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Oct
23
comment what are the Rosser Turquette axioms of Lukasiewicz 3 valued propositional logic?
@DougSpoonwood I guess it all depends on what is your intention in axiomatizing a logic. In my earlier response I have pointed out that this can now be done for any finite-valued logic, using a uniform classic-like setting not unlike the one used by Rosser and Turquette, and giving rise to tableau systems with associated proof strategies and corresponding decision procedures. These systems would seem almost incomparable, in terms of usefulness, if you place them side by side traditional axiomatic systems.
Oct
15
answered What was the first bit of mathematics that made you realize that math is beautiful? (For children's book)
Sep
30
awarded  Explainer
Sep
21
answered A knights and knave problem involving a native with a speech disorder
Sep
21
revised A knights and knave problem involving a native with a speech disorder
corrected spelling, corrected comment about the statement done by C
Sep
21
suggested approved edit on A knights and knave problem involving a native with a speech disorder
Sep
21
comment A knights and knave problem involving a native with a speech disorder
Further, there is obviously a number of things that C could have meant to say, but some of them would not be enough for the kind of D to be determined. So, for deterministic solutions to be available to your problem, I understand you should postulate that the statement of the problem conveys enough information for one to uncover what kind of native D is (you might add to it the information that there was some logician present at the scene who overheard what C said and was then able to determine the kind of D). Do you accept that? If you do, the problem will turn out to be indeed solvable!
Sep
21
comment A knights and knave problem involving a native with a speech disorder
When you write about C that "You can't tell if he is identifying B or C as knights or knaves", just after having written that C "is saying something about B and D", I assume you meant to write about C that "You can't tell if he is identifying B or D as knights or knaves". Right?
Sep
2
answered Having hard time understanding proofs by contradiction.
Sep
2
comment Proof by Contradiction with Multiple Axioms
Your "second format" is a particular case of your "standard form". Note indeed that (i) $\neg Q$ is equivalent to $\top\land\neg Q$, and that (ii) $\top\Rightarrow Q$ is equivalent to $Q$. Thus, your "standard form" $((P\land\neg Q)\Rightarrow \bot)\Rightarrow(P\Rightarrow Q)$ instantiates into $((\top\land\neg Q)\Rightarrow \bot)\Rightarrow(\top\Rightarrow Q)$, and this is turn equivalent, given (i) and (ii), to $(\neg Q\Rightarrow \bot)\Rightarrow Q$, a slight variation of your "second format".
Feb
9
comment What is the “correct” reading of $\bot$?
@MauroALLEGRANZA No, in both cases you could use it: rule $\lor$-intro allows for any formula to be introduced, be it a propositional parameter, or a nullary connective, or a complex well-formed formula with many propositional parameters and connectives of all kinds. The point in this case is that the natural deduction rule is formulated with schematic letters, and such strategy automatically embeds substitution into the formalism.
Feb
9
answered What is the “correct” reading of $\bot$?
Feb
9
comment Is Paraconsistent Negation Really Negation?
@Willemien Start with $LP$, a weakly expressive well-known three-valued logic having $\{0,\frac{1}{2},1\}$ as truth-values, having $0$ as its sole undesignated value, and where $\neg x=1-x$ and $x\vee y=\max(x,y)$. Now upgrade $LP$ by adding an implication such that $x\supset y=0$ if $x\neq 0$ and $y=0$, and $x\supset y=1$ otherwise. Note now, in particular, that $A\supset A\models A\lor\neg A$ yet $\neg(A\lor\neg A)\not\models \neg(A\supset A)$. The phenomenon is not uncommon.
Feb
8
awarded  Yearling
Feb
8
comment Is Paraconsistent Negation Really Negation?
Saying this is 'the most basic definition of negation' is really arguable. The literature on paraconsistent logics and the literature on many-valued logics both abound with examples of negations failing such global version of contraposition, and with good reason. However, paraconsistent negations with modal semantics (such as the ones that Restall is interested upon), may well satisfy this property. In fact, any normal modal logic may be rewritten in a language that extends the classical language by the addition of a paraconsistent negation satisfying global contraposition.
Feb
8
answered Is Paraconsistent Negation Really Negation?
Jan
28
comment Is there a name for this property of a binary relation? $\forall x\forall y(x\mathsf{R}y\to\exists z(x\mathsf{R}z\land y\mathsf{R}z)))$
@Willemien Indeed, from the comments I received so far it seems that the axiom G(0,1,1,1) ---or, equivalently, G(1,1,0,1)--- has not been baptized, or even studied for its own sake. It does seem to correspond, anyway, to a qualified version of the notion of joinability, in ARS, where one demands $x\mathsf{R}y$ as a necessary condition for $x\downarrow y$.
Jan
28
comment Is there a name for this property of a binary relation? $\forall x\forall y(x\mathsf{R}y\to\exists z(x\mathsf{R}z\land y\mathsf{R}z)))$
In the study of abstract rewriting systems, such diamond property is a particular case of the notion of confluence. You get convergence if you add termination to confluence.
Jan
21
awarded  Student
Jan
21
revised Is there a name for this property of a binary relation? $\forall x\forall y(x\mathsf{R}y\to\exists z(x\mathsf{R}z\land y\mathsf{R}z)))$
added 5 characters in body