reserve f,g,h for Function,
  A for set;
reserve F for Function,
  B,x,y,y1,y2,z for set;
reserve x,z for object;
reserve X for non empty set,
  Y for set,
  F for BinOp of X,
  f,g,h for Function of Y,X,
  x,x1,x2 for Element of X;
reserve Y for non empty set,
  F for BinOp of X,
  f,g,h for Function of Y,X,
  x,x1,x2 for Element of X;
reserve Y for set,
  F for BinOp of X,
  f,g,h for Function of Y,X,
  x,x1,x2 for Element of X;
reserve Y for non empty set,
  F for BinOp of X,
  f,g,h for Function of Y,X,
  x,x1,x2 for Element of X;
reserve Y for set,
  F for BinOp of X,
  f,g,h for Function of Y,X,
  x,x1,x2 for Element of X;
reserve Y for non empty set,
  F for BinOp of X,
  f,g,h for Function of Y,X,
  x,x1,x2 for Element of X;
reserve Y for set,
  F for BinOp of X,
  f,g,h for Function
  of Y,X,
  x,x1,x2 for Element of X;
reserve y for Element of Y;

theorem
  F is associative implies F[:](F[;](x1,f),x2) = F[;](x1,F[:](f,x2))
proof
  assume
A1: F is associative;
  per cases;
  suppose
    Y = {};
    hence thesis;
  end;
  suppose
A2: Y <> {};
    now
      let y;
      reconsider x3 = f.y as Element of X by A2,FUNCT_2:5;
      thus (F[:](F[;](x1,f),x2)).y = F.((F[;](x1,f)).y,x2) by A2,Th48
        .= F.(F.(x1,x3),x2) by A2,Th53
        .= F.(x1,F.(x3,x2)) by A1
        .= F.(x1,(F[:](f,x2)).y) by A2,Th48;
    end;
    hence thesis by A2,Th54;
  end;
end;
