(P v Q) & ~P & ~Q, is a contradiction.
((T v T) & F & F) = F
((F v T) & T & F) = F
((T v F) & F & T) = F
((F v F) & T & T) = F.

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

No. C2 and P2 are inconsistent, but R could still be true for all that. C3 doesn't follow.

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 the following a fallacy ?
x ∧ y
~y ;
hence ~x — Philarete

Yes it is a fallacy. Once you know either of the conjuncts is false, you know the conjunction is false without ever looking at the other conjunct or knowing its truth value. (Disjunction behaves similarly with truth rather than falsehood.)

In particular, here you have learned nothing at all about the truth or falsehood of R, though you have both P and Q being false.

Better still do this:

1. x & y (premise)
2. ~y (premise)
3. y (from 1)
contradiction.

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 ?

Would something like this work ?


P ˅ Q
P ⊃~R
Q ⊃ ~R
Hence: ~R

((P V Q) & (P -> ~R) & (Q -> ~R)) -> ~R, is tautologous.
T T T F T F F
F T F F T F F
T F T F F F F
F F F F F F F

Because ((P V Q) & (P -> ~R) & (Q -> ~R)) is a contradiction.

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)

There's no particular use here for P1. Just use P2-P4 to derive C1.

The general form here is sometimes referred to as "argument by cases".