請注意這是由另外的假設體係導出交換律，　與此同時，　算術公理體係也可以參考這裏 http://en.wikipedia.org/

來源: littlecat8 於 2014-07-11 10:41:39 [博客] [舊帖] [給我悄悄話] 本文已被閱讀：次

There are many different, but equivalent, axiomatizations of Peano arithmetic. While some axiomatizations, such as the one just described, use a signature that only has symbols for 0 and the successor, addition, and multiplications operations, other axiomatizations use the language of ordered semirings, including an additional order relation symbol. One such axiomatization begins with the following axioms that describe a discrete ordered semiring.^[8]

$\forall x, y, z \in N$ . $(x + y) + z = x + (y + z)$ , i.e., addition is associative.
$\forall x, y \in N$ . $x + y = y + x$ , i.e., addition is commutative.
$\forall x, y, z \in N$ . $(x \cdot y) \cdot z = x \cdot (y \cdot z)$ , i.e., multiplication is associative.
$\forall x, y \in N$ . $x \cdot y = y \cdot x$ , i.e., multiplication is commutative.
$\forall x, y, z \in N$ . $x \cdot (y + z) = (x \cdot y) + (x \cdot z)$ , i.e., the distributive law.
$\forall x \in N$ . $x + 0 = x \and x \cdot 0 = 0$ , i.e., zero is the identity element for addition.
$\forall x \in N$ . $x \cdot 1 = x$ , i.e., one is the identity element for multiplication.
$\forall x, y, z \in N$ . $x < y \and y < z \Rightarrow x < z$ , i.e., the '<' operator is transitive.
$\forall x \in N$ . $\neg (x < x)$ , i.e., the '<' operator is irreflexive.
$\forall x, y \in N$ . $x < y \or x = y \or y < x$ .
$\forall x, y, z \in N$ . $x < y \Rightarrow x + z < y + z$ .
$\forall x, y, z \in N$ . $0 < z \and x < y \Rightarrow x \cdot z < y \cdot z$ .
$\forall x, y \in N$ . $x < y \Rightarrow \exists z \in N$ . $x + z = y$ .
$0 < 1 \and \forall x \in N$ . $x > 0 \Rightarrow x \geq 1$ .
$\forall x \in N$ . $x \geq 0$ .

The theory defined by these axioms is known as PA^–; PA is obtained by adding the first-order induction schema.