reserve d,i,j,k,m,n,p,q,x,k1,k2 for Nat,
  a,c,i1,i2,i3,i5 for Integer;

theorem Th11:
  1 < m & (n*k) mod m = k mod m & k,m are_coprime implies n mod m = 1
proof
  assume that
A1: 1 < m and
A2: (n*k) mod m = k mod m and
A3: k,m are_coprime;
  consider t2 being Nat such that
A4: k = m*t2 + (k mod m) and
  (k mod m) < m by A1,NAT_D:def 2;
  consider t1 being Nat such that
A5: n*k = m*t1 + ((n*k) mod m) and
  ((n*k) mod m) < m by A1,NAT_D:def 2;
  n*k - 1*k = m*(t1 - t2) by A2,A5,A4;
  then
A6: m divides k*(n - 1);
  reconsider n,m as Integer;
  m divides (n - 1) by A3,A6,INT_2:25;
  then consider tt being Integer such that
A7: n - 1 = m*tt;
  n = m*tt + 1 by A7;
  then n mod m = (1 qua Integer) mod m by NAT_D:61
    .= 1 by A1,NAT_D:14;
  hence thesis;
end;
