I have 2 clauses of cardinality 3

$$(a \lor b \lor c)\land(a \lor \lnot b \lor\lnot c)$$

or, in set notation

$$\{a, b, c\}, \{a, \lnot b, \lnot c\}$$

I applied resolution incorrectly and got clause


Now, someone told me that if we apply resolution on these 2 clauses we get a tautology, but I don't understand why/how.
Can someone please explain? Thank you.


You only resolve on one literal. This was your mistake: you resolved on both $b$ and $c$, but you can only resolve on one of them.

Now, if you resolve on $b$, the result is $\{a,c,\neg c\}$, which is indeed a tautology, since whether $c$ is true or false, this clause will always be satisfied: it is a tautologous clause.

Likewise, if you resolve on $c$, the result is $\{a , b , \neg b\}$, which is a tautologous clause as well.

  • $\begingroup$ I'm confused on how you get a tautology here. If $a$ is false and either $b$ and $c$ are true or $b$ and $c$ are both false, won't the statement be false: $(F \vee T \vee T) \wedge (F \vee F \vee F)$? It would seem to me that this simplifies to $a \vee (b \oplus c)$. $\endgroup$
    – Jared
    Feb 9 '19 at 2:53
  • $\begingroup$ But then I may be confused on what is meant by "applying resolution". $\endgroup$
    – Jared
    Feb 9 '19 at 2:58
  • 1
    $\begingroup$ @Jared You are right that the original statement is not a tautology ... all that is being shown here is that, through the resolution rule, you can derive a tautology from this statement. But of course, any statement implies a tautology! So, the fact that a tautology can be derived from some statement means absolutely nothing ... it certainly does not mean that the original statement is a tautology itself, if that's what you're thinking. $\endgroup$
    – Bram28
    Feb 9 '19 at 3:54
  • $\begingroup$ @Bram28 Oooh, so clear... Thanks a lot 👍 $\endgroup$
    – Grasta
    Feb 9 '19 at 12:57
  • $\begingroup$ @Grasta You're welcome! :) $\endgroup$
    – Bram28
    Feb 9 '19 at 12:58

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