Fell swoop method I'm new to logic, and I'm really confused about the fell swoop method. I have to determine the validity of $$((P \lor Q) \land (\lnot P \land \lnot Q)) \to \lnot P)$$ using the fell swoop method. So far, I have managed to come up with this. I'm thinking that the formula is valid, but I'm having trouble illustrating this properly. I hope someone can correct and guide me please. Thanks
fell swoop table
 A: The "fell swoop" method (see WVO Quine, Methods of Logic (4th ed., 1982))is a method of "fast checking" that can be used to verify (in an important range of cases, also if it is not valid in general) if a formula is satisfiable or falsifiable.
In a nutshell, it amounts to build a single (suitably chosen) row of the truth table for the formula to be checked.
Regarding the example :

$[(P∨Q) ∧ (¬P ∧ ¬Q)] → ¬P$,

the method try to check if the formula is not valid.
Being the main connective the conditional, this can happen only if the antecedent is TRUE and the consequent is FALSE.
Thus, we start with a valuation $v$ such that $v(¬P)= \text F$, i.e. $v(P)= \text T$.
This is enough to have $v(P∨Q)= \text T$.
Regarding $(¬P ∧ ¬Q)$, we have that $v(¬P)= \text F$ implies : $v(¬P ∧ ¬Q))= \text F$.
Thus $v([(P∨Q) ∧ (¬P ∧ ¬Q)]= \text T \land \text F = \text F$.
In conclusion : we have shown that if the consequent is FALSE also the antecedent must be, i.e. that it is impossible that the antecedent is TRUE and the consequent is FALSE, and this means that the formula is valid.

The method works when the main connective is a conditional because in the truth table for the conditional there is only one row that falsifies the formula.
