# 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.