Here is the infamous Skeptical Paradox: (1) One cannot know that one is not being deceived in a skeptical scenario, such as a dream or a simulation (i.e. one cannot know that the skeptical hypothesis is false); (2) if one knows some everyday things about reality are true (call this E), then one knows that one is not in a skeptical scenario; (3) one knows some everyday things about reality are true. Clearly, if all three premises are true, then one must conclude that one knows the skeptical hypothesis is false and one does not know the skeptical hypothesis is false, a contradiction.
First, one would be right to suspect that the "cannot" in the first premise is rather strong. Some (e.g. Pritchard, 2013) have argued that the weaker formulation “one does not know one is not in a skeptical scenario” (call this SH) works well enough to get the paradox. However, it is not clear that one could get the paradox from the weaker version. If the skeptic is not claiming that one cannot know, then her interlocutor, for all she knows, does know, since it is possible to know. In order for the skeptic to say that any given person does not know, she must, ipso facto, also claim that one cannot know. Also, as long as a potential resolution is available, we do not have a strict paradox, but only a temporary obstruction. Thus, it seems that to get the paradox per se, we do need the strong version of the first premise. And it is in the strength of this premise, which actually imposes a necessity condition, that we may look for a solution to the paradox.
Here is my modal
solution to the skeptical paradox: The statement, "One cannot know one is not in some skeptical
scenario (q)" is actually very strong. In epistemic modal logic, it would
be:
~◊KS(~q)
By De Morgan's law for modal operators, this gives us:
□~KS(~q)
I.e. "It is necessarily the case that one does not
know one is not in some skeptical scenario (q)."
Why is it
necessarily the case? It could be that there would always be a way I could
know, in all possible worlds. It could even be that I would always
know, if I were. It hasn't been proven to me that these counter claims are
false, so I can't assent to the first premise. I have no basis for believing it
is necessarily the case. That is, to dispense with the entire problem,
one doesn't even have to deny that it is necessarily the case that one
doesn’t know. One just has to say that one has no reason to assent to this
claim. There is no compelling reason to believe it. This makes the entire
problem only a hypothetical problem, a problem only when one assents to
something one has no grounds for believing, and not a real epistemic
problem.
Perhaps one might then claim that if I do assent to the third
premise and the whole closure principle, then I can deduce that I am not in a
skeptical scenario and thereby know it, given that I have no reason to dispute
this. So, I am only defaulting back to the neo-Moorean position by a different
route. However, in saying that I am neither denying nor affirming that it is
necessarily the case that I don’t know I’m not in a skeptical scenario, I also
am affirming that it is possible for me to not have everyday
knowledge, though at present I do, contingently. What this contingency
is is not asserted or given, as this is a separate question, independent of my
assertion of present contingent everyday knowledge (of some everyday fact p).
That is, we avoid the simple Moorean rejection of skeptical scenarios here by asserting ◊~KS(p), together with the assertion of E, which is the same as asserting that it is not necessary that one possesses everyday knowledge. Call this the contingent everyday knowledge concession to radical skepticism. What is one’s knowledge contingent on? That at some point one could have a reason to assert that ~KS(~q). If some evidence were presented that caused one to assert this, then one could no longer uphold the everyday knowledge claim. This keeps the first conjunct in ◊~KS(~q) & ◊KS(~q) active.
This fragility of our everyday knowledge, contingent on the possibility of some justifying evidence for not knowing if we are being deceived, seems right. This is the truth that skeptical scenarios reveal. Through our modal solution, we can retain this truth without getting ourselves stuck in a paradox.
We can state this solution more concisely, as a reformulation of all three premises. First, from two premises, the Epistemic Necessitation Rule and the Contingent Everyday Knowledge assumption, we can prove that the first premise of the Skeptical Paradox is false:
P1. □(KS(p) Ã KS(~q)) Epistemic
Necessitation Rule
P2. (KS(p) & ◊~KS(p)) v
(~KS(p) & ◊KS(p)) Contingent Everyday
Knowledge
1. ~◊KS(~q) Supposition,
Strong SH
2. □~KS(~q) 1,
Modal De Morgan’s Law
3. ~KS(~q) 2,
Axiom T
4. ~KS(p) 3, P1, Modus Tollens (with Axiom T)
5. (KS(p) & ◊~KS(p)) Supposition,
first disjunct P2
6. KS(p) 5, Simplification
7. KS(p) & ~KS(p) 4,
6, Conjunction
8. ~(KS(p) & ◊~KS(p)) 5-7, Indirect Proof
9. (~KS(p) & ◊KS(p)) 8, P2, Disjunctive Syllogism
10. □(~KS(~q) Ã ~KS(p)) P1, Transposition
11. □~KS(~q) Ã □~KS(p) 10, Axiom K
12. □~KS(p) 2, 11, Modus Ponens
13. ~◊KS(p) 12, Modal De Morgan’s Law
14. ◊KS(p) 9, Simplification
15. ◊KS(p)
& ~◊KS(p) 13, 14, Conjunction
16. ~~◊KS(~q) 1-15,
Indirect Proof
17. ◊KS(~q) 16,
Double Negation
QED
The Contingent Everyday
Knowledge (CE) premise (KS(p) & ◊~KS(p)) v (~KS(p)
& ◊KS(p)) by Axiom D gives us (KS(p) & ◊~KS(p)
& ◊KS(p)) v (~KS(p) & ◊~KS(p) & ◊KS(p)) and by Distribution gives us (◊~KS(p) & ◊KS(p)) &
(KS(p) v ~KS(p)), so that by Tautology Elimination we get:
◊KS(p) & ◊~KS(p), as another form of CE.
Meanwhile, instead of the usual use of the closure principle on knowledge of the implication p à ~q, we can say that if p necessarily (by definition) excludes q, then knowledge of p necessarily implies knowledge of the negation of q, since the de re knowledge referred to here is the same, or includes the same. We state this as the Epistemic Necessitation Rule (EN), given p ⊨ ~q: □(KS(p) à KS(~q)). We need EN above (in P1-17, QED) to show that CE & EN ⊢ ◊KS(~q) (i.e. to disprove the first premise of the Skeptical Paradox). If we also grant that ◊~KS(~q), as the sine non qua of the entire paradox, i.e. as the Skeptical Background Assumption (SB), then we can conjoin this to get ◊~KS(~q) & ◊KS(~q), which we can call the Weak Skeptical Hypothesis (WSH). Then we can conjoin CE and WSH to get (◊KS(p) & ◊~KS(p)) & (◊KS(~q) & ◊~KS(~q)), which we will call general Epistemic Contingency (EC). Likewise, given the Epistemic Background Assumption (EB) of the possibility of knowledge ◊KS(p), together with EN & WSH, we can derive ◊~KS(p) & ◊KS(p), that is, CE.
P1. ◊KS(~q) & ◊~KS(q) Weak Skeptical Hypothesis (WSH)
P2. □(KS(p) Ã KS(~q)) Epistemic Necessitation (EN)
P3. ◊KS(p) Epistemic Background (EB)
1. □KS(p) Ã □KS(~q) P2, Axiom K
2. □KS(p) Supposition
3. □KS(~q) 1, 2, Modus Ponens
4. ◊~KS(q) P1, Simplification
5. ~□KS(~q) 4, Modal De Morgan’s Rule
6. □KS(~q) & ~□KS(~q) 3, 5, Conjunction
7. ~□KS(p) 2-6, Indirect Proof
8. ◊~KS(p) 7, Modal De Morgan’s Rule
9. ◊KS(p) & ◊~KS(p) P3, 8, Conjunction
QED
If we then combine SB & EB as the Background Assumption (BA), then given EN & BA, we can show that CE and WSH are materially equivalent.
The Modal Solution:
◊KS(p) & ◊~KS(p) Contingent Everyday Knowledge (CE)
◊~KS(~q) & ◊KS(~q) Weak Skeptical Hypothesis (WSH)
◊KS(p) & ◊~KS(~q) Background Assumption (BA)
□(KS(p) Ã KS(~q)) Epistemic Necessitation (given p ⊨ ~q) (EN)
_______________
(EN & BA) Ã (CE ↔ WSH)
_______________
_______________
Where E := KS(p) and SH := ~KS(~q),
(◊E & ◊~E) & (◊SH & ◊~SH) Epistemic Contingency (EC)
(◊E & ◊~E) ↔ (◊SH & ◊~SH) Epistemic Equivalence (given EN & BA) (EE)
That is, we have derived from our solution an important result: Allow the necessary background assumptions of the problem, together with the concession that if it is a priori derived that everyday facts exclude a skeptical scenario, then it is necessary that if one knows everyday facts, then one knows one is not in a skeptical scenario. Now, from this, we find that the contingency of everyday knowledge is materially equivalent to the contingency of the skeptical hypothesis. This in turn shows that there is a strong correlation and mutual dependence between everyday knowledge and the skeptical hypothesis in their epistemic contingency.