Mancosu - Philosophy of Mathematical Practice (Oxford, 2008).pdf

understanding proofs 339integers in a language with 0, 1, +, and< is decidable, and in the early1930s, Tarski showed that the theory of the real numbers in a language with0, 1, +, ×, and< is decidable (the result was not published, though, until1948, and was soon reprinted as Tarski, 1951). These decision procedureswork for the full first-order language, not just the quantifier-free fragments,and have been implemented in a number of systems. But decision proceduresdo not tell the whole story. Thanks to Gödel, we know that decidability isthe exception rather than the rule; for example, the theory of the integersbecomes undecidable in the presence of multiplication, and the theory of thereals becomes undecidable in the presence of, say, the sine function. Even inrestricted settings where problems are decidable, full decision procedures tendto be inefficient or even infeasible. In fact, for the types of inferences thatarise in practice, short justifications are often possible where general decisionprocedures follow a more circuitous route.There has therefore been a fair amount of interest in ‘heuristic procedures’,which search for proofs by applying a battery of natural inferences in asystematic way (for some examples in the case of real arithmetic, see Beeson,1998; Hunt et al., 2003; Tiwari, 2003). One strategy is simply to workbackwards from the goal inequality. For example, one can prove an inequalityof the form 1/(1 + s) 1 + t > 0, and onecandothat,inturn,byprovings > t > −1. This amounts to using basicrules likex > y, y > 0 ⇒ 1/x < 1/yto work backwards from the goal, a technique known as ‘backchaining’.Backchaining has its problems, however. For example, one can also prove1/x < 1/y by proving that x and y are negative and x > y, orthatx isnegative and y is positive. Trying all the possibilities can result in dreaded casesplits. Search procedures can be nondeterministic in more dramatic ways; forexample, one can prove q + r < s + t + u,say,byprovingq < t and r ≤ s + u,or by proving q < s + t + u and r ≤ 0. Similarly, one can prove s + t < 5 + uby proving, say, s < 3andt ≤ 2 + u.So, simply working backwards is insufficient on its own. In the proof ofTheorem 1, we used bounds on ( N ) and N to bound composite expressionskusing these terms, and in the proof of Lemma 1, we used the bounds |z| ≤N/2and N < n to bound the terms |z i /n i |. Working forwards in this way to amassa store of potentially useful inequalities is also a good idea, especially whenthe facts may be used more than once. For example, when all the terms insight are positive, noting this once and for all can cut down the possibilitiesfor backwards search dramatically. But, of course, deriving inequalities blindly