Yeah. My solution is basically the same as yours. Setting A=B=C makes F(T,T) = T. But setting A=B AND C → ~A makes F(T,T) = F (warning: unfortunate notation collision here).
Ah OK. You’re right. I guess I was taking the ‘extension of logic’ thing a little too far there. I had it in my head that ({any prop} | {any contradiction}) = T since contradictions imply anything. Thanks.
That’s legit so far as it goes—it’s just that every proposition is also false at the same time, since every proposition’s negation is also true, and the whole enterprise goes to shit. There’s no point in trying to extend logic to uncertain propositions when you can prove anything.
Yeah. My solution is basically the same as yours. Setting A=B=C makes F(T,T) = T. But setting A=B AND C → ~A makes F(T,T) = F (warning: unfortunate notation collision here).
Given C → ~A, ({any proposition} | AC) is undefined. That’s why I couldn’t follow your argument all the way.
Ah OK. You’re right. I guess I was taking the ‘extension of logic’ thing a little too far there. I had it in my head that ({any prop} | {any contradiction}) = T since contradictions imply anything. Thanks.
That’s legit so far as it goes—it’s just that every proposition is also false at the same time, since every proposition’s negation is also true, and the whole enterprise goes to shit. There’s no point in trying to extend logic to uncertain propositions when you can prove anything.