
theorem Th6:
  for m,n being Nat holds n = (m-'(m-'n))+(n-'m)
  proof
    let m,n be Nat;
    per cases;
    suppose m <= n;
      then
A1:   m-'n = 0 & n-'m = n-m by NAT_2:8,XREAL_1:233;
      then m-'(m-'n) = m by NAT_D:40;
      hence thesis by A1;
    end;
    suppose m > n;
      then n-'m = 0 & m-'(m-'n) = n & n+0 = n by NAT_D:58,NAT_2:8;
      hence thesis;
    end;
  end;
