# Is $(p \to q) \to r$ logically equivalent to $p \to (q \to r)$?

Is $(p \to q) \to r$ logically equivalent to $p \to (q \to r)$?

I try to simply each one, I got $\lnot ( \lnot p \lor q) \lor r$ and $\lnot p \lor ( \lnot q \lor r)$ respectively,

then I am stuck. Does that mean they aren't equivalent?

How do I know if expression cannot be further simply? Any tips to solve such questions?

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You don't even need to do any simplification if you want a proof of equivalence; just work out the truth tables for each statement and see if they're the same. –  David Zhang Sep 5 '13 at 13:56

Since you've gotten to:

$$\lnot ( \lnot p \lor q) \lor r\tag{1}$$ $$\lnot p \lor ( \lnot q \lor r)\tag{2},$$

we can indeed simplify further. First, by associativity, note that $$\lnot p \lor ( \lnot q \lor r)\equiv (\lnot p \lor \lnot q) \lor r\tag{2}$$

Next, using DeMorgan's Laws,

$$\lnot (\lnot p \lor q) \lor r \equiv (p \land \lnot q) \lor r\tag{1}$$ $$(\lnot p \lor \lnot q) \lor r\equiv \lnot(p \land q) \lor r\tag{2}$$

Finally, we need only see that the clauses $$p \land \lnot q \not\equiv \lnot (p \land q)$$ to see that $(1)$ and $(2)$ are not equivalent,

Indeed, when $p$ is F(alse), $p \land \lnot q \equiv F$ and $\lnot(p \land q) \equiv T$, as we can see nicely via the corresponding truth tables (source: Wolfram Alpha):

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Such a nice write up +1 –  Amzoti Sep 6 '13 at 0:16

No, compare $(p\implies p)\implies r$, which is equivalent to $r$, and $p\implies (p\implies r)$, which is equivalent to $p\implies r$.

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It is not. Suppose $p$, $q$, $r$ are false. Then $p \rightarrow q$ and $q \rightarrow r$ are true, so $(p\rightarrow q)\rightarrow r$ is false and $p\rightarrow (q\rightarrow r)$ is true.

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Good question!

No.

The second statement, namely $$p \rightarrow (q \rightarrow r)$$ can be read: "In the presence of $p$, we have that $q$ implies $r$."

Turns out this is the same as saying: "In the presence of both $p$ and $q$, we have $r$." That is

$$(p \wedge q) \rightarrow r.$$

The first statement is a little harder put words to. In the comments, Lord Farin suggests the phrasing: "If $p$ necessitates $q$, then $r$."

A bit of symbol pushing shows its equivalent to:

$$(p \vee r) \wedge (q \rightarrow r)$$

which might shed some light on its meaning. Substituting in $q = \bot$ (a contradiction) we obtain

$$(p \vee r).$$

So if $p$ necessitates a contradiction implies $r$, then either $p$ or $r$. And vice versa.

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How about "If $p$ necessitates $q$, it follows that $r$?" A typical case is when $q$ is a contradiction and $r = \neg p$, the negation of $p$. –  Lord_Farin Sep 5 '13 at 10:43
@Lord_Farin, thanks I edit my answer in line with your comment. Still not totally satisfied though. Feel free to edit it as you see fit. –  goblin Sep 5 '13 at 10:49

The second formula is "if $p$, then: if $q$ then $r$" which is equivalent to "if $p$ and $q$, then $r$". So it is equivalent to $$(p\land q) \implies r.$$

On the other hand, the first is "if $p$ implies $q,$ then $r$". As we know any implication "$p$ implies $q$" is equivalent to "not $p$ or $q$", so our formula is "if not $p$ or $q$, then $r$", which written with symbols is: $$(\neg p \lor q) \implies r.$$

Because of course $\neg p \lor q \not\equiv p \land q$, the two formulas are not equivalent.

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