1
$\begingroup$

To show that $[(p \rightarrow q) \rightarrow r] \Rightarrow [p \rightarrow (q \rightarrow r)]$ without using a truth table. That is, using logical laws.

$\endgroup$
3
  • 1
    $\begingroup$ Hint: Try using a truth table. $\endgroup$ – anakhro Jul 6 '15 at 13:05
  • $\begingroup$ @anakhronizein. Sorry, can you tell me another way not using a truth table? $\endgroup$ – PHPIsTheBestLanguage Jul 6 '15 at 13:08
  • 1
    $\begingroup$ Why is one of your arrows a $\Rightarrow$ and the others are $\rightarrow$? Do they all mean the same? $\endgroup$ – Ian Jul 6 '15 at 13:09
1
$\begingroup$

With Natural Deduction :

1) $(p→q)→r$ --- premise

2) $q$ --- assumed [a]

3) $p→q$ --- from 2) by $\to$-introduction

4) $r$ --- from 1) and 3) by $\to$-elimination

5) $q \to r$ --- from 2) and 4) by $\to$-introduction, discharfging [a]

6) $p \to (q \to r)$ --- from 5) by $\to$-introduction

$[(p→q)→r] \vdash [p \to (q \to r)]$ --- from 1) and 6).

$\endgroup$
1
$\begingroup$

Hint: $p \Rightarrow q \equiv \neg p \vee q$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.