• Self-explanatory facts
    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.
  • Is this argument form valid ? (contradiction through disjunctive syllogism)
    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)
  • Is this argument form valid ? (contradiction through disjunctive syllogism)
    Would something like this work ?


    P ˅ Q
    P ⊃~R
    Q ⊃ ~R
    Hence: ~R
  • Is this argument form valid ? (contradiction through disjunctive syllogism)
    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 ?
  • Is this argument form valid ? (contradiction through disjunctive syllogism)
    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
  • Is this argument form valid ? (contradiction through disjunctive syllogism)
    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