Solve this by converting sentences to clausal form?

Question

If:

$E\land R \implies B$
$E\implies R\lor P\lor L$
$K\implies B$
$\lnot(L\land B)$
$P\implies \lnot K$

Which of the following can't be deducted?

1. $E\land P$
2. $K \land E \implies R$
3. $L \land P \implies\lnot K$
4. $L \implies\lnot(K \land E)$

Answer 1

Let's use the following resolution rule

$\frac{a \vee b, \quad \neg a \vee c} {b \vee c}$

multiple times to deduce 2, 3, 4 from the following (given) knowledgebase

(a) $\lnot E\lor \lnot R \lor B$
(b) $\lnot E \lor R\lor P\lor L$
(c) $\lnot K \lor B$
(d) $\lnot L \lor \lnot B$
(e) $\lnot P \lor \lnot K$

From (c) and (d) we can deduce (f), i.e.,

$B \lor \lnot K, \quad \lnot B \lor \lnot L$

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(f) $\lnot K \lor \lnot L$

From (f) we can deduce $\lnot K \lor \lnot L \lor \lnot P$, which means we can deduce $\lnot (L \land P) \lor \lnot K$, i.e.,

3. $L \land P \implies\lnot K$

Also, From (f) we can deduce $\lnot K \lor \lnot L \lor \lnot E$, which means we can deduce $\lnot L \lor \lnot (K \land E)$, i.e.,

4. $L \implies\lnot (K \land E)$

Again, from (b) and (f) we can deduce (g)

$L\lor \lnot E \lor R\lor P, \quad \lnot L \lor \lnot K$

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(g) $\lnot E \lor R\lor P\lor \lnot K$

Now, from (g) and (e)

$P\lor \lnot E \lor R\lor \lnot K, \quad \lnot P \lor \lnot K$

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$\lnot E \lor R\lor \lnot K$

which means we have deduced $\lnot (K \land E) \lor R$, i.e.,

2. $K \land E \implies R$

Thus we have proved that 2, 3, 4 can be deduced from the knowledgebase provided.

We can see 1 is not necessarily TRUE even if (a)-(e) holds TRUE, e.g., Let K=False, L=False, E=False, P=False, then (a)-(e) holds TRUE but 1 does not hold TRUE.