I have the following:

$(\neg A \lor B) \rightarrow (\neg A \lor B) \\ (\neg A \lor B) \\ \vdash \neg A \lor B $

And in my mind this seems like a legitimate use of the Modus Ponens rule. But the textbook I'm using disagrees.

Why is this wrong?


The textbook has the following line in it as a part of a larger exercise:

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The exercise asks to state whether the use of a giving rule is valid in each case. In the answer section it is marked as an invalid use.

  • $\begingroup$ Why do you say the textbook disagrees? $\endgroup$ – Git Gud Feb 5 '15 at 22:35
  • $\begingroup$ That's fine. Please post the context in which your textbook disagrees more precisely. $\endgroup$ – Kevin Arlin Feb 5 '15 at 22:36
  • $\begingroup$ The shot you posted agrees with what you typed before the edit. I repeat: why do you say the textbook disagrees? $\endgroup$ – Git Gud Feb 5 '15 at 22:41
  • $\begingroup$ @GitGud the textbook JUST disagrees with the statement, that you can use MP for this example. It does not elaborate as to why. Sorry if I didn't make that more apparent. $\endgroup$ – Morgan Wilde Feb 5 '15 at 22:43
  • $\begingroup$ What is the textbook? $\endgroup$ – Rob Arthan Feb 5 '15 at 23:05

Note that $$(\lnot A \lor B) \rightarrow (\lnot A \lor B) \equiv \lnot(\lnot A \lor B) \lor (\lnot A \lor B) \equiv \top$$ In other words, the first premise is a tautology. It says nothing more than "either $\lnot(\lnot A \lor B)$ or else $(\lnot A \lor B)$ holds. And in assuming the law of the excluded middle, one of the two disjuncts must be true, and thus the entire statement is tautologically true.

The second premise is $\lnot A \lor B$.

The argument then can be stated as follows:

$\quad \top\tag{premise 1} $
$\quad \lnot A \lor B\tag{premise 2}$
$\therefore \lnot A \lor B\tag{repetition of premise 2}$

So the conclusion becomes a reiteration of the second premise.

Perhaps that is what your book was attempting to convey?

  • $\begingroup$ I see what you mean, but the use is still valid, even if it is redundant in this case, right? $\endgroup$ – Morgan Wilde Feb 5 '15 at 22:41
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    $\begingroup$ Well, trivially so, perhaps. But it is like no different than the argument: $$p\rightarrow p$$ $$p$$ $$\therefore p$$ $\endgroup$ – amWhy Feb 5 '15 at 22:43
  • $\begingroup$ Gotcha, well this settles it then. Thanks! $\endgroup$ – Morgan Wilde Feb 5 '15 at 22:44
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    $\begingroup$ Inferring $p$ from $p \rightarrow p$ and $p$ is a perfectly good application of modus ponens in all the presentations of first-order logic that I have ever seen. To say it is illegitimate or invalid seems bizarre. $\endgroup$ – Rob Arthan Feb 5 '15 at 23:09
  • $\begingroup$ @Rob I never said it was invalid to do so. $\endgroup$ – amWhy Feb 5 '15 at 23:11

To infer $p$ from $p \rightarrow p$ and $p$ is a legitimate application of modus ponens. It isn't a particularly useful inference, but it is a correct one.


It doesn't appear to be wrong. If $C=C'=\neg A\vee B$, then a correct use of modus ponens is simply $(C'\to C)\wedge C'$ gives $C$, which is what you have.

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    $\begingroup$ What you're doing, essentially, is proving that $((C'\to C)\wedge C')\to C$ is a tautology. $\endgroup$ – Shaun Mar 8 '17 at 20:03
  • $\begingroup$ $C, C'$ cannot both be equivalent to $\lnot A \lor B$. Settle on your assignment. In any case $((\lnot C \to C) \land \lnot C) = ((C\lor C) \land \lnot C)\equiv C\land \lnot C\equiv \bot$, But of course, anything and everything follows from a contradiction. But how instructive is this post, posted more than a year after the question was posted? $\endgroup$ – amWhy Mar 8 '17 at 20:26
  • $\begingroup$ @amWhy Why not? I'm using both $C$ and $C'$ to better illustrate modus ponens. Forgive me if I'm wrong in that. I didn't notice the time of postage: I must have had active rather than new questions displayed. $\endgroup$ – Shaun Mar 8 '17 at 20:34
  • $\begingroup$ That's fine...I was just confused, because the original question was of the form $C \lor \ $\endgroup$ – amWhy Mar 8 '17 at 20:38
  • $\begingroup$ oops, I clicked on the add comment button when I hadn't finished with it. Actually, it would be correct if you had said $((C\to C) \land C )\to C$, in your first comment above, (that is, setting $C:= \lnot A \lor B$. That is equivalent to saying $$(\lnot C \lor C)\land C \equiv C$$ $\endgroup$ – amWhy Mar 8 '17 at 20:41

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