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  "body": "\n## Example\n\n*Presburger arithmetic* is the first-order theory of the [[natural numbers|0MWKJ44V5A]] with $0$, $1$ and addition.\nIt is the set of all sentences in the signature $\\set{0, 1, +}$ that are true in $(\\N, 0, 1, +)$.\nIt can be axiomatized by the axioms $x + 1 \\ne 0$, $x + 1 = y + 1 \\to x = y$, $x + 0 = x$, $x + (y + 1) = (x + y) + 1$, and the induction scheme for all formulas of this signature.\n\nIt does not have quantifier elimination in $\\set{0, 1, +}$.\nIt has quantifier elimination after adding $\u003c$ and, for each $n \\ge 2$, a unary predicate $D_n(x)$ for \"$n$ divides $x$\".\nCheck on one formula: in $\\N$, the formula $\\exists y \\, (x = y + y)$ says that $x$ is even.\nIt is equivalent to the quantifier-free formula $D_2(x)$.\nIt is not equivalent to any quantifier-free formula in $\\set{0, 1, +}$, since those define finite or cofinite subsets of $\\N$.\nThe set of even numbers is neither finite nor cofinite.\n\n## Remarks\n\nQuantifier elimination gives an algorithm that decides, for every sentence of the signature, whether it is true in $\\N$.\nSo Presburger arithmetic is [[complete|655K8Y7VS8]] and decidable.\nMojżesz Presburger proved this in 1929.\nIn contrast, the theory of $(\\N, 0, 1, +, \\cdot)$ is undecidable, and the [[Peano axioms|0QFT7Z76N5]] do not decide all its sentences, by Gödel's incompleteness theorems.\nMultiplication of two variables is therefore not definable from addition in $\\N$.\nThe same method applies to the [[integers|0C4JS9RF5M]] with addition and order, where divisibility by $n$ is [[congruence modulo $n$|0WAY5WX8NY]] to $0$.\n\n## Explanations\n\n### L1 human\n\nPresburger arithmetic is arithmetic of the natural numbers with addition but without multiplication.\nEvery statement in this language can be decided by a computer, unlike full arithmetic with multiplication.\nThe trick is that \"there exists\" can be removed if one also allows the order and statements such as \"$x$ is divisible by $3$\".\n",
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