reserve A for non empty closed_interval Subset of REAL;

theorem Th12:
for f,g be Function of REAL,REAL st
f is_integrable_on A & f | A is bounded &
g is_integrable_on A & g | A is bounded
holds
min(f,g) is_integrable_on A & min(f,g) | A is bounded &
integral(min(f,g),A)
 = (1/2)*( integral(f,A) + integral(g,A) - integral(abs(f-g),A) )
proof
 let f,g be Function of REAL,REAL;
 assume that
 A1A:  f is_integrable_on A & f | A is bounded and
 A2A:  g is_integrable_on A & g | A is bounded;
 reconsider fm = min(f,g) as PartFunc of REAL,REAL;
 reconsider f as PartFunc of REAL,REAL;
 reconsider g as PartFunc of REAL,REAL;
 A1:  f is_integrable_on A & f | A is bounded by A1A;
 A2:  g is_integrable_on A & g | A is bounded by A2A;
 Df: REAL = dom f by FUNCT_2:def 1;
 Dg: REAL = dom g by FUNCT_2:def 1;
 Dm: REAL = dom (f-g) by FUNCT_2:def 1;
 Dp: REAL = dom (f+g) by FUNCT_2:def 1;
 Da: REAL = dom abs(f-g) by FUNCT_2:def 1;
 DD: REAL = dom ((f+g) - abs(f-g)) by FUNCT_2:def 1;
 i1:  (f+g) is_integrable_on A by A1,A2,INTEGRA6:11,Df,Dg;
 B1: (f+g) | (A /\ A) is bounded by A1A,A2A,RFUNCT_1:83;
 B2: (f-g) | (A /\ A) is bounded by A1A,A2A,RFUNCT_1:84;
 (f-g) is_integrable_on A by A1,A2,INTEGRA6:11,Dg,Df; then
 i3:  (abs(f-g)) is_integrable_on A by B2,INTEGRA6:7,Dm;
 B3:  (abs(f-g)) | (A /\ A) is bounded by B2,RFUNCT_1:82;
 i4: ((f+g) - abs(f-g)) is_integrable_on A
          by i1,i3,B1,B3,INTEGRA6:11,Dp,Da;
 B4: ((f+g) - abs(f-g)) | (A /\ A) is bounded by B1,B3,RFUNCT_1:84;
 i5: (1/2)(#)((f+g) - abs(f-g)) is_integrable_on A
          by i4,B4,INTEGRA6:9,DD;
 B5: ( (1/2)(#)((f+g) - abs(f-g)) ) | (A /\ A) is bounded
          by B4,RFUNCT_1:80;
 integral(fm,A) = integral(( (1/2)(#)((f+g) - abs(f-g)) ),A) by Th13
 .= (1/2)*integral( (f+g) - abs(f-g),A) by INTEGRA6:9,i4,B4,DD
 .= (1/2)*( integral(f+g,A) - integral(abs(f-g),A) )
    by INTEGRA6:11,i1,B1,i3,B3,Da,Dp;
 hence thesis by Th13,i5,B5,INTEGRA6:11,A1,A2,Df,Dg;
end;
