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Propositional versions of resolution:
Just because of so far we've only looked at propositional versions of resolution. However in first-order logic we require to also deal with such variables and quantifiers. And need to worry just about quantifiers: that we are going to be working with sentences like only contain free variables. So that recall that we treat these variables as implicitly universally quantified so by that they can take any value. Moreover this allows us to state a more generally first-order binary resolution inference rule as:
A ? B, ¬ C? D
Subst(θ, B) = Subst(&theta, C)
Subst(θ, A ? D)
For such rule has the side condition Subst(θ, B) = Subst(&theta, C), that uses there to be a substitution θ that makes B and C the same just before we can apply the rule. Notice there θ can substitute fresh variables when making B and C equal. But there it doesn't have to be a ground substitution! Like if we can find such a θ and we can make the resolution step and apply θ to the outcome. In generally we say that the first-order binary rule is simply equivalent to applying the substitution to the original sentences, and then applying the propositional binary rule.
Prove the equations A + A‾ .B + A.B‾ = A + B using the Boolean algebraic theorems ? Ans. The equation is A + A‾.B + A.B‾ = A + B L.H.S. = A + A‾ .B + A.B‾ = (A + A.B‾) + A‾.B
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