# Lukasiewicz logic, deduction theorem and complete calculi

3-valued Lukasiewicz logics is known to lack a “semantical" deduction theorem, i.e. $$A\vDash B\not\Leftrightarrow\vDash A\supset B$$ if we define $A\vDash B$ as $\forall v(v(A)=1\Rightarrow v(B)=1)$ with $1$ being the designated value. For example $$p\wedge\neg p\vDash q\text{ but }\nvDash(p\wedge\neg p)\supset q$$

There are Hilbert style and hypersequent calculi for $L_3$ (cf. e.g. Natural 3-valued logics — characterization and proof theory by Avron) which are complete w.r.t. all valid formulas of $L_3$, i.e. $$\vdash A\Leftrightarrow\vDash A$$ They, of course, have “syntactical” deduction theorem, i.e. $$A\vdash B\Leftrightarrow\vdash A\supset B$$

They are, because of this, incomplete w.r.t. $\vDash$ in the following way $$A\vDash B\not\Leftrightarrow A\vdash B$$ since, obviously $\nvdash(p\wedge\neg p)\supset q$ and therefore $p\wedge\neg p\nvdash q$.

The question is, therefore, as follows.

Is it possible to construct such a calculus for $L_3$ that $\Gamma\vdash A\Leftrightarrow\Gamma\vDash A$ for any $\Gamma$?

• Your example isn't right. $p \land \lnot p \supset q$ is valid in $L_3$. A better example would be that $p \models p \land p$ is true but $p \supset p \land p$ is not valid. Apr 2 '18 at 19:51
• I disagree… Set $p$ as $\frac{1}{2}$ and $q$ as $0$ and we get $\frac{1}{2}$ as a result, no? Furthermore, if we take $\wedge$ as $\min$, we have that $p\supset(p\wedge p)$… Or are we talking about different logics? Apr 2 '18 at 20:56
• We are talking about the same logics, but I didn't know that you were using $\land$ for the weak conjunction rather than the strong conjunction. What symbol are you using for the strong conjunction? Apr 2 '18 at 21:12
• Strong conjunction is the same as $\neg(A\supset\neg B)$? All formulations of $L_3$ I have so far encountered do not use any specified symbol for this… I guess, we can use $\circ$ since $\neg(A\supset\neg B)$ is a common definition of fusion… Apr 2 '18 at 21:23
• I think $\otimes$ is standard for the strong conjunction if you are using $\land$ for the weak conjunction. If you define $\Gamma \vdash A$ to mean $A$ is derivable from $\Gamma$ in Avron's Hilbert-style system, then you will get the calculus you want. The deduction theorem will hold in a weaker form: if $\Gamma, A \vdash B$, then $\Gamma \vdash A^n \supset B$ for some $n$ where $A^n$ denotes the strong conjuction of $n$ copies of $A$. Apr 2 '18 at 21:30

Lewis and Langford (Symbolic Logic, 1932, p219) observed that modus ponens in its conventional form CApCpqq fails. Using p $\rightarrow$ q for the Lukasiewiz conditional Cpq,

(p & (p $\rightarrow$ q)) $\rightarrow$ q

is not a tautology in L3.

However, examination of the truth table reveals that this is for good reason. Where p has the third value U, and Cpq has also has the third value U, CApCpqq has a value of U when q has the value F.

This means that if p and Cpq are both doubtful, then applying unrestricted modus ponens can result in a false conclusion q. This is not and should not be valid reasoning, and the truth table correctly identifies it as problematic.

However, if the conditional is asserted as certain or necessary, LCpq, matters are different. CApLCpqq is valid. Using the symbolism of modal logic, we have

(p & $\Box$ (p $\rightarrow$ q)) $\rightarrow$ q

which is a tautology.

There are a great many theorems of classical two-valued logic which are not considered valid in L3 that would be if this version of the strict conditional were used instead. The strict conditional used by Lewis $\Box$ (p $\supset$ q) where $\supset$ signifies the ordinary material conditional is not adequate.

Given these conditions, it appears that the answer to the question would be yes.

• Am I right that $v(\Box p)=1\Leftrightarrow v(p)=1$ and $v(\Box p)=0$ otherwise? Apr 4 '18 at 8:03
• @Daniil Koszemiachenko. That is correct. Apr 4 '18 at 12:25