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 commutative BCK-Algebra_with_Condition(S), a,b,c being
  Element of X st Condition_S(a,b) c= Initial_section(c) holds for x being
  Element of Condition_S(a,b) holds x <= c\((c\a)\b)
proof
  let X be commutative BCK-Algebra_with_Condition(S);
  let a,b,c be Element of X;
  assume
A1: Condition_S(a,b) c= Initial_section(c);
  for x being Element of Condition_S(a,b) holds x <= c\((c\a)\b)
  proof
    set u = c\((c\a)\b);
    let x be Element of Condition_S(a,b);
A2: (c\(c\x)) \ x = (c\x)\(c\x) by BCIALG_1:7
      .= 0.X by BCIALG_1:def 5;
    x in {t2 where t2 is Element of X: t2 <= c} by A1,TARSKI:def 3;
    then consider x2 being Element of X such that
A3: x=x2 and
A4: x2 <= c;
A5: x \ (c\(c\x)) = x\(x\(x\c)) by Def9
      .= x2\c by A3,BCIALG_1:8
      .= 0.X by A4;
    then
A6: ((c\(c\x)) \ (c\((c\a)\b))) \ (((c\a)\b)\(c\x)) = 0.X & x\u = (c\(c\x
    )) \ (c \((c\a)\b)) by A2,BCIALG_1:1,def 7;
    x in {t1 where t1 is Element of X: t1\a <= b};
    then
A7: ex x1 being Element of X st x=x1 & x1\a <= b;
    (((c\a)\b)\(c\x)) = ((c\a)\(c\x))\b by BCIALG_1:7
      .= ((c\(c\x))\a)\b by BCIALG_1:7
      .= (x\a)\b by A5,A2,BCIALG_1:def 7
      .= 0.X by A7;
    then x\u = 0.X by A6,BCIALG_1:2;
    hence thesis;
  end;
  hence thesis;
end;
