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

IMPLICIT:

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

OR

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

$$ \frac{PO\ OR\ P1\ P0\vdash R\ P1\vdash\ R}{R} $$ 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.

  • FYI: Here's an example of an answer which I know is correct . http://imagebin.org/287822 – peter_gent Jan 20 '14 at 18:45
  • do you have rules for "!" ? also Q13 looks not wellformed (contains no "|-") see also http://en.wikipedia.org/wiki/Sequent_calculus – Willemien Jan 22 '14 at 07:27
  • I believe the rule for not is P |- False / NOT(P) – peter_gent Jan 23 '14 at 12:45
  • 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? – Willemien Jan 23 '14 at 13:04
  • alas there is none – peter_gent Jan 24 '14 at 18:07
  • I guess you read http://math.stackexchange.com/questions/649053/good-textbook-for-learning-sequent-calculus? I am planning to add a bounty to my question. The only book I found that has a reasonable amount of sequent calculus is David Bostock's "Intermediate Logic" (Oxford University press, 1997). but sequent calculus is just the fourth way to do logic (after Tableaux, Natural deduction, axiomatic deductions). – Willemien Jan 24 '14 at 19:39
  • I did look at the book suggested by unwisdom in http://math.stackexchange.com/questions/649053/good-textbook-for-learning-sequent-calculus , looks like a good book using this system, have a look at it (did give my bonus to unwisdom) – Willemien Jan 31 '14 at 22:44

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