On the Circularity in the Sorites Paradox

I begin by highlighting the importance of the step size in the induction step of the sorites paradox. A careful analysis reveals that the step size can be characterised as a proper instance of the concept very small. After having accurately described the structure of sorites-susceptible predicates, I argue that the structure of the induction step in the Sorites Paradox is inherently circular. This circularity emerges in the structure of Wang's paradox and also of the classical variations of the paradox with the young, bald, etc. predicates. 1. Elements of the Paradox The sorites paradox (thereafter, SP) is one of the most ancient and unresolved paradoxes, which is attributed to Eubulides. It runs as follows: (B1) a collection with 100000 grains of sand is a heap (I1+) if a collection with n grains of sand is a heap then a collection with n 1 grains of sands is a heap (C1) ∴ a collection with 1 grain of sand is a heap Given the two prima facie indisputable premises, one is then led, after a seemingly legitimate repetition of modus ponens, to the unpalatable conclusion that a collection with 1 grain of sand is a heap. Other variations of SP involve vague predicates such as young, small, bald, etc. Let us proceed, for the sake of accuracy, to highlight some elements of the internal structure of SP. To begin with, it is worth drawing a distinction between the incremental and decremental versions of SP. P being a vague predicate, we have then the incremental version (SP+) of SP: (B2) P(1) (I2+) if P(n) then P(n + 1) (C2) ∴ P(10) Let also Q be a vague predicate such as Q = ~P. We have then the decremental version (SP-): (B3) Q(10) (I3-) if Q(n) then Q(n 1) (C3) ∴ Q(1) On the other hand, it is worth considering the step size notion within the induction step (I). Let us denote this step size by s. Indeed, the induction step has the following structure: (I4+) if P(n) then P(n + s) This leads to consider a hierarchy of versions of the induction step: (I1) if P(n) then P(n + 1) (I2) if P(n) then P(n + 2) 1 One considers here (SP+). Concerning (SP-), one has then: (I-) if P(n) then P(n s).