Logic for Computer Scientists/Induction

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Induction plays a crucial role at least in two aspect throughout this book. Firstly, it is one of main proof principles in mathematics and of course in logics. In particular it can be used to investigate properties of infinite sets. Very often it is used as natural induction, namely over the natural numbers. We will introduce it as a more general principle over well founded partial orders, which is called structural induction. The second aspect is, that it can be used as well to define infinite structures, as the set of well formed formulae in a particular logic or the set of binary trees.

We start with a very general structure over arbitrary sets, namely partial orders. A relation   over   is a partial order iff   is reflexive, transitive and anti-symmetric (i.e.  ) . Partial ordered sets (p.o. sets)   are usually written as  .

Definition 1Edit

The necessary structure for our induction principle is a partial order, such that there exist minimal elements. Given a p.o. set  , we define:

  •   iff   and  .
  •   is called well-founded iff there is no infinite sequence   and  .
  •   is called a chain iff   or  
  •   is a total ordering iff   is a chain.

Lemma 1Edit

  is well-founded, if every non-empty subset of   has a minimal Element. Proof can be done by contradiction and will be given as an exercise.

We finally have the machinery to introduce the principle of complete induction:

Definition 2 (Complete (structural) induction)Edit

Given a well-founded p.o. set   and a predicate  , i.e.  . The principle of induction is given by the following (second order) formula

Lemma 2Edit

The induction principle holds for every well-founded set.

Proof: The proof is given by contradiction: Assume the principle is wrong; i.e. the implication is wrong, which means, that we have to assume the premise as true:


and the conclusion as wrong:


Hence we can assume that the set   is not empty.

Since   is a subset of a well-founded set, it has a minimal element, say  . From assumption (premise) we conclude


Now we can distinguish two cases:

  •   is minimal in  : Hence there is no  , such that  . Hence the premise   of the implication in (inst) is true, which implies that the conclusion   is true. This is a contradiction to the assumption that
  •   is not minimal in  :   holds and it must be that   is true, because otherwise it would be that   and   not minimal in  . Hence, again the premise   of the implication in (inst) is true, which implies that the conclusion   is true. This is a contradiction to the assumption that  !

An ExampleEdit

In this subsection we will carry out a proof with induction in detail. For this we need the extension of p.O. sets:

Definition 3 (Lexicographic Ordering)Edit

A p.O. set   induces an ordering   over  :   iff

  •   and   or
  •   or
  •   and  

Lemma 3Edit

If   is a well-founded set, then   is well-founded as well.

Theorem 1Edit

The Ackermann-function   defined by the following recursion is a total function over  

  ACK(x,y) = 
    if x=0  then y+1 else
                     if y=0 then ACK(x-1,1) 
                            else ACK(x-1,ACK(x, y-1))

Proof: For the induction start we take the minimal element   of the well-founded set,
 , where   is the lexicographic ordering induced by  . Hence, assume  . By definition of
 , we conclude   and hence defined. Assume for an an arbitrary  , that

  is defined for all  , if  

We distinguish the following cases:

  •  : i.e.   and hence defined.
  •   and  : We know that   and  . From the induction hypothesis we know that   is defined, and hence   as well.
  •   and  : According to the definition of   we have two cases to consider:
    •   and  : From the induction hypothesis we conclude immediately that   is defined.
    •   and  : Independent from the values of   and  . If we assume
        and  , we again can conclude from the hypothesis, that   is defined and hence   as well

Altogether we proved, that   is defined for all  .


Problem 1 (Induction)Edit

Prove the following lemma: If   is well founded also  .

Note: The lexicographical Order   is defined as follows:


Problem 2 (Induction)Edit

How many points of intersection could   straight lines have at most? Find a recursive and explicit formula and show

Problem 3 (Induction)Edit

Prove that a number   is even if and only if   is even.

Problem 4 (Induction)Edit

Point by an indirect proof that there is not any greatest prime number!

Problem 5 (Induction)Edit

Which prerequisite do you need that the following order


  1. partial ordered
  2. total ordered
  3. well-founded?

Problem 6 (Induction)Edit

An example of a well founded set is the power set   over a finite set   which is comparable over the relation of the subset  . If   then is e.g.   but   and   are not comparable. Give a definition of a relation   in this way that   is total ordered and well founded.

Problem 7 (Induction)Edit

Examine which of the following partial order are total and which are well founded!

  1.   with   is the power set for natural numbers.
  2.   with   marks the relation "`is factor of"'.
  3.   with   iff   or (  and  ).
  4.   with   is the lexicographical .
  5. for  , i.e.  

Problem 8 (Induction)Edit

Give for the natural numbers   an order relation, that is

  1. both well-founded and total,
  2. total but not well-founded,
  3. well-founded but not total and
  4. neither well-founded nor total.

Problem 9 (Induction)Edit

Prove, that a partial order   is well-founded iff every non-empty partial set of   (at least) contains a minimum element.

Problem 10 (Induction)Edit

A root tree consists (a) of a single node or (b) of a node - that's the root of the tree - and at least one, but only at most finite many (part)trees, this one is bandaged over an edge with the root. Point formally by means of induction, that in every tree the number of the nodes   around   is taller than the number of the edges  , i.e.  .

Problem 11 (Induction)Edit

Prove: If ε is a quality of the natural numbers   and it is valid

  1. ε(0) and
  2.   : [ε(n)   ε(n+1)].

then   : ε(n) is valid.

Note: The proof can be done by the fact that the principle of the complete induction in   (which should be proved) can be reduced to the principle of the transfinite induction for well founded orders.