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 Th11:
  for X being BCI-Algebra_with_Condition(S) holds for x,y,z being
  Element of X holds (x\y)\z = x\(y*z)
proof
  let X be BCI-Algebra_with_Condition(S);
  let x,y,z be Element of X;
  (x\((x\y)\z))\y = (x\y)\((x\y)\z) by BCIALG_1:7
    .= ((x\y)\0.X)\((x\y)\z) by BCIALG_1:2;
  then ((x\((x\y)\z))\y)\(z\0.X) = 0.X by BCIALG_1:11;
  then (x\((x\y)\z))\y <= z\0.X;
  then (x\((x\y)\z))\y <= z by BCIALG_1:2;
  then x\((x\y)\z) <= y*z by Lm2;
  then (x\((x\y)\z))\(y*z) = 0.X;
  then
A1: (x\(y*z))\((x\y)\z) = 0.X by BCIALG_1:7;
  (y*z)\y <= z by Lm2;
  then ((x\y)\(x\(y*z)))\((y*z)\y) = 0.X & ((y*z)\y)\z = 0.X by BCIALG_1:11;
  then ((x\y)\(x\(y*z)))\z = 0.X by BCIALG_1:3;
  then ((x\y)\z)\(x\(y*z)) = 0.X by BCIALG_1:7;
  hence thesis by A1,BCIALG_1:def 7;
end;
