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I have a couple of questions about the distinction between the so-called "basic rules" and the so-called "derived rules" in logic. I have been told that there is nothing substantial about this distinction, i.e., that it is a mere convention: for instance, modus ponens is usually regarded as a basic rule, whereas modus tollens is generally regarded as a derived rule, but actually one can derive modus ponens from modus tollens, with suitable additional rule. I have two questions: (1) how exactly you derive modus ponens from modus tollens in propositional logic? (2) Is it always the case (i.e., is it true for any logic) that modus ponens and modus tollens can derive each other or is this only true for propositional logic?

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  • $\begingroup$ Already asked some hours ago $\endgroup$ Jul 15, 2021 at 6:03
  • $\begingroup$ Actually, it's not exactly the same question. $\endgroup$
    – Mijito
    Jul 15, 2021 at 9:39
  • $\begingroup$ I believe it is considered a good practice to link to your previous related question, and write something like, "As a follow up to this post..." $\endgroup$
    – Joe
    Jul 15, 2021 at 10:42

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If we use material equivalence, that is $$(p\implies q) \equiv \lnot p \vee q$$, we can derive modus ponens and modus tollens only from disjunctive syllogism : $$(P \vee Q) \wedge \lnot Q \implies P$$

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Long comment

The issue is that there is an interplay between axioms and rules.

With suitable axioms, MP is enough and we can easily derive additional rules, but with the same set of axioms other "derivable rules" cannot be enough, for example to derive MP.

Conclusion: the question above is not well defined: if we have a complete axioms+rules set we can derive all tautologies, and thus also Modus Tollens:

$ [(p\rightarrow q)\land \neg q]\rightarrow \neg p]$.

From it - using MP - we have the MT derived rule:

$\dfrac {(p\rightarrow q) \ \ \neg q}{\neg p}$.

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How exactly you derive modus ponens from modus tollens in propositional logic?

You will need the equivalence: $$P \to Q ~~\equiv ~~ \neg [P \land \neg Q]$$

Using a form of natural deduction, here is a proof by contradiction. (Screenshot from my proof checker)

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