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Sentenial Logic homework? Hi everyone.. So I have sentenial Logic problems... I can't figure this out can anyone help?

Given the following four sets:

  • $A=\{\neg P\to (R \to Q), \neg(\neg R \land P), R \lor\neg Q\}$
  • $B=\{R \leftrightarrow (R \lor (P \land\neg P)), R \leftrightarrow\neg P, \neg P \to (P \leftrightarrow (Q \to Q)), P \to Q\}$
  • $\Gamma=\{\neg(P \land Q), \neg R \to Q, R \to\neg P,\neg P \to P\}$
  • $\Delta=\{R \leftrightarrow (\neg(P \to (Q \land\neg R)) \to P), P, Q \lor R\}$

    1. State which of them is consistent and which inconsistent.

    2. From one of the consistent sets $R$ follows, from one of them $\neg R$ follows, and from one of them neither $R$ nor $\neg R$ follows. State which is which.

    3. Give a derivation of $\neg R$ from the consistent set from which it follows.

    4. Give a derivation of $R$ from the consistent set from which it follows.

    5. For each inconsistent set, give a proof which will show that it is inconsistent

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closed as off-topic by Lord Soth, Daniel Fischer, user61527, Kirill, azimut Oct 31 '13 at 22:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

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If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ What are your thoughts? You don't expect random people on the internet to do your homework for you just like that, do you? $\endgroup$ – Henning Makholm Oct 31 '13 at 20:59
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To figure out (1), make four truth-tables $T_A$, $T_B$, $T_G$, $T_D$ and then use this equivalence:

Set X is consistent $=_{df}$ there exists a row in $T_X$ that assigns 'T' to all $y \in X$.

Three will turn out to be consistent, because in the truth-tables for three of them you will find a row that assigns 'T' to all member formulas. But don't throw away the truth-tables just yet, because you can use them to figure out (2).

To figure out (2), take the truth-tables for the three consistent sets and use this definition:

R follows from set X $=_{df}$ all rows that assign 'T' to all $y \in X$, assign 'T' to variable R.

In other words, R follows from X just in case no assignment of truth-values to the propositional variables makes all formulas in X true and propositional variable R false. Once you follow that procedure you will have identified three sets, $S_R$, $S_{\lnot R}$, and $S$, from which $R$, $\lnot R$, and (neither $R$ nor $\lnot R$) follow, respectively.

To figure out (3), you need to prove $S_{\lnot R} \vdash \lnot R$, so start by assuming $R$. I don't want to spoil your fun by giving the answer, so let me just hint that at some point you will get $\lnot P$, which will soon take you to an absurdity. Having reached a contradiction at that point you can conclude that $\lnot R$.

To figure out (4) all you need to do is focus on a single premise in whichever set it is that you found to be the $S_R$ set. Let me just hint that when you use the definition of '$\rightarrow$' and apply a couple of De Morgans you will reach an equivalence of the following form: $(\phi \leftrightarrow \top)$, which will give you $\phi$!

To figure out (5), choose a conditional (i.e. a formula of form $\phi \rightarrow \psi$) in the inconsistent set $S_\bot$, turn that conditional into a disjunction using the equivalence $(\phi \rightarrow \psi) \equiv (\lnot \phi \lor \psi)$, then proceed to derive a contradiction by case analysis: assume $\phi$, use the premises to get a contradiction, then assume $\psi$ and also derive a contradiction. Conclude by $\lor$-elimination that $S_\bot$ is inconsistent.

If you're having trouble executing any of these steps, leave a comment.

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