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$(p \land (p \implies q) \land (q \implies r)) \implies r$

It is written slightly different in the text book, but this should be the equivalent form. The book is a bit unclear but I think the author wants to transform the previous eqution into

$r \implies r$ which is always true.

The proof:

1) $p$

2) $p \implies q$

3) $q \implies r$

4) $q$ (modus ponens 1 and 2)

5) $r$ (modus ponens 4 and 3)

I don't understand step 4 and 5. The author uses modus ponens with 1 and 2 to get to $q$. But they are not the same thing.

$(p \implies q) \land p$ to me is the same thing as $p \land q$ and not just $q$

What am I missing here?

Maik Klein
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  • You aren't missing anything. You are correct in pointing out that modus ponens can be stated as $p \wedge (p \Rightarrow q) \Rightarrow p \wedge q$. However, $p \wedge q \Rightarrow q$ and therefore $p \wedge (p \Rightarrow q) \Rightarrow q$. – Archaick Jul 09 '16 at 11:31
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    @Archaick: This is somewhat misleading. The "from this you can conclude that" of an inference rule is not the same as the $\Rightarrow$ connective. What the modus ponens rule in particular says is that if you have a $\Rightarrow$ then you can convert that to an inference, but if you represent that inference symbolically as a $\Rightarrow$ too, then all you get is that $\Rightarrow$ gives you $\Rightarrow$, which doesn't tell you anything. See http://math.stackexchange.com/questions/351685/whats-the-problem-this-logic, in particular Peter Smith's answer, for more discussion. – hmakholm left over Monica Jul 09 '16 at 13:24
  • @HenningMakholm You are quite correct! I totally overlooked this subtlety. Thanks for pointing this out and providing the reference. – Archaick Jul 09 '16 at 23:02

1 Answers1

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What you're looking at is a series of inferences, but you're confusing yourself because you think it is an algebraic rewriting to an equivalent formula.

Modus ponens says: if you know $p$ and you know $p\to q$, then you can also know $q$.

You also still know $p$ and $p\to q$, but the inference rule doesn't need to mention that. In ordinary logic, using an inference rule never makes you not know something you already knew.

Your error is that you assume that premises and the conclusion of an inference rule have to be "the same thing", but they don't. The rule is valid if only the conclusion follows from the premises. This is a one-way relation: If you know the premises are true then the conclusion must be true too -- but merely knowing that the conclusion is true does not in itself imply anything about the premises.

It appears that you view using logical inferences as a kind of equation solving. That's probably a poor way to look at it, but even if you insist on this viewpoint, what's going on here is not new. In algebra you can have $x=y$ and therefore conclude $x^2=y^2$ -- but $x^2=y^2$ is not "the same as" $x=y$; instead $x^2=y^2$ is true more often than $x=y$ is.

hmakholm left over Monica
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