Exercise 1.5 from Arnold Milner Logic Notes:
While disjunction is easily defined via implication (
p v q = p->(q->p)) I have trouble defining conjunction and guess this is impossible. I've examined truth tables for expressions with 3 terms and need an insight why this exhausts the search. Or, perhaps, I need to invoke some more advanced method of logic inexpressibility?
The problem reduces to "smaller" one: if I express false constant
0 in terms of implication, then it will allow negation (via
-p = p->0) and, consequently, conjunction (via De Morgan's law). This would give full set of connections, which we know only Sheffer connective and its dual enjoy. Therefore, neither
0, nor negation is expressed in terms of implication as well?