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I wonder about two things. First, how do we prove that entailment in some logic is monotonic? The second one - What is the relationship between monotonicity of logic and deduction theorem? It seems rather clear that deduction theorem doesn't hold in nonmonotonic logics. But do we have deduction theorem in every possible monotonic logic?

I will be extremely thankful for answers!

Trapszan
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2 Answers2

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Take your favourite Hilbert-style axiomatization of propositional logic. Mendelson's for example. We can prove that the deduction theorem holds with respect to that theory.

Now kill e.g. all instances of Mendelson's first axiom schemata. The proof of the deduction theorem won't go through any more. Damn!

But the weakened logic is still monotonic. That is to say, if in the weakened logic you have $A \vdash C$, then you'll have $A, B \vdash C$, for any $B$. That's because a proof is still defined to be a sequence of wffs such that each one is either an axiom or follows from early wffs in the sequence by modus ponens -- a definition which ensures monotonicity but says nothing about which axioms are actually available.

Peter Smith
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    Perhaps more simply: declare that $A \vdash C$ never holds unless $A$ is identical with $C$. This is certainly a monotonic logic in the sense of this answer and the deduction theorem fails. – Carl Mummert Aug 24 '13 at 21:17
  • Simpler indeed! – Peter Smith Aug 24 '13 at 22:01
  • @CarlMummert The scope of the question concerned logical systems in general (it talks about logics, and it says "some logic" in a way that it seems to mean "a logic" indicating a question about logical systems in general). If two theses come as identical within a logic, they have to at least come as inter-derivable. Can you prove any two theses as inter-derivable within a logical system if you don't have the law of simplification "CpCqp"? If not, how can you meaningfully assert that A |- C necessarily implies that A is identical with C? – Doug Spoonwood Aug 25 '13 at 03:02
  • @Doug Spoonwood: every definition of "$\vdash$" gives some sort of "general logic". We could define that $A\vdash B$ never holds, and that would still be a logic in the most general sense. My suggestion was to define a logic with a single inference rule: from a formula $A$, derive $A$ itself. The resulting logic has properties relevant to the question. – Carl Mummert Aug 25 '13 at 11:20
  • @CarlMummert How have you derived A from itself? I don't see it... a derivation of any sort has to take place on a piece of paper or a computer screen. You have a symbol "A" and another symbol "A" which come as equiform up to our ordinary way of seeing things. That and no more. If they came as identical, then they would have to have all of the same properties. But, they don't. They exist in different places. – Doug Spoonwood Aug 25 '13 at 23:28
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The received Tarskian notion of entailment is often explicated using models, through the idea of truth-preservation: $\Gamma\models A$, given a set of formulas $\Gamma\cup\{A\}$, if every valuation satisfying all premises in $\Gamma$ also satisfies the conclusion $A$. An entailment relation is called monotonic if $\Delta\models A$ whenever $\Gamma\models A$ and $\Gamma\subseteq\Delta$. It is easy to check that the Tarskian notion of entailment is monotonic: assume $\Gamma\subseteq\Delta$ and suppose by contraposition that you have a counter-model for $\Delta\models A$, that is, suppose that there is a valuation that satisfies $\Delta$ but does not satisfy $A$; it follows from the assumption that $\Gamma$ is also satisfied, thus you have indeed a counter-model for $\Gamma\models A$. To put it otherwise, if you're dealing with some non-monotonic notion of consequence, this notion will fail to meet the requirements to be characterized by way of the Tarskian notion of entailment.

It should be noted that the deduction theorem is neither sufficient nor necessary for a consequence relation to be monotonic. Furthermore, that there are some Tarskian logics which respect the deduction theorem, and other Tarskian logics that disrespect this theorem.

J Marcos
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  • @PeterSmith Now I'm a little bit confused. I was reading about non-monotonic logic and in the article on the Stanford Encyclopedia of Philosophy was a statement that said that in non-monotonic logic deduction theorem doesn't hold. The informal proof that was given seems pretty convincing to me; however, I'm at very early stage at learning logic, so I will quote it here, in the next comment. – Trapszan Aug 25 '13 at 10:06
  • "Consider, for example, the left-to-right direction: suppose that Γ ∪ {p} proves q. Should it follow that Γ proves (p ⇒ q)? Not at all. It may be that, normally, if p then ¬q, but Γ may contain defaults and information that defeat and override this inference. For instance, it might contain the fact r and the default ((r & p) ⇒ q). Similarly, consider the right-to-left direction: suppose that Γ proves (p ⇒ q). Should it follow that Γ ∪ {p} proves q? Again, clearly not. Γ might contain both r and a default ((p & r) ⇒ ¬q), in which case Γ ∪ {p} proves ¬q." – Trapszan Aug 25 '13 at 10:07
  • So, as i understand, if I apply the law of contraposition to the statement "If logic is nonmonotonic, then deduction theorem doesn't hold" i will aquire statement "If deduction theorem holds, then logic is monotonic". Am i right? Or am I reading more than it really is in the earlier argument? – Trapszan Aug 25 '13 at 10:07
  • @Trapszan: yes, that is a correct application of contraposition. Of course, it takes more than just monotonicity for the deduction theorem to hold, so monotonicity is necessary, but not sufficient. – Carl Mummert Aug 25 '13 at 11:22