ISTM that you’re using the word contradiction in a non-standard way: in the usual sense they are logical falsehoods. What do you actually mean? (ETA: I guess paradoxes such as “This sentence is false”?)
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.
Logical falsehoods, and disjoint events, all reduce to contradictions in boole. There is no difference in propositional logic. Where are you getting this from?
ISTM that you’re using the word contradiction in a non-standard way: in the usual sense they are logical falsehoods. What do you actually mean? (ETA: I guess paradoxes such as “This sentence is false”?)
I use contradiction in the completely ordinary sense as seen in propositional logic. (P & ~P)
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.
Logical falsehoods, and disjoint events, all reduce to contradictions in boole. There is no difference in propositional logic. Where are you getting this from?
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