reserve a,b,m,x,y,i1,i2,i3,i for Integer,
  k,p,q,n for Nat,
  c,c1,c2 for Element of NAT,
  z for set;

theorem Th19:
  n>0 & p>0 implies p*q mod p|^n = p*(q mod p|^(n-'1))
proof
  assume that
A1: n>0 and
A2: p>0;
A3: n>=0+1 by A1,NAT_1:13;
  p*(q mod p|^(n-'1)) = p*(q-(q div p|^(n-'1))*p|^(n-'1)) by A2,NEWTON:63
    .= p*q-(q div p|^(n-'1))*(p*p|^(n-'1))
    .= p*q-(q div p|^(n-'1))*(p|^(n-'1+1)) by NEWTON:6
    .= p*q-(q div p|^(n-'1))*p|^n by A3,XREAL_1:235;
  then
A4: p*q=(q div p|^(n-'1))*p|^n+p*(q mod p|^(n-'1));
  p*(q mod p|^(n-'1)) < p*p|^(n-'1) by A2,NAT_D:1,XREAL_1:68;
  then p*(q mod p|^(n-'1)) < p|^(n-'1+1) by NEWTON:6;
  then p*(q mod p|^(n-'1)) < p|^n by A3,XREAL_1:235;
  hence thesis by A4,NAT_D:def 2;
end;
