Question

State the basic principle of resolution method for both proposition and predicates

Answer 1

1. Implications (I):

 φ ⇒ ψ → ¬φ ∨ ψ

 φ ⇐ ψ → φ ∨ ¬ψ

 φ ⇔ ψ → (¬φ ∨ ψ) ∧ (φ ∨ ¬ψ)

2. Negations (N):

¬¬φ → φ

¬(φ ∧ ψ) → ¬φ ∨ ¬ψ

¬(φ ∨ ψ) → ¬φ ∧ ¬ψ

3. Distribution (D):

φ ∨ (ψ ∧ χ) → (φ ∨ ψ) ∧ (φ ∨ χ)

(φ ∧ ψ) ∨ χ → (φ ∨ χ) ∧ (ψ ∨ χ)

φ ∨ (φ1 ∨ ... ∨ φn) → φ ∨ φ1 ∨ ... ∨ φn

(φ1 ∨ ... ∨ φn) ∨ φ → φ1 ∨ ... ∨ φn ∨ φ

φ ∧ (φ1 ∧ ... ∧ φn) → φ ∧ φ1 ∧ ... ∧ φn

(φ1 ∧ ... ∧ φn) ∧ φ → φ1 ∧ ... ∧ φn ∧ φ

4. Operators (O):

 φ1 ∨ ... ∨ φn → {φ1, ... , φn}

 φ1 ∧ ... ∧ φn → {φ1}, ... , {φn}

As an example, consider the job of converting the sentence (g ∧ (r ⇒ f)) to clausal form. The conversion process is shown below.

g ∧ (r ⇒ f)

I g ∧ (¬r ∨ f)

N g ∧ (¬r ∨ f)

D g ∧ (¬r ∨ f)

O {g}

{¬r, f}

As a slightly more complicated case, consider the following conversion. We start with the same sentence except that, in this case, it is negated.

¬(g ∧ (r ⇒ f))

I ¬(g ∧ (¬r ∨ f))

N ¬g ∨ ¬(¬r ∨ f)

¬g ∨ (¬¬r ∧ ¬f)

¬g ∨ (r ∧ ¬f)

D (¬g ∨ r) ∧ (¬g ∨ ¬f)

O {¬g,r}

{¬g, ¬f}

Answer 2

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