
theorem Th1:
  for n,m,k being Nat holds (n+k)-'(m+k)=n-'m
proof
  let n,m,k be Nat;
A1: n+k-(m+k)=n-m;
  per cases;
  suppose
    n-m>=0;
    then n-'m=n-m by XREAL_0:def 2;
    hence thesis by A1,XREAL_0:def 2;
  end;
  suppose
A2: n-m<0;
    then n-'m=0 by XREAL_0:def 2;
    hence thesis by A1,A2,XREAL_0:def 2;
  end;
end;
