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In a 4-valued logic with values 0, 1/2, 2/3 and 1, three connectives are defined:

$V(\varphi \wedge \psi)$ = min $V((\varphi), V(\psi))$
$V(\varphi \vee \psi)$ = max $V((\varphi), V(\psi))$
$V(\diamond \varphi) = 1$ if $V(\varphi) \geq 2/3$, else 0

On formulas in the language composed of proposition letters and these three connectives we define a function $C$ that places the connective $\diamond$ in front of every proposition letter in the formula.

E.g.: $C((p \vee \diamond q) \vee q) = ((\diamond p \vee \diamond \diamond q)\vee \diamond q)$

The question:

Give an inductive definition of $C$

Inductive definitions have always given me troubles! I have the following versions of this inductive definition, and I just can't decide which version is better.:

Version 1:

base case:

$C(p) = \diamond p$

I never know what to write here, I got a plethora of possibilities and I never know which one I should take:

  • If $\varphi$ is an atomic formula, then $C(\varphi) = \diamond \varphi$
  • for all atomic formulas $\varphi$, $C(\varphi) = \diamond \varphi$
  • for all atoms $\varphi$, $C(\varphi) = \diamond \varphi$
  • If $p$ is an atomic formula, then $C(p) = \diamond p$
  • For all atomic formulas $p$, $C(p) = \diamond p$
  • For all atoms $p$, $C(p) = \diamond p$

Inductive step:

$C(\varphi \wedge \psi) = (C(\varphi) \wedge C(\psi))$
$C(\varphi \vee \psi) = (C(\varphi) \vee C(\psi))$
$C(\diamond \varphi) = \diamond (C(\varphi))$

Version 2:

With this one, I moved a step in the base case.

base case:

$C(p) = \diamond p$ and $C(\diamond p) = \diamond \diamond C(p)$

And the same trouble of which version to write like in version one

Inductive step:

$C(\varphi \wedge \psi) = (C(\varphi) \wedge C(\psi))$
$C(\varphi \vee \psi) = (C(\varphi) \vee C(\psi))$

  • What is $V$? is it a function? 1-arity? 2-arity? – DanielV May 03 '14 at 10:23
  • Yes. It's just a unary valuation function. – Hunan Rostomyan May 03 '14 at 10:30
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    Must be my years in front of a compiler, the parenthesis in the first two equations set my internal signals off. V is used as both unary and binary. I guess it is something like "the value of the expression". I'm guessing they were supposed to be $V(\varphi \land \psi) = \min{V(\varphi), V(\psi)}$. – DanielV May 03 '14 at 10:34
  • Exactly. Forget about what I said about min. – Hunan Rostomyan May 03 '14 at 10:39
  • @DanielV, HunanRostomyan

    The V is used in the other questions, but I needed that part to tell about the connectives that were used.

    – Garth Marenghi May 03 '14 at 13:41
  • are you really using a logic without negation? – Willemien May 03 '14 at 17:47
  • @Willemien Yes I am. I never really thought about that - it was "just" a question. I guess in the context of the question it really doesn't matter, as long as the questions at hand get an answer. And with that I mean, the question I have as homework, not the question I posted here ;-) – Garth Marenghi May 03 '14 at 18:18
  • I find the whole question a bit strange, normally a translation ( and C is a translation) does not make a difference between atomic and non-atomic formulas, In the 4 valued logic you describe the difference also would not matter (the <> just makes the formula bivalued and <>(<>p & <>Q ) is equivalent to <>p & <>q (also for C(p v q) so why not use normal translation rules? – Willemien May 03 '14 at 20:29
  • @Willemien I guess you're right, but I do not have any insight into the hows and whys of the questions, so I can not offer you any explanation of why this question was posited this way. There were more questions to answer though, e.g. to proof with induction that V$C(\varphi) = V(\diamond \varphi)$. – Garth Marenghi May 04 '14 at 05:58
  • Which course are you on,? , but just make (4 valued) truth tables – Willemien May 04 '14 at 07:25
  • Het is een cursus logica, onderdeel van computer science van de OU, en maakt gebruik van het boek Logica voor Informatica. Als je wilt, kan ik je de vraag opsturen, niet zozeer voor mijzelf, maar meer dat je begrijpt waar het over gaat. Ik heb wat informatie hier en daar weggelaten namelijk. – Garth Marenghi May 04 '14 at 07:33
  • This question is about syntax and does not depend on any interpretation V. So it is not specific to a 4 valued logic. – miracle173 May 10 '14 at 14:54

2 Answers2

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Your first version looks fine. Here is one, albeit not very explicit (because the language details are implicit) way of presenting the definition. $C(\phi)$ is defined by induction on the structure of $\phi$:

  • If $\phi \in \mathsf{Props}$, then $C(\phi) = \diamond\phi$;

  • If $\phi$ is of form $\psi \circ \chi$ and $\circ \in \mathsf{Cons_2}$, then $C(\phi) = C(\psi) \circ C(\chi)$;

  • If $\phi$ is of form $\diamond \psi$, then $C(\phi) = \diamond C(\psi)$;

where $\mathsf{Prop}$ and $\mathsf{Cons_2}$ are the sets of propositional letters and binary connectives respectively.

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I also think your first version is perfectly fine. Note that as a definition by recursion on the complexity of $\phi$ (which is the more appropriate term) your second candidate is ill-defined.

The base clause of a recursive definition of some function $f: \mathcal{F} \rightarrow A$ (where $\mathcal{F}$ is the set of formulas and $A$ some set) says that $f(\phi) = f_{At}(\phi)$, where $f_{At}$ is some function from $At$ (the set of atoms from $\mathcal{F}$) to $A$ and $\phi \in At$.

The base clause of your second version, however, uses a function defined on the set $At \cup \lbrace \diamond \phi : \phi \in At \rbrace \not = At$. In other words, the base clause must be specified independently of the values of $f$.

One might try to repair your second version by putting $C(\diamond \phi) = \diamond \diamond C(\phi)$ to the recursive clauses, where $\phi \in \mathcal{F}$. But this repaired version yields wrong results. According to that definition $C((p∨\diamond q)∨q) = ((\diamond p \lor \diamond \diamond C(q)) \lor \diamond p) = ((\diamond p \lor \diamond \diamond \diamond q) \lor \diamond p) \not = ((\diamond p \lor \diamond \diamond q) \lor \diamond p)$.

Jon
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