Presburger Arithmetic is a decidable theory but if multiplication is added to it would that theory remain decidable?
UPDATE: I began to write out the axioms that would distinguish Presburger Arithmetic from Peano Arithmetic and realized that adding the Peano axioms which give semantics for multiplication (E.g., commutativity of multiplication, the distributive law -- essentially, all of the peano axioms which include the multiplication symbol -- etc) to Presburger Arithmetic result in Peano Arithmetic. And the decidability of Peano Arithmetic is already widely known due to Gödel. So I've decided to change my question: why does a definition of multiplication in Presburger Arithmetic result in an undecidable theory? In other words, why would the addition of the multiplication symbol and axioms for multiplication result in an undecidable theory?