reserve X for BCI-algebra;
reserve I for Ideal of X;
reserve a,x,y,z,u for Element of X;
reserve f,f9,g for sequence of  the carrier of X;
reserve j,i,k,n,m for Nat;

theorem
  x` is minimal iff for y,z holds (((x\z)\(y\z))`)` = y`\x`
proof
  thus x` is minimal implies for y,z holds (((x\z)\(y\z))`)` = y`\x`
  proof
    assume
A1: x` is minimal;
    let y,z;
    y`\(x\y)\x`=0.X by BCIALG_1:def 3;
    then
A2: y`\(x\y)<=x`;
    (((x\z)\(y\z))`)` = ((x\z)`\(y\z)`)` by BCIALG_1:9
      .= ((x`\z`)\((y\z)`))` by BCIALG_1:9;
    then (((x\z)\(y\z))`)` = ((y`\(x\y)\z`)\(y\z)`)` by A1,A2
      .= ((y`\z`\(x\y))\(y\z)`)` by BCIALG_1:7
      .= (((y\z)`\(x\y))\(y\z)`)` by BCIALG_1:9
      .= (((y\z)`)\((y\z)`)\(x\y))` by BCIALG_1:7
      .= ((x\y)`)` by BCIALG_1:def 5
      .= (x`\y`)` by BCIALG_1:9
      .= ((y`)`\x)` by BCIALG_1:7
      .= ((y`)`)`\x` by BCIALG_1:9
      .= y`\x` by BCIALG_1:8;
    hence thesis;
  end;
  thus (for y,z holds (((x\z)\(y\z))`)` = y`\x`) implies x` is minimal
  proof
    assume
A3: for y,z holds (((x\z)\(y\z))`)` = y`\x`;
    now
      let x1 be Element of X;
      assume x1<=x`;
      then
A4:   x1\x`=0.X;
      then (x1`\x`)\(x1`\x1)=0.X by BCIALG_1:4;
      then ((((x\0.X)\(x1\0.X))`)`)\(x1`\x1)=0.X by A3;
      then ((x1`\x1)`)\(((x\0.X)\(x1\0.X))`)=0.X by BCIALG_1:7;
      then ((x1`\x1)`)\((x\(x1\0.X))`)=0.X by BCIALG_1:2;
      then ((x1`\x1)`)\((x\x1)`)=0.X by BCIALG_1:2;
      then (((x1`)`)\x1`)\((x\x1)`)=0.X by BCIALG_1:9;
      then (((x1`)`)\x1`)\(x`\x1`)=0.X by BCIALG_1:9;
      then (((x1`)`)\x`)`=0.X by BCIALG_1:def 3;
      then (((x1`)`)`)\((x`)`)=0.X by BCIALG_1:9;
      then x1`\((x`)`)=0.X by BCIALG_1:8;
      then (((x`)`)`)\x1=0.X by BCIALG_1:7;
      then x`\x1=0.X by BCIALG_1:8;
      hence x1=x` by A4,BCIALG_1:def 7;
    end;
    hence thesis by Lm1;
  end;
end;
