I have to find, whether a CNF formula is satisfiable. I got multiple sets of clauses and of them is empty: $T(d)_{~} = \emptyset.$
Why isn't the $T(d)$ empty clause, if it contains no clauses, but it is an empty set of clauses?
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embe99
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The empty clause is a contradiction, i.e. it is unsatisfiable. – Mauro ALLEGRANZA Mar 20 '20 at 12:53
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1The empty clause is produced in Resolution proof procedure when we conjoin two clauses containing respectively a literal $l$ and its negation $\lnot l$. The result of the application of the Resolution rule is $l \land \lnot l$. – Mauro ALLEGRANZA Mar 20 '20 at 12:56
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So again, the question is: What would happen if I try to conjoin $(a \vee b \vee c)$ with $(\neg a \vee \neg b)$? Will I get $(b \vee \neg b \vee c)$ or this is just an empty clause because of the part $(b \vee \neg b)$ ? – embe99 Mar 20 '20 at 13:24
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1What happens conjoining $(a∨b∨c)$ with $(¬a∨¬b)$ ? See Resolution: "The resolution rule is applied to all possible pairs of clauses that contain complementary literals. After each application of the resolution rule, the resulting sentence is simplified by removing repeated literals. If the clause contains complementary literals, it is discarded (as a tautology). " – Mauro ALLEGRANZA Mar 20 '20 at 13:38
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1Thus, applying the rule twith regard to $a$ and $\lnot a$, the result will be $(b \lor \lnot b \lor c)$. But $b \lor \lnot b$ is $\text T$ and $\text T \lor c \equiv \text T$. – Mauro ALLEGRANZA Mar 20 '20 at 13:44
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Great, now I understand, how it is. Thank you very much :) . – embe99 Mar 20 '20 at 13:46
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1What does it mean ? That we have not derived the empty clause (i.e. a contradiction) and thus the initial set of clauses is satisfiable (check with $v(c)=\text T$ and $v(a)=\text F$). – Mauro ALLEGRANZA Mar 20 '20 at 13:46
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Refer to this answer: In Satisfiability, what is the difference between the empty clause and the empty set? – x.projekt Sep 25 '20 at 14:29
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Remember how you convert a CNF formula to clauses. For example, something like:
$(A \lor B) \land (\neg A \lor C \lor D) \land (\neg B \lor \neg C \lor \neg D)$
becomes the clause set:
$\{ \{ A , B \}, \{\neg A , C , D \} , \{ \neg B , \neg C , \neg D\} \}$
OK, but this means that an individual clause corresponds to a generalized disjunction, while the clause set as a whole corresponds to a generalized conjunction.
Now, a generalized disjunction is true iff at least one of its disjuncts is true. But, an empty clause, which corresponds to a generalized disjunction with $0$ disjuncts, clearly cannot have at least one true disjunct. Hence, an empty clause corresponds to a contradiction.
On the other hand, a generalized conjunction is true iff all of its conjuncts are true. Well, this would trivially be the case if there are no conjuncts at all: all zero of them are true! So, a generalized conjunction with $-$ conjuncts corresponds to a tautology. Hence, an empty clause set, which corresponds to a generalized conjunction with $-$ conjuncts, is a tautology.
