On 24 Mrz., 11:19, William Hughes <wpihug...@gmail.com> wrote: > On Mar 24, 11:04 am, WM <mueck...@rz.fh-augsburg.de> wrote: > > Your proof covers all lines. We have for all lines l > of the list. > > if l and all its predecessors are removed > and no other line is removed, > then the union of all lines is not changed" > > However, there is no information about what will > happen if you try to apply this to two > lines e.g. l along with all its predecessors > and m along with all its predecessors.
Nothing will "happen". Induction holds for every line, so it holds for all lines of the set of finite lines. > > Now it is easy to see what will happen in this > case. Since we can replace l and m with > one of either l or m, we know what will happen > if we remove two lines. > > Since we can replace l,m and p with > one of either l or m or p, we know what will happen > if we remove three lines.
This way is not necessary, since the proof holds for every line including all its predecessors. Therefore it holds for l, m, and p (because they belong to a set of predecessors. This is a property of the natural numbers and dos not require any further attention.
> > It is easy to see we know what > will happen if we remove a natural > number of finite lines. > > However, we do not know what will happen > if we remove an infinite number of > finite lines.
That's why we use induction. Induction holds for all elements of the infinite set of natural numbers.