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Assume $$C_1 = (x_1 \lor x_2 \lor \lnot x_3)\hspace{0.2cm} C_2 = (x_4 \lor x_5 \lor x_3) \hspace{0.2cm} C_3 = (x_3 \lor x_5 \lor x_6)$$ Let

$$ \phi_1 = C_1 \land C_2 \land C_3$$ and $$ \phi_2 = (x_1 \lor x_2 \lor x_4 \lor x_5) \land (x_3 \lor x_5 \lor x_6)$$

Question

Are the two formulas $\phi_1, \phi_2$ equisatisfiable? Basically I have merged two clauses ($C_1$ and $C_2$) because they had a literal appearing in one and also in the other one negated. The same literal appears in the third clause $C_3$ which I have simply left untouched ...

what I did

set $x_3 = 1$ and observed that both can be satisfied then set the variable to $x_3 = 0$ and observe as well that both can satisfied. So ... at least if $\phi_1$ is satisfiable then $\phi_2$ is as well ... For the reverse I can do the same here, but is it always the case ?

follow up

Is this merging always possible? i.e merge two clauses in a CNF like so, if they share a variable which negated in one as opposed to another.

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  • $\begingroup$ Merging a conjunction of two disjunctive clauses that contain complementary literals eg: $(x \lor y) \land (\neg x \lor z) \Leftrightarrow (x \lor y) \land (\neg x \lor z) \land (y \lor z)$. The third clause $y \lor z$ is the resolvent of the first two clauses and this is called the resolution rule. In your example, $\phi_1 \Leftrightarrow C_1 \land C_2 \land \C_3 \Leftrightarrow C_1 \land C_2 \land C_3 \land resolvent(C_1, C_2) \Leftrightarrow C_1 \land C_2 \land \phi_2 C_3$. $\endgroup$
    – vvg
    Commented Mar 9, 2023 at 9:54
  • $\begingroup$ I am not necessarily asking for logical equivalence ... I am "satisfied" with just equisatisfiability with the advantage that I have reduced the number of clauses. I am not necessarily interested in their size. $\endgroup$
    – C Marius
    Commented Mar 9, 2023 at 9:59
  • $\begingroup$ If you use the resolution rule, you get equivalence. Equivalence gives you equisatisfiability (not vice versa). $\endgroup$
    – vvg
    Commented Mar 9, 2023 at 10:02
  • $\begingroup$ I think two formulas can be equisatisfiable but not equivalent see en.wikipedia.org/wiki/…. Equivalence is too much for me here $\endgroup$
    – C Marius
    Commented Mar 9, 2023 at 10:07

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