
theorem Th26:
  for C being non empty set, f being PartFunc of C,ExtREAL, c be
  Real st 0 <= c holds max+(c(#)f) = c(#)max+f & max-(c(#)f) = c(#)max-f
proof
  let C be non empty set;
  let f be PartFunc of C,ExtREAL;
  let c be Real;
  assume
A1: 0 <= c;
A2: dom max+(c(#)f) = dom(c(#)f) by MESFUNC2:def 2
    .= dom f by MESFUNC1:def 6
    .= dom max+f by MESFUNC2:def 2
    .= dom(c(#)max+f) by MESFUNC1:def 6;
  for x be Element of C st x in dom max+(c(#)f) holds (max+(c(#)f)).x = (c
  (#)max+f).x
  proof
    let x be Element of C;
    assume
A3: x in dom max+(c(#)f);
    then
A4: x in dom(c(#)f) by MESFUNC2:def 2;
    then x in dom f by MESFUNC1:def 6;
    then
A5: x in dom max+ f by MESFUNC2:def 2;
A6: (max+(c(#)f)).x = max((c(#)f).x,0) by A3,MESFUNC2:def 2
      .= max( c * f.x,0) by A4,MESFUNC1:def 6;
    (c(#)max+f).x =  c * max+f.x by A2,A3,MESFUNC1:def 6
      .= c * max(f.x,0) by A5,MESFUNC2:def 2
      .= max(c *(f.x), c * (0 qua ExtReal)) by A1,Th6;
    hence thesis by A6;
  end;
  hence max+(c(#)f) = c(#)max+f by A2,PARTFUN1:5;
A7: dom(max-(c(#)f)) = dom(c(#)f) by MESFUNC2:def 3
    .= dom f by MESFUNC1:def 6
    .= dom max-f by MESFUNC2:def 3
    .= dom(c(#)max-f) by MESFUNC1:def 6;
  for x be Element of C st x in dom max-(c(#)f) holds (max-(c(#)f)).x = (
  c(#)max-f).x
  proof
    let x be Element of C;
    assume
A8: x in dom max-(c(#)f);
    then
A9: x in dom(c(#)f) by MESFUNC2:def 3;
    then x in dom f by MESFUNC1:def 6;
    then
A10: x in dom max- f by MESFUNC2:def 3;
A11: (max-(c(#)f)).x = max(-(c(#)f).x,0) by A8,MESFUNC2:def 3
      .= max(-( c)*f.x,0) by A9,MESFUNC1:def 6;
    (c(#)max-f).x =  c * max-f.x by A7,A8,MESFUNC1:def 6
      .= c * max(-f.x,0) by A10,MESFUNC2:def 3
      .= max(c * (-f.x),c * (0 qua ExtReal)) by A1,Th6
      .= max(-(c)*f.x, c * (0 qua ExtReal)) by XXREAL_3:92;
    hence thesis by A11;
  end;
  hence thesis by A7,PARTFUN1:5;
end;
