Axiomatisation of propositional logic using $\land$ and $\neg$ I am looking for a simple axiomatisation of a particular version of propositional logic that is defined in terms of $\land$ and $\neg$ only.
I am guessing that it only needs one rule of inference: $p, \neg(p \land \neg q) \vdash q$.
Can you give me a short set of axioms to add to this inference rule?
(I did look around on the googly internet, and found this http://en.wikipedia.org/wiki/List_of_logic_systems but it did not answer my question).
thanks,
Richard
 A: A query rather than a straight answer. It is something of an unfortunate historical accident that early formal systems of logic (Frege, Russell/Whitehead, Hilbert) went for many-axioms/few-rules. These systems are pretty unnatural, and indeed in the background there is arguably a mistake about what logic is about. For the early founding fathers tended to speak as if logic was about cataloguing logical truths, rather than valid proofs -- take the latter view, and you'll conceive of the natural way of presenting a logic as a system of rules of inference which you can use in constructing mathematical and other proofs, rather than as a system of logically true axioms from which more logically true propositions can be derived.
So I do wonder why, other than for a somewhat pointless technical exercise, we should nowadays  be interested in presenting a many-axioms/one-rule system of logic for negation/conjunction, rather than a much more natural no-axioms/many-rules system? And for the latter, lots of standard textbooks deliver the goods for free. To get a complete natural deduction system for negation and conjunction just take the system for the usual four or five connectives and leave out the rules for disjunction and the (bi)conditional.  
A: I found a suitable axiomatisation in Mendelson (he cites Rosser 1953).
It uses Modus Ponens ($p, \neg(p \land \neg q) \vdash q$) plus three axioms:


*

*$\neg (p \land \neg (p \land p))$

*$\neg ((p \land q) \land \neg p)$

*$\neg (\neg (p \land \neg q) \land \neg (q \land r) \land r \land p)$

