0

Working on P.D. Magnus. forallX: an Introduction to Formal Logic (pp. 154, exercise D. 7). It is the last exercise of Chapter 16. This is what I got so far.

Is my $IP$ strategy correct ?

$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}} \fitch{}{ \fitch{¬(((A \to B) \to B) \to B)}{ \fitch{¬((A \to B) \to B) }{ \fitch{¬(A \to B)}{ \fitch{B}{ \fitch{A}{ B } \\ A \to B \\ \bot } } } } }$

P.S: Here is a link to the open-source book: forallX

F. Zer
  • 2,559
  • 1
  • 10
  • 23
  • 2
    I couldn't locate this exercise in the downloadable version of the book. It would be more helpful to give a chapter and section reference, because the page references may vary in different formats of the book. – Rob Arthan Mar 06 '20 at 22:58
  • Ok, I’ll edit my question to include the information you suggest. – F. Zer Mar 06 '20 at 23:08
  • I think you need to give a URL. The online version of the book I can find only doesn't have a chapter 16. – Rob Arthan Mar 07 '20 at 00:08
  • 3
    Maybe what they intended the exercise to be is Peirce's law: $((A \rightarrow B) \rightarrow A) \rightarrow A$. – Daniel Schepler Mar 07 '20 at 00:09
  • @RobArthan, just added a link as a P.S. – F. Zer Mar 07 '20 at 00:45
  • Thanks for providing the link. You have misquoted the proposition to be proved. As Daniel conjectured you are being asked to prove Peirce's law. – Rob Arthan Mar 07 '20 at 00:51
  • @RobArthan In my edition, it is as I wrote it. Being open source, perhaps it was recently corrected. – F. Zer Mar 07 '20 at 01:11
  • 1
    My commiserations to you on having to work from an unstable text. – Rob Arthan Mar 07 '20 at 01:15
  • I just found two or three typos. Other than that, the book is simply awesome. – F. Zer Mar 07 '20 at 01:20
  • 2
    As far as I see, the exercise D.7 in Ch. 16 of the book you mentioned ask for a proof in natural deduction of the formula $((A \to B) \to A) \to A$ (aka Pierce law), not of the formula $((A \to B) \to B) \to B$ you mentioned in the title. – Taroccoesbrocco Mar 07 '20 at 02:05
  • @Taroccoesbrocco, as I said to Rob, they probably changed it in a recent update. It was as I wrote it when I downloaded the book (probably one month ago). – F. Zer Mar 07 '20 at 12:12

2 Answers2

4

Use a truth table.

$$\begin{array}{|c|c|c|c|c|} \hline A & B & A \to B & (A \to B) \to B & ((A \to B) \to B) \to B\\ \hline 0 & 0 & 1 & 0 & 1\\ \hline 0 & 1 & 1 & 1 & 1\\ \hline 1 & 0 & 0 & 1 & 0\\ \hline 1 & 1 & 1 & 1 & 1\\ \hline \end{array}$$

Since the last column is not identically true (i.e., "$1$"), then you can conclude that $ ((A \to B) \to B) \to B$ is not a tautology.

the_candyman
  • 14,234
  • 4
  • 37
  • 65
2

Hint: $(A \to B) \to B$ is logically equivalent to $A \lor B$. So $((A \to B) \to B) \to B$ is logically equivalent to $(A \lor B) \to B$, which is not a tautology: if $A$ is true and $B$ is false, $(A \lor B) \to B$ is false.

Rob Arthan
  • 51,538
  • 4
  • 53
  • 105