A Modal Solution to the Skeptical Paradox

       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: 

            ~◊K_S(~q) 

By De Morgan's law for modal operators, this gives us: 

            □~K_S(~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 ◊~K_S(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 ~K_S(~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 ◊~K_S(~q) & ◊K_S(~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. □(K_S(p) à K_S(~q))                                                Epistemic Necessitation Rule

            P2. (K_S(p) & ◊~K_S(p)) v (~K_S(p) & ◊K_S(p))        Contingent Everyday Knowledge

                        1. ~◊K_S(~q)                                                        Supposition, Strong SH

                        2. □~K_S(~q)                                                        1, Modal De Morgan’s Law

                        3. ~K_S(~q)                                                           2, Axiom T

                        4. ~K_S(p)                                                            3, P1, Modus Tollens (with Axiom T)

                                    5. (K_S(p) & ◊~K_S(p))                           Supposition, first disjunct P2

                                    6. K_S(p)                                                  5, Simplification

                                    7. K_S(p) & ~K_S(p)                               4, 6, Conjunction

                        8. ~(K_S(p) & ◊~K_S(p))                                    5-7, Indirect Proof

                        9. (~K_S(p) & ◊K_S(p))                                      8, P2, Disjunctive Syllogism

                        10. □(~K_S(~q) à ~K_S(p))                               P1, Transposition

                        11. □~K_S(~q) à □~K_S(p)                               10, Axiom K

                        12. □~K_S(p)                                                      2, 11, Modus Ponens

                        13. ~◊K_S(p)                                                      12, Modal De Morgan’s Law

                        14. ◊K_S(p)                                                         9, Simplification

                        15. ◊K_S(p) & ~◊K_S(p)                                    13, 14, Conjunction

            16. ~~◊K_S(~q)                                                               1-15, Indirect Proof

            17. ◊K_S(~q)                                                                  16, Double Negation

                QED

     The Contingent Everyday Knowledge (CE) premise (K_S(p) & ◊~K_S(p)) v (~K_S(p) & ◊K_S(p)) by Axiom D gives us (K_S(p) & ◊~K_S(p) & ◊K_S(p)) v (~K_S(p) & ◊~K_S(p) & ◊K_S(p)) and by Distribution gives us (◊~K_S(p) & ◊K_S(p)) & (K_S(p) v ~K_S(p)), so that by Tautology Elimination we get: ◊K_S(p) & ◊~K_S(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: □(K_S(p) à K_S(~q)). We need EN above (in P1-17, QED) to show that CE & EN ⊢ ◊K_S(~q) (i.e. to disprove the first premise of the Skeptical Paradox). If we also grant that ◊~K_S(~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  ◊~K_S(~q) & ◊K_S(~q), which we can call the Weak Skeptical Hypothesis (WSH). Then we can conjoin CE and WSH to get (◊K_S(p) & ◊~K_S(p)) & (◊K_S(~q) & ◊~K_S(~q)), which we will call general Epistemic Contingency (EC). Likewise, given the Epistemic Background Assumption (EB) of the possibility of knowledge ◊K_S(p), together with EN & WSH, we can derive ◊~K_S(p) & ◊K_S(p), that is, CE. 

P1. ◊K_S(~q) & ◊~K_S(q)             Weak Skeptical Hypothesis (WSH)

P2. □(K_S(p) à K_S(~q))            Epistemic Necessitation (EN) 

P3. ◊K_S(p)                                   Epistemic Background (EB) 

1. □K_S(p) à □K_S(~q)               P2, Axiom K  

        2. □K_S(p)                                Supposition  

        3. □K_S(~q)                            1, 2, Modus Ponens  

        4. ◊~K_S(q)                               P1, Simplification  

        5. ~□K_S(~q)                           4, Modal De Morgan’s Rule  

        6. □K_S(~q) & ~□K_S(~q)          3, 5, Conjunction 

7. ~□K_S(p)                                    2-6, Indirect Proof 

8. ◊~K_S(p)                                      7, Modal De Morgan’s Rule 

9. ◊K_S(p) & ◊~K_S(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:

◊K_S(p) & ◊~K_S(p)             Contingent Everyday Knowledge (CE)

◊~K_S(~q) & ◊K_S(~q)         Weak Skeptical Hypothesis (WSH)

◊K_S(p) & ◊~K_S(~q)           Background Assumption (BA)

□(K_S(p) à K_S(~q))           Epistemic Necessitation (given p ⊨ ~q) (EN)

 _______________

(EN & BA) à  (CE ↔ WSH)

 _______________

 _______________

Where E := K_S(p) and SH := ~K_S(~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.

The Rationality of Knowledge: On the Logical Necessity of the Epistemic Closure Principle

The Specificity of Skepticism: On the Irrelevance of Hinge Commitments to the Skeptical Paradox

Measurement by Model & Model as Measurement: On the Variety of Relations between Scientific Models and Measurements

Sketch 2, "Theatre," 4.2020

Sketch 1, 4.2020