I think that the answer is no, in this particular case : a brute fact is a fact which isn't grounded (explained) by anything at all, even itself. A self-explanatory fact is grounded and explained by itself, hence it is not brute. A self-explanatory fact implies to drop the irreflexivity of grounding and explanation; while a brute fact doesn't.

Thanks a lot ! I take it, then, that this is the proper logical form of the reductio I want to make :

(P1) Assume R
(P2) (P ˅ Q)
(P3) P → ~R
(P4) Q → ~R
(C1) ~R (reductio)

Would something like this work ?


P ˅ Q
P ⊃~R
Q ⊃ ~R
Hence: ~R

I understand better, thanks for your help.

Here's what I am struggling with, then :

Take principle P. Consider two philosophical doctrines, X and Y, which are supposedly exhaustive and such that one or the other must be true. Suppose we can show that P inconsistent with both X and Y.

Formally, how can we arrive, from this, at the conclusion that ~P ?

Hi,

Thanks for your help. This is probably because I am not so good with logic, but there is something I don't understand with your answer. The the truth table for (p ∧ (q ∨ r)), which is equivalent to (P2) above, seems to support the point that if both disjuncts are false, the whole conjunction is false :

p q r (p ∧ (q ∨ r))
F F F F
F F T F
F T F F
F T T F
T F F F
T F T T
T T F T
T T T T

Is the following a fallacy ?
x ∧ y
~y ;
hence ~x

Many thanks for your reply. I wanted to make sure of that simple point before asking something more complex. Here's the final form of my argument. Does it seem right by you ?

(P1) Assume R for reductio
(P2) R ^ (P ˅ Q)
(P3) ~P
(C1) Q (P2, P3, disjunctive syllogism)
(P4) ~Q
(C2) ~(P ˅ Q) (P3, P4)
(C3) ~R (reductio, P2, C2)

Philarete