
theorem MT:
  for a be Nat, b be non trivial Nat,c be non zero Nat holds
    a mod (b*c) = 1 implies a mod b = 1
  proof
    let a be Nat, b be non trivial Nat, c be non zero Nat;
    reconsider k = b - 1 as non zero Nat;
    per cases;
    suppose c is trivial; then
      c = 1 by NAT_2:def 1;
      hence thesis;
    end;
    suppose c is non trivial; then
    reconsider l = c - 1 as non zero Nat;
    assume a mod (b*c) = 1; then
    a mod (b*c) = 1 mod (b*c); then
    A2: b*c divides a - 1 by INT_4:23;
    b divides b*c; then
    b divides (a - 1) by A2,INT_2:9; then
    a mod b = 1 mod (1 + k) by INT_4:23;
    hence thesis;
  end;
end;
