# Propositional Logic, Resolution - derived tautology from 2 clauses of cardinality 3…

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

$$\{a\}$$

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.

• 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 $a \vee (b \oplus c)$. – Jared Feb 9 at 2:53
• But then I may be confused on what is meant by "applying resolution". – Jared Feb 9 at 2:58
• @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. – Bram28 Feb 9 at 3:54
• @Bram28 Oooh, so clear... Thanks a lot 👍 – Grasta Feb 9 at 12:57
• @Grasta You're welcome! :) – Bram28 Feb 9 at 12:58