 Here \(n\) and \(m\) range over natural numbers, \(\alpha\) and \(\beta\) over choice sequences, and \(\beta\in\alpha(\overline{m})\) means that the first \(m\) elements of \(\alpha\) and \(\beta\) are equal. Although until now there has never been given a completely satisfactory justification of most continuity axioms for arbitrary choice sequences, not even by Brouwer, when restricted to the class of lawless sequences arguments supporting the validity of the weak continuity axiom run as follows. When could a statement of the form \(\forall\alpha\exists n A(\alpha,n)\) be established by the intuitionist? By the very nature of the notion of lawless sequence, the choice of the number \(n\) for which \(A(\alpha,n)\) holds has to be made after only a finite initial segment of \(\alpha\) is known. For we do not know how \(\alpha\) will proceed in time, and we therefore have to base the choice of \(n\) on the initial segment of \(\alpha\) that is known at that point in time where we wish to fix \(n\). This implies that for every lawless sequence \(\beta\) with the same initial segment as \(\alpha\), \(A(\beta,n)\) holds as well. 