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$$((q → p) → q) → q$$ There are no premises which makes this problem very difficult for me. What I tried was assuming that the whole left side $((q → p) → q)$ then I assumed $¬q$ and tried to find a contradiction so I can conclude $q$, then do a conditional proof. However, it doesn't seem to be working. Any hints/help?

Bumblebee
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user737163
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    But your approach is correct; assume $\lnot q$ and $q$ and derive $p$ from which $(q \to p)$ (discharging assumption $q$). With it and premise derive $q$ and you have the contradiction needed to conclude with $\lnot \lnot q$ (discharging assumption $\lnot q$) – Mauro ALLEGRANZA Nov 10 '21 at 15:52
  • Hint: use the truth table. – xpaul Nov 10 '21 at 16:10
  • Like what MJD said, use the fitch system! This is the way to do these kinds of proofs by natural deduction. – James Nov 10 '21 at 16:31

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You can freely download my intro logic book from https://www.logicmatters.net/ifl -- and you'll find that it does propositional logic natural deduction proofs Fitch-style over a number of chapters. (I like to think it is reasonably clear! ;-)

And the proof of this theorem is on pp. 212--213.

This is so-called Pierce's Law, which is rather interesting as it only involves the conditional, but it can't be proved using the standard inference rules for the conditional alone. You need also to appeal to the law of excluded middle or to an equivalent like the double negation law (as in the proof I give).

Peter Smith
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Using the truth table, one has \begin{array}{|c|c|c|c|c|} \hline q & p & q\to p&(q\to p)\to q&((q\to p)\to q)\to q \\ \hline 1 & 0 & 0 & 1 & 1\\ \hline 1 & 1 & 1 & 1 & 1\\ \hline 0 & 0 & 1 & 0 & 1\\ \hline 0 & 1 & 1 & 0 & 1\\ \hline \end{array} and hence $$ ((q\to p)\to q)\to q. $$

xpaul
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