I've been given the following introduction proofs and I need to solve the following questions. There aren't yet any answers to the solutions so I was hoping for some help, to compliment the progress that I've made so far. I've shared the question sheet here. At the moment I don't have clear answers for 11, 12 and 13 and others. As it's frowned upon to post whole sheets on here I thought Id just add a few related questions.

Introduction proofs:

AND: $$ \frac{P \ Q}{P\ AND\ Q} $$


$$ \frac{P0 \Rightarrow P1\ P0}{P1} $$


$$ \frac{P0\ OR\ P1}{P0\ P1} $$

$$ \frac{PO\ OR\ P1\ P0\vdash R\ P1\vdash\ R}{R} $$ NOT $$ \frac{P\ \vdash\ FALSE}{NOT(P)} $$

Q11) $$ \frac{R}{\frac{P\ \vdash Q\ , Q\vdash\ R\ ,\ \ P\ AND\ Q\ \Rightarrow\ R}{(P\ \Rightarrow\ Q)\ OR\ (Q\ \Rightarrow\ R)\ \Rightarrow\ (P\ AND \ Q\ \Rightarrow\ R)}} $$

1) Make assumption that left side can be proven ( assume p then q, assume q then R)
2) prove right sides inference with R, assume R can be proven.

Q12) $$ \frac{\frac{Q\ \vdash\ FALSE}{NOT(Q)},\ \frac{R\ \vdash\ FALSE}{NOT(R)}}{P\ \Rightarrow\ !Q\ AND\ !R} $$

Q13) $$ \frac{Q,P\ \vdash\ NOT(Q)\ , NOT(P)}{(P\ \Rightarrow\ Q)\Rightarrow\ !Q\Rightarrow\ !P} $$

Start with right most P - P can be assumed, as can Q.
2) P=>Q assume p, if you can prove Q. If you can assume Q you can prove ! Q and ! D.

  • $\begingroup$ FYI: Here's an example of an answer which I know is correct . imagebin.org/287822 $\endgroup$ – peter_gent Jan 20 '14 at 18:45
  • $\begingroup$ do you have rules for "!" ? also Q13 looks not wellformed (contains no "|-") see also en.wikipedia.org/wiki/Sequent_calculus $\endgroup$ – Willemien Jan 22 '14 at 7:27
  • $\begingroup$ I believe the rule for not is P |- False / NOT(P) $\endgroup$ – peter_gent Jan 23 '14 at 12:45
  • $\begingroup$ It is all to bad: a lecturer who comes with not wellformed formula's and your inference rules are (very) incomplete, you are studying logic using sequent calculus unfortunedly i don't know any good textbook that uses it (most use the siimpler natural deduction method) which book are you using? $\endgroup$ – Willemien Jan 23 '14 at 13:04
  • $\begingroup$ alas there is none $\endgroup$ – peter_gent Jan 24 '14 at 18:07

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