reserve X for non empty BCIStr_1;
reserve d for Element of X;
reserve n,m,k for Nat;
reserve f for sequence of  the carrier of X;

theorem
  for X being non empty BCIStr_1 holds (X is positive-implicative
  BCK-Algebra_with_Condition(S) iff X is semi-Brouwerian-algebra)
proof
  let X be non empty BCIStr_1;
A1: X is semi-Brouwerian-algebra implies X is positive-implicative
  BCK-Algebra_with_Condition(S)
  proof
    assume
A2: X is semi-Brouwerian-algebra;
A3: for x,y being Element of X holds (x\y=0.X & y\x=0.X implies x = y)
    proof
      let x,y be Element of X;
      assume that
A4:   x\y=0.X and
A5:   y\x=0.X;
A6:   x = x*x by A2,Def12
        .= (x\x)*x by A2,Def14
        .= 0.X * x by A2,BCIALG_1:def 5
        .= y*x by A2,A5,Def14;
      y = y*y by A2,Def12
        .= (y\y)*y by A2,Def14
        .= y*(y\y) by A2,Def13
        .= y*0.X by A2,BCIALG_1:def 5
        .= (x\y)*y by A2,A4,Def13
        .= x*y by A2,Def14;
      hence thesis by A2,A6,Def13;
    end;
A7: for x,y,z being Element of X holds (x\y)\z = (x\z)\y
    proof
      let x,y,z be Element of X;
      (x\y)\z = x\(y*z) by A2,Def2
        .= x\(z*y) by A2,Def13;
      hence thesis by A2,Def2;
    end;
A8: for x,y,z being Element of X holds ((x\y)\z)\((x\z)\y)=0.X
    proof
      let x,y,z be Element of X;
      ((x\y)\z)\((x\z)\y) = ((x\y)\z)\((x\y)\z) by A7;
      hence thesis by A2,BCIALG_1:def 5;
    end;
A9: for x,y being Element of X holds x*y = x*(y\x)
    proof
      let x,y be Element of X;
      x*(y\x) = (y\x)*x by A2,Def13
        .= y*x by A2,Def14;
      hence thesis by A2,Def13;
    end;
A10: for x being Element of X holds x`=0.X
    proof
      let x be Element of X;
      0.X \ x = (x\x)\x by A2,BCIALG_1:def 5
        .= x\(x*x) by A2,Def2
        .= x\x by A2,Def12
        .= 0.X by A2,BCIALG_1:def 5;
      hence thesis;
    end;
    for x,y,z being Element of X holds ((x\y)\(z\y))\(x\z)=0.X
    proof
      let x,y,z be Element of X;
      ((x\y)\(z\y))\(x\z) = (x\(y*(z\y)))\(x\z) by A2,Def2
        .= (x\((z\y)*y))\(x\z) by A2,Def13
        .= (x\(z*y))\(x\z) by A2,Def14
        .= ((x\z)\y)\(x\z) by A2,Def2
        .= ((x\z)\(x\z))\y by A7
        .= y` by A2,BCIALG_1:def 5;
      hence thesis by A10;
    end;
    hence thesis by A2,A10,A3,A8,A9,Th47,BCIALG_1:def 3,def 4,def 7,def 8;
  end;
  X is positive-implicative BCK-Algebra_with_Condition(S) implies X is
  semi-Brouwerian-algebra
  proof
    assume
A11: X is positive-implicative BCK-Algebra_with_Condition(S);
A12: for x,y being Element of X holds (x\y) * y = x * y
    proof
      let x,y be Element of X;
      y*x = y*(x\y) by A11,Th47;
      then x*y = y*(x\y) by A11,Th6;
      hence thesis by A11,Th6;
    end;
    ( for x being Element of X holds x*x = x)& for x,y being Element of X
    holds x * y = y * x by A11,Th6,Th44;
    hence thesis by A11,A12,Def12,Def13,Def14;
  end;
  hence thesis by A1;
end;
