reserve x,y,y1,y2 for set;
reserve C for non empty set;
reserve c for Element of C;
reserve f,h,g,h1 for Membership_Func of C;

theorem Th10:
  1_minus(max(f,g)) = min(1_minus(f),1_minus(g)) & 1_minus(min(f,g
  )) = max(1_minus(f),1_minus(g))
proof
A1: C = dom 1_minus(min(f,g)) & C = dom max(1_minus(f),1_minus(g)) by
FUNCT_2:def 1;
A2: for x being Element of C st x in C holds (1_minus(max(f,g))).x = min(
  1_minus(f),1_minus(g)).x
  proof
    let x be Element of C;
A3: (1_minus(max(f,g))).x =1 - max(f,g).x by Def5
      .=1 - max(f.x,g.x) by Def4
      .=1 - (f.x + g.x + |.f.x - g.x.|)/2 by COMPLEX1:74;
    min(1_minus(f),1_minus(g)).x =min((1_minus(f)).x,(1_minus(g)).x) by Def3
      .=min((1 - f.x),(1_minus(g)).x) by Def5
      .=min((1 - f.x),(1- g.x)) by Def5
      .=((1-f.x) + (1-g.x) - |.(1-f.x) - (1-g.x).|)/2 by COMPLEX1:73
      .=(2-f.x -g.x- |.-(f.x-g.x).|)/2
      .=(2-(f.x + g.x) - |.f.x-g.x.|)/2 by COMPLEX1:52
      .=1 - ((f.x + g.x) + |.f.x-g.x.|)/2;
    hence thesis by A3;
  end;
A4: for x being Element of C st x in C holds (1_minus(min(f,g))).x = max(
  1_minus(f),1_minus(g)).x
  proof
    let x be Element of C;
A5: (1_minus(min(f,g))).x =1 - min(f,g).x by Def5
      .=1 - min(f.x,g.x) by Def3
      .=1 - (f.x + g.x - |.f.x-g.x.|)/2 by COMPLEX1:73;
    max(1_minus(f),1_minus(g)).x =max((1_minus(f)).x,(1_minus(g)).x) by Def4
      .=max((1 - f.x),(1_minus(g)).x) by Def5
      .=max((1 - f.x),(1- g.x)) by Def5
      .=((1-f.x) + (1-g.x) + |.(1-f.x) - (1-g.x).|)/2 by COMPLEX1:74
      .=(2-f.x -g.x + |.-(f.x-g.x).|)/2
      .=(2-(f.x + g.x) + |.f.x-g.x.|)/2 by COMPLEX1:52
      .=1 - ((f.x + g.x) - |.f.x-g.x.|)/2;
    hence thesis by A5;
  end;
  C = dom 1_minus(max(f,g)) & C = dom min(1_minus(f),1_minus(g)) by
FUNCT_2:def 1;
  hence thesis by A1,A2,A4,PARTFUN1:5;
end;
