reserve a,b,m,x,n,l,xi,xj for Nat,
  t,z for Integer;

theorem Th12:
  for a,m,n being natural Number holds
  a <> 0 & m > 1 & a*n mod m = n mod m & m,n are_coprime
  implies a mod m = 1
proof
  let a,m,n be natural Number;
  assume that
A1: a <> 0 and
A2: m > 1 and
A3: a*n mod m = n mod m and
A4: m,n are_coprime;
  consider k2 be Nat such that
A5: n = m*k2+(n mod m) and
  n mod m < m by A2,NAT_D:def 2;
  consider k1 be Nat such that
A6: a*n = m*k1+(a*n mod m) and
  (a*n mod m) < m by A2,NAT_D:def 2;
  (a-1)*n = m*(k1-k2) by A3,A6,A5;
  then
A7: m divides (a-1)*n by INT_1:def 3;
  reconsider t = (a-1),m as Integer;
  m divides t by A4,A7,INT_2:25;
  then consider tt be Integer such that
A8: (a-1) = m*tt by INT_1:def 3;
  a-1 >= 0 by A1,Lm2;
  then
A9: tt >= 0 by A2,A8;
A10: a = m*tt+1 by A8;
  reconsider tt,m as Element of NAT by A9,INT_1:3;
  a mod m = (((m*tt) mod m)+1) mod m by A10,NAT_D:22
    .= (0+1) mod m by NAT_D:13
    .= 1 by A2,NAT_D:24;
  hence thesis;
end;
