 reserve i,j,k,m,n,m1,n1 for Nat;
 reserve a,r,r1,r2 for Real;
 reserve m0,cn,cd for Integer;
 reserve x1,x2,o for object;

theorem Th2:
  r is irrational implies
    rfs(r).n is irrational
  proof
    assume
A1: r is irrational;
    set rn = rfs(r).n;
A3: scf(rn).0 = [\ rfs(rn).0 /] by REAL_3:def 4
    .= [\ rfs(r).n /] by REAL_3:def 3
    .= scf(r).n by REAL_3:def 4;
A4: scf(rn).m = scf(r).(n+m) & rfs(rn).m = rfs(r).(n+m)
    proof
      defpred P[Nat] means
      scf(rn).$1 = scf(r).(n+$1) & rfs(rn).$1 = rfs(r).(n+$1);
A5:   P[0] by REAL_3:def 3,A3;
A6:   for m be Nat st P[m] holds P[m+1]
      proof
        let m be Nat;
        assume
A7:     P[m];
A9:     rfs(rn).(m+1) = 1 / frac(rfs(rn).m) by REAL_3:def 3
            .= rfs(r).(n+m+1) by A7,REAL_3:def 3;
        scf(rn).(m+1)= [\ rfs(rn).(m+1) /] by REAL_3:def 4
            .= scf(r).(n+m+1) by A9,REAL_3:def 4;
        hence thesis by A9;
      end;
      for m be Nat holds P[m] from NAT_1:sch 2(A5,A6);
      hence thesis;
    end;
    assume
A10:rfs(r).n is rational;
    consider n1 such that
A12:for m1 st m1 >= n1 holds scf(rn).m1 = 0 by A10,REAL_3:42;
A13:for m1 st m1 >= n1 holds scf(r).(n+m1) = 0
    proof
      let m1;
      assume
A14:  m1 >= n1;
      scf(r).(n+m1)=scf(rn).m1 by A4 .= 0 by A12,A14;
      hence thesis;
    end;
    for m st m >= n1+n holds scf(r).m = 0
    proof
      let m;
      assume
A15:  m >= n1+n;
A16:  m -n >= n1+n - n by A15,XREAL_1:9;
      m - n in NAT by A16,INT_1:3; then
      reconsider l = m - n as Nat;
      scf(r).(n+l) = 0 by A13,A16;
      hence thesis;
    end;
    hence contradiction by A1,REAL_3:42;
  end;
