# Section 1.5: Problem 4 Solution

Working problems is a crucial part of learning mathematics. No one can learn... merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.
James R. Munkres
Let M be the ternary minority connective. (Thus always disagrees with the majority of , , and .) Show the following:
(a) is complete.
(b) is not complete.
(a) is true iff no more than one of and is true, i.e. it is equivalent to .
(b) However, neither not can be realized using only. Indeed, similar to Exercise 3, satisfies the property that “if we change all truth assignments to their opposites, the resulting value changes to its opposite as well“. By induction, any wff based on only, satisfies this property as well. Indeed, all sentence symbols satisfy it, and if , and satisfy it, then so does . Note, that unlike in Exercise 3, can realize , namely, is tautologically equivalent to . So that, similar to Exercise 3, is also not complete.