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By Mark V. Lawson

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Let A be a wff. First convert A to NNF and then if necessary use the distributive laws to convert to a wff which is in DNF. 5. We show how to convert ¬(p → (p ∧ q)) into DNF using a sequence of logical equivalences. The first step is to replace →. We use the fact that x → y ≡ ¬x∨y. This gives us ¬(¬p∨(p∧q)). Now use de Morgan’s laws to push negation inside the brackets. This yields ¬¬p ∧ ¬(p ∧ q) and then ¬¬p ∧ (¬p ∨ ¬q). We now apply double negation to get p ∧ (¬p ∨ ¬q). This is in NNF. Finally, we apply one of the distributive laws to get the ∨ out of the brackets.

E) p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r) and p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r). Distributivity. (f) ¬(p ∧ q) ≡ ¬p ∨ ¬q and ¬(p ∨ q) ≡ ¬p ∧ ¬q. De Morgan’s laws. 5. Let F stand for any wff which is a contradiction and T stand for any wff which is a tautology. Prove the following. (a) p ∨ ¬p ≡ T . (b) p ∧ ¬p ≡ F . (c) p ∨ F ≡ p. (d) p ∨ T ≡ T . (e) p ∧ F ≡ F . (f) p ∧ T ≡ p. 6. Prove the following by using known logical equivalences (rather than using truth tables). (a) (p → q) ∧ (p ∨ q) ≡ q. (b) (p ∧ q) → r ≡ (p → r) ∨ (q → r).

To model this problem we shall need 27 atomic statements cijk where 1 ≤ i ≤ 3 and 1 ≤ j ≤ 3 and 1 ≤ k ≤ 3. The atomic statement cijk is defined as follows cijk = the cell in row i and column j contains the number k. For example, the atomic statement c113 is true when the grid is as follows: 3 ? ? ? ? s mean that we don’t know what is in that cell. In the above case, the atomic statements c111 and c112 are both false. We shall now construct a wff A from the above 27 atoms such that A is satisfiable if and only if the above problem can be solved and such that a satisfying truth assignment can be used to read off a solution.

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