Imagine there is an unlimited supply of balls with us, with positive integers painted on each. It is very much possible that a particular positive integer is repeated many times on various balls. We do not know anything else about the distribution of numbers on the balls. Further imagine that there is a box with a ball initially within it.
To play the game we may do either of two things:
(1) We may remove a ball from the box and replace it with any finite number of balls, arbitrarily chosen, but constrained to have numbers painted on them which are less then the number of the ball removed. For example, if we have a number 1000 ball we may remove it and replace it by 6000 balls: 3000 balls on which 657 is painted, and 3000 balls on which 999 is painted.
(2) We may remove a ball from the box and not do any replacement.
After playing the game once we can continue playing provided there are balls left in the box.
The question is: Is it possible to play the ball game indefinitely? Can we ensure that the box will never be empty?
Smullyan himself answered this problem in the negative by using Konig’s lemma.
Theorem: The Smullyan game terminates after a finite number of steps.
Proof: For any play of the Smullyan game define a (rooted) tree as follows:
1. The vertices are the painted balls used during the game.
2. The root is the painted ball intially within the box.
3. The edges are given by the following criteria: There is an edge between the vertices and iff during some move of the game, the ball is among the balls which replaces the ball (or vice-versa).
By the definition of the game this is a rooted tree where each level is finite. Can there be an infinite path in the tree? No there can’t. This is because, by definition an infinite path would correspond to a infinite strictly decreasing sequence of positive integers which is absurd. By the contrapositive form of Konig’s lemma we can therefore conclude that the tree is finite. In other words only finitely many balls were used and so the game terminated after finitely many steps.
An alternate proof without using Konig’s lemma is available here.