It is not a logical falsehood for several reasons. What I actually mean by using the traditional notation of propositional calculus is that the statement A is a true statement. Were it a false statement I would write ~A.
Similarly i write (P & ~P) to mean “It is true that both P and not P” while I write ~(P & ~P) to mean “It is true that not both P and not P.”
Solving the latter as the equation ~(P & ~P) = TRUE for the variable P gives the trivial solution set of {TRUE, FALSE}, solving the former equation (P & ~P) = TRUE for the variable P gives the empty solution set {}
This is simple convention of notation, I am sorry if that wasn’t clear. Yes, evaluating the logic arithmetic statement (P & ~P) for any given boolean value of P yields false.
How is that not a logical falsehood?
It is not a logical falsehood for several reasons. What I actually mean by using the traditional notation of propositional calculus is that the statement A is a true statement. Were it a false statement I would write ~A. Similarly i write (P & ~P) to mean “It is true that both P and not P” while I write ~(P & ~P) to mean “It is true that not both P and not P.”
Solving the latter as the equation ~(P & ~P) = TRUE for the variable P gives the trivial solution set of {TRUE, FALSE}, solving the former equation (P & ~P) = TRUE for the variable P gives the empty solution set {}
This is simple convention of notation, I am sorry if that wasn’t clear. Yes, evaluating the logic arithmetic statement (P & ~P) for any given boolean value of P yields false.