By Stephen Pollard

This booklet is predicated on premises: one can't comprehend philosophy of arithmetic with out knowing arithmetic and one can't comprehend arithmetic with out doing arithmetic. It attracts readers into philosophy of arithmetic by means of having them do arithmetic. It deals 298 routines, masking philosophically vital fabric, offered in a philosophically knowledgeable method. The workouts supply readers possibilities to recreate a few arithmetic that might light up vital readings in philosophy of arithmetic. themes contain primitive recursive mathematics, Peano mathematics, Gödel's theorems, interpretability, the hierarchy of units, Frege mathematics and intuitionist sentential good judgment. The publication is meant for readers who comprehend easy houses of the common and genuine numbers and feature a few heritage in formal logic.

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**Extra resources for A Mathematical Prelude to the Philosophy of Mathematics**

**Example text**

42 2 Peano Arithmetic, Incompleteness Let C be the set of all these sentences. Each member of C says something compatible with our conception of the natural numbers. For example, c = SSSSS0 makes the innocent claim that a certain (so far unidentified) natural number is distinct from five. Given any model of PA, we could make the above sentence true by assigning to ‘c’ an object in the range of our bound variables. Just let c be anything other than the object named by ‘SSSSS0’. 8 Suppose C ∃ is a proper subset of C.

1. We can let φ(x, y) be any PA-formula with free occurrences only of ‘x’ and ‘y’. If we let φ(x, y) be ‘(y + x) = x’, we obtain the axiom ⇐y((y + 0) = 0 ⇒ (⇐x((y + x) = x ⇒ (y + Sx) = Sx) ⇒ ⇐x (y + x) = x)) This says the formula ((y + 0) = 0 ⇒ (⇐x((y + x) = x ⇒ (y + Sx) = Sx) ⇒ ⇐x (y + x) = x)) will be true no matter what y is. So, in particular, we are free to replace each free occurrence of ‘y’ with an occurrence of ‘0’: ((0 + 0) = 0 ⇒ (⇐x((0 + x) = x ⇒ (0 + Sx) = Sx) ⇒ ⇐x (0 + x) = x)) This sentence says that you can show ⇐x (0 + x) = x by first showing (0 + 0) = 0 (to get through the first ‘⇒’) and then showing ⇐x((0 + x) = x ⇒ (0 + Sx) = Sx) (to get through the last ‘⇒’).

Can we identify such a formula φ(x)? Well, does the vocabulary of PA allow us to say that a number x is odd? Of course it does. We just say that x is not a multiple of two: ⇐y (y · SS0) = x. As it turn out, this formula really does represent f because PA proves each of the sentences ⇐y (y · SS0) = S0, ⇐y (y · SS0) = SSS0, ⇐y (y · SS0) = SSSSS0, . . and disproves each of the sentences ⇐y (y · SS0) = 0, ⇐y (y · SS0) = SS0, ⇐y (y · SS0) = SSSS0, . . So, when we offer f the number n and f responds “yes”, we do not just learn that n is odd: we learn that PA thinks n is odd.