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Is there an axiomatic system for (classical) propositional logic that does not use modus ponens as a (primitive) rule?

I would be particularly interested in a derivation of it from some other set of rules. I would assume that this is possible with rules only for, e. g., $\wedge$ and $\neg$, but I've never seen it done before.

susypeti
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    Any system with the following rules (as primitive or derivable) as well as the usual monotonicity and transitivity rules doesn't need MP as primitive:

    $${a\rightarrow b}\vdash \neg a\vee b\quad\mbox{and}\quad {a, \neg a\vee b}\vdash b.$$ And the first of those can be completely ignored if we agree to treat "$a\rightarrow b$" as an abbreviation for "$\neg a\vee b$" in the first place. Does that count?

     – Noah Schweber Feb 23 '22 at 05:07
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    The cut-free ${\wedge,\neg}$ fragment of the sequent calculus does the job. Just define $A \vee B$ as an abbreviation for $\neg(\neg A \wedge \neg B)$, and $A \rightarrow B$ as $\neg (A \wedge \neg B)$. – Z. A. K. Feb 23 '22 at 05:41
  • See Joseph Shoenfield, Mathematical Logic (Addison-Wesley, 1967) (page 21) – Mauro ALLEGRANZA Feb 23 '22 at 07:13
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    @Z.A.K. If I understood you correctly, in this answer: https://math.stackexchange.com/questions/3894022/is-there-any-finite-deductive-system-for-propositional-logic-which-only-uses-una?rq=1 you gave an axiom system, which doesn't use modus ponens as a primitive, since modus ponens has two premises, and the rules you gave were all had one premise. The primitive connectives, were → and ¬. – Doug Spoonwood Feb 24 '22 at 03:35

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