I am studying theoretical computer science and I am in the part about resolution in propositional calculus. I was reading a theorem (and its proof) that propositional resolution is refutation complete, but I could not fill a particular detail.

The theorem is:

Refutational Completeness of Propositional Resolution. The resolution method in propositional calculus is refutation complete, i.e., a set of propositional clauses $S$ is unsatisfiable if and only if the empty clause can be deduced from $S$ by application of the cut rule and the simplification rule.

The sketch of the proof given is:

Proof Sketch.

($\Leftarrow$). By correction we have that if $\square$ is deduced from $S$ then $S$ is unsatisfiable.

($\Rightarrow$). First you need to show that if $S$ is inconsistent and the set of propositional symbols that occur in the clauses of $S$ are $\{p_1, \ldots, p_n \}$, then it is possible to obtain from $S$ by resolution a set of clauses $S'$ such that only the propositional symbols $\{p_1, \ldots, p_{n-1} \}$ occur in $S'$. The proof will then follow by induction on the number of propositional symbols that occur in $S$. Details are left as exercise.

The cut rule and the simplification rule were defined previously as:

Cut Rule. Let $p \lor l_1 \lor \ldots \lor l_n$ and $\lnot p \lor l'_1 \lor \ldots \lor l'_m$ be two propositional clauses. Then the clause $l_1 \lor \ldots \lor l_n \lor l'_1 \lor \ldots l'_m$ is inferred from the two previous ones by the cut rule.

Simplification Rule. Let $p \lor p \lor l_1 \lor \ldots \lor l_n$ be a propositional clause. Then, the clause $p \lor l_1 \lor \ldots \lor l_n$ is inferred from the one before by the simplification rule. Equally, $\lnot p \lor l_1 \lor \ldots \lor l_n$ is inferred from $\lnot p \lor \lnot p \lor l_1 \lor \ldots \lor l_n$ by the simplification rule.

To obtain the set $S'$ from $S$ my idea was to first notice that since $S$ is unsatisfiable, there is at least one propositional symbol $p_i$ such that both $p_i$ and $\lnot p_i$ appear in clauses of $S$. Without loss of generality, we can assume $p_i = p_n$. Then, let $p_n \lor l_1 \lor \ldots \lor l_n$ and $\lnot p_n \lor l'_1 \lor \ldots \lor l'_m$ be two clauses of $S$ and notice that we can apply the cut rule to obtain only $l_1 \lor \ldots l_n \lor l'_1 \lor \ldots l'_m$. We can proceed in this way as many times as necessary and obtain a set $S'$ such that only the propositional symbols $\{p_1, \ldots, p_{n-1}\}$ appear in $S'$.

However, I could not prove that the set $S'$ obtained this way is inconsistent. For instance, if $S = \{p_1 \lor p_2, p_1 \lor \lnot p_2 \}$, then $S' = \{p_1\}$ and I don't see how I can derive $\square$.

What am I missing? How can I fill the details in this proof? Thanks in advance!


1 Answer 1


There is an error in the example: $S = \{p_1 \lor p_2, p_1 \lor \lnot p_2 \}$ is actually satisfiable. Every interpretation $I$ that sends $p_1$ to $true$ actually works.

With this in mind, notice that if $S$ is unsatisfiable, then for every proposition symbol $p_i$ we may have both $p_i$ and $\lnot p_i$ occurring in $S$ or only one of them. Construct the unsatisfiable subset $S_n$ of $S$ that omits the propositional symbols $p_i$ such that only one of $p_i$ or $\lnot p_i$ occur in $S$. For instance, if $S = \{p_1 \lor p_2, p_1 \lor \lnot p_2, \lnot p_1, p_3 \lor p_1 \}$, then $S_n = \{p_1 \lor p_2, p_1 \lor \lnot p_2, \lnot p_1\}$.

We can then proceed as you described, using the cut rule to combine every pair $p \lor l_1 \lor \ldots \lor l_n$ and $\lnot p \lor l'_1 \lor \ldots \lor l'_m$ and replace them by $l_1 \lor \ldots \lor l_n \lor l'_1 \lor \ldots \lor l'_m$. After that we need to apply the simplification rule to the new rules. We can then form the set $S_{n-1}$ that only contains the propositional symbols $\{p_1, \ldots, p_{n-1} \}$ and also satisfies the property that we have both $p_i$ and $\lnot p_i$ occuring in the clauses of $S_{n-1}$, for every $1 \leq i \leq n-1$. This property will allow us to continue this process of eliminating one propositional symbol at a time until we obtain the empty clause $\square$.


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