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In its usual, implicative form, modus ponens says:

$$A \wedge (A \rightarrow B) \rightarrow B$$

There's also an equational version, namely:

$$A \wedge (A \rightarrow B) = A \wedge B$$

Now define that the implicative knockout principle is: $$\neg A \wedge (A \vee B) \rightarrow B$$

And the equational knockout principle is $$\neg A \wedge (A \vee B) = \neg A \wedge B$$

So basically, they let you knock-out some possibilities from a disjunction, based on knowledge of the falseness of those possibilities. We do this all the time, of course, though these principles aren't often made explicit, perhaps due to ease of derivation. In particular, the two implicative statements can be easily inter-derived, by replacing each copy $A$ with $\neg A$. Similarly with the two equational statements. In any event, I think its pedagogically useful to state the two "knockout" principles explicitly, since they're actually pretty useful.

The word "knockout" is my own, so I'm wondering:

In logic and/or math pedagogy, do these principles have an accepted name?

goblin GONE
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  • Short partial answer: I don't know of an accepted name in math pedagogy. – Ethan Bolker Feb 08 '17 at 16:08
  • I've seen some presentations of natural deduction (I don't like the approach but..) which define $\lor$ Elimination using disjunctive syllogism, or as you call it, "implicative knockout". It is a lot simpler looking than proof by cases, but requires axioms about negation. – DanielV Feb 09 '17 at 10:11
  • Modus ponens usually refers to a rule of inference, not an implied well-formed formula. – Doug Spoonwood Feb 09 '17 at 16:02
  • @DougSpoonwood, I agree, but we also need a name for the corresponding tautology, and it seems easiest to just call them the same thing. That is, I'm claiming that if $$A_0,\ldots,A_{n-1} \vdash B$$ has a name, then $$\vdash A_0 \wedge \ldots \wedge A_{n-1} \rightarrow B$$ can be safely referred to by the same name w/o causing too much confusing. And, of course, I've left the turnstile implicit in my question. – goblin GONE Feb 10 '17 at 02:39
  • @goblin Well, I might call what you've written conjunctive modus ponens. That's distinct from what we could call implicative modus ponens (A$\rightarrow$((A$\rightarrow$B)$\rightarrow$B)). The rule of inference modus ponens doesn't involve an object level conjunction or implication in the same way, and it's not necessary to think to of the formulas as having some implicit conjunction. You can just think of all formulas on the left-side of the turnstile as belonging to the same set of formulas. In other words, the rule of inference is {A, (A$\rightarrow$B)} $\vdash$ B, which is... – Doug Spoonwood Feb 10 '17 at 22:57
  • the same set-theoretically as {(A$\rightarrow$B) A} $\vdash$ B, since both involve the same set. That differs from (A$\land$(A$\rightarrow$B)) and ((A$\rightarrow$B)$\land$A), which though logically equivalent, are not logically the same. So, we might even want to specific which conjunctive modus ponens we refer to by referencing (A→((A→B)→B)). – Doug Spoonwood Feb 10 '17 at 22:59

3 Answers3

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The implicative form

$$\neg A \wedge (A \vee B) \rightarrow B$$

is the conditional version of Disjunctive Syllogism ... which is better known as the inference

$$\neg A$$

$$A \vee B$$

$$\therefore B$$

The equational form

$$\neg A \wedge (A \vee B) \Leftrightarrow \neg A \wedge B$$

is sometimes referred to as Reduction (in the context of $\neg A$, the term $A \vee B$ 'reduces' to just $B$). But few texts do, and it is clearly not as 'standard' as, say, Absorption.

... which is too bad! Indeed, yes, I totally agree with you, Reduction should be more prominently taught and be part of our standard logic toolbox, as it is a very useful little short-cut to doing Distribution, Complement, and Identity. Here are a few posts where I used Reduction:

How to find logical equivalence?

Simplify the following expression using Boolean Algebra into sum-of-products (SOP) expressions

Also, Reduction is a special case of the more general Resolution, whose implicate form would be:

$$((\neg A \lor C) \wedge (A \vee B)) \rightarrow (C \lor B)$$

and whose equational form would be:

$$(\neg A \lor C) \wedge (A \vee B) \Leftrightarrow (\neg A \lor C) \wedge (A \vee B) \land (C \lor B)$$

Community
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Bram28
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This is an instance of resolution which is so useful for calculation that is the basis for many (classical) logic provers. The general resolution rule takes your "knockout" idea to the logical conclusion.

Derek Elkins left SE
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In classical logic, your "knockout" principles are exactly tje modus ponens, because of the following :

$A \iff \neg \neg A$

$(A\implies B) \iff (\neg A \lor B)$

So your implicative knockout principle becomes : $\neg A \land (\neg \neg A \lor B) \rightarrow B$, which then becomes $\neg A \land (\neg A \implies B) \rightarrow B$, which is indeed a special case of modus ponens. Same thing for the "equational knockout principle"

Maxime Ramzi
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