Boolean operators and Induction

Question

Let @ denote the binary boolean operator defined by the right-hand side below: p @ q = (p ^ ¬q)

(b) Is the set of operators {@, ¬} complete? Explain in detail.

(c) Prove by induction that any propositional formula in a single propositional variable p that uses only the boolean operator @ (or no operators at all) is equivalent to either the truth value False or to the single propositional variable p. Explain.

Answer 1

(b) Yes.

• ~(p @ q) = ~(p & ~q) = ~p | q = p -> q.
• p @ ~q = p & ~~q = p & q.
• ~(~p @ q) = ~(~p & ~q) = ~~p | ~~q = p | q.

(c) An inductive proof might look like the following:

Base case: p is equivalent to p, while p @ p is False since p & ~p is a contradiction.

Induction hypothesis: Assume all propositions up to length k operations containing only p and @ are equivalent either to p or to False.

Induction step: Every proposition containing only p and @ can be decomposed into a formula of the form ((x) @ (y)), where x and y are formulas of length less than or equal to k. By the induction hypothesis, both x and y are equivalent to either p or False. There are four cases:

• x = p, y = p; then x @ y = False, as required;
• x = p, y = False; then x @ y = p, as required;
• x = False, y = p: then x @ y = False, as required;
• x = False, y = False: then x @ y = False, as required.

In all four cases we find that ((x) @ (y)) must be equivalent either to p or to False.