reserve a,b,c,k,l,m,n for Nat,
  i,j,x,y for Integer;

theorem Th7:
  a gcd b = 1 implies (a + b) gcd b = 1
proof
  assume
A1: a gcd b = 1;
  set n = (a + b) gcd b;
A2: n divides b by NAT_D:def 5;
  n divides (a + b) by NAT_D:def 5;
  then n divides a by A2,NAT_D:10;
  then n divides a gcd b by A2,NEWTON:50;
  then
A3: n <= 1 + 0 by A1,NAT_D:7;
  per cases by A3,NAT_1:9;
  suppose
    n = 1;
    hence thesis;
  end;
  suppose
    n = 0;
    then b = 0 by INT_2:5;
    hence thesis by A1;
  end;
end;
