How to distribute this logic expressions

Question

I want to prove the binary resolution proof that if I have $a \vee b$ and $ \lnot b \vee c$ then this will imply $$ a \vee c$$

Now I want to do it this way $$(a \vee b) \wedge ( \lnot b \vee c) \implies (a \vee c)$$

So i want to distribute this and among the expression to arrive at the conclusion.

I tried to apply the distribute property and hence I have $$(a \wedge (\lnot b \vee c) \vee (b \wedge ( \lnot b \vee c))$$

and then I will have $$((a \wedge \lnot b) \vee (a \wedge c)) \vee (b \vee \lnot b) \vee (b \wedge c)) $$

I know that $$b \wedge \lnot b$$ will cancel but then How would I simplify

$$((a \wedge \lnot b) \vee (a \wedge c)) \vee (b \wedge c)) $$

? any suggestions please ! ?

Answer 1

We have to prove that :

$((a∨b)∧(¬b∨c)) \to (a∨c)$

is a tautology.

After your application of distributivity, we have :

$[(a∧¬b)∨(a∧c)∨(b∧c)] \to (a∨c)$.

We can rewrite it as :

$\lnot [(a∧¬b)∨(a∧c)∨(b∧c)] \lor (a∨c)$

that is equivalent to :

$[(\lnot a \lor b) \land (\lnot a \lor \lnot c) \land (\lnot b \lor \lnot c)] \lor (a∨c)$.

By distributivity again, we get a conjunction of three disjuncts :

(1) $[(\lnot a \lor b) \lor (a \lor c)] \equiv [(\lnot a \lor a) \lor (b \lor c)] \equiv T \lor (b \lor c) \equiv T$

(2) $[(\lnot a \lor \lnot c) \lor (a \lor c)] \equiv [(\lnot a \lor a) \lor (\lnot c \lor c)] \equiv T \lor T \equiv T$

(3) $[(\lnot b \lor \lnot c) \lor (a \lor c)] \equiv [(\lnot b \lor a) \lor (\lnot c \lor c)] \equiv (\lnot b \lor a) \lor T \equiv T$.

Thus, the original formula is equivalent to :

$T \land T \land T \equiv T$

i.e. it is a tautology, and thus we can conclude (again) that the Resolution rule is sound.