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By Mark V. Lawson

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The connectives ¬ and ∧ together form an adequate set. 2. The connectives ¬ and ∨ together form an adequate set. 3. We show that the following wff are equivalent to wff using only the connectives ¬ and ∧. 1. p ∨ q ≡ ¬(¬p ∧ ¬q). 2. p → q ≡ ¬p ∨ q ≡ ¬(¬¬p ∧ ¬q) ≡ ¬(p ∧ ¬q). 3. p ↔ q ≡ (p → q) ∧ (q → p) ≡ ¬(p ∧ ¬q) ∧ ¬(q ∧ ¬p). At this point, you might wonder if we can go one better. Indeed, we can but we have to define some new binary connectives. Define p ↓ q = ¬(p ∨ q) called nor. Define p | q = ¬(p ∧ q) called nand.

This is left as an exercise. 30 CHAPTER 1. 1. Don’t be perturbed by the number of atoms in the above examples nor by the amount of labour needed to write down the wff A. The point is that the problem can be faithfully represented by a wff in PL and that the solution of the problem is achieved via a satisfying assignment of the atoms. ’ later on. In many ways, PL is like an assembly language and it is perfectly adapted to studying a particular class of problems that are widespread and important.

We construct truth function. T T T T F F F F a wff that has as truth table the following T T F F T T F F T F T F T F T F T (1) F F T (2) T (3) F F F We need only consider the rows that output T , which I have highlighted. I have also included a reference number that I shall use below. The basic conjunction corresponding to row (1) is p ∧ q ∧ r. The basic conjunction corresponding to row (2) is p ∧ ¬q ∧ ¬r. The basic conjunction corresponding to row (3) is ¬p ∧ q ∧ r. The disjunction of these basic conjunctions is A = (p ∧ q ∧ r) ∨ (p ∧ ¬q ∧ ¬r) ∨ (¬p ∧ q ∧ r).

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