Topics Mathematics

Question What does it mean when a certain axiom is neither provable nor deniable? Does it imply that such axiom is self-evident and can't be doubted? I don't think that "real skeptics"(a skeptic who is so deep in doubt that he doubts his own existence and even his own doubt) like Pyrrho would be happy with that.

Accepted:
May 2, 2013

Comments

Let's consider, for example, what philosopher Hilary Putnam has called "the minimal principle of contradiction":

(MPC) Not every contradiction is true.

Arguably, MPC is unprovable because whichever premises and inference rules we might use to try to prove MPC are no better-known by us, and no more securely correct, than MPC itself is. But MPC would also appear to be undeniable, since in standard logic to deny MPC is to imply that every contradiction is true, and it's hard (for me, anyway) to make any sense of the notion of denying something in circumstances in which every contradiction is true.

So, arguably, MPC is self-evident and can't be doubted: that is, the notion of MPC's being doubted makes no sense. You suggest that this result would bother a skeptic who doubts his own existence and even doubts (the fact of) his own doubt. Such a skeptic seems committed to the possibility that (i) he doubts something even while he himself fails to exist and (ii) he entertains a doubt even when no doubt exists to be entertained. Both (i) and (ii) look self-inconsistent to me. If the skeptic would be willing to accept inconsistencies such as (i) and (ii), then I don't see why he should be bothered by the undeniability of MPC: he could concede that MPC is undeniable and then (claim to) deny it anyway!

Descartes famously explored these issues: see this link.