max(0, b - |. (-1)*(c*b/c) .|) by XCMPLX_1:74
 .= max(0, b - |. (-c*b/c) .|)
 .= max(0, b - |. -b .|) by A2,XCMPLX_1:89
 .= max(0, b - -(- b)) by ABSVALUE:def 1,A1
 .= 0;
 f.(a+c) = max(0, b - |. b*((a+c)-a)/c .|) by A3
 .= max(0, b - |. b .|) by A2,XCMPLX_1:89
 .= max(0, b - b) by ABSVALUE:def 1,A1
 .= 0;
 hence thesis by A6,A8,INTEGRA1:5;
end;
