有限左分配代数与嵌入代数\endtitle
Finite left-distributive algebras and embedding algebras\endtitle
摘要：
本文考虑具有一个二元运算$\cdot$和一个生成元（{ \it单演}）并满足左分配律$a\cdot (b\cdot c)=(a\cdot b)\cdot (a\cdot c)$的代数.定义一个有限左分配代数序列$A_n$，然后取一个极限得到一个无限单演左分配代数$A_\infty$ . Laver和Steel假设一个强的大基数公理的结果意味着$A_\infty$是自由的; $A_\infty$的自由性是否可以在没有大基数假设的情况下证明，甚至在Peano算法中证明，这是开放的。本文的主要结果是这个问题与自然数上的增函数的某种代数的存在性等价，称为{ \it嵌入代数}。利用这一点和第一作者的结果，我们得出结论，$A_\infty$的自由性是不可证明的原始递归算术。
英文摘要：
We consider algebras with one binary operation $\cdot$ and one generator ({\it monogenic}) and satisfying the left distributive law $a\cdot (b\cdot c)=(a\cdot b)\cdot (a\cdot c)$. One can define a sequence of finite left-distributive algebras $A_n$, and then take a limit to get an infinite monogenic left-distributive algebra~$A_\infty$. Results of Laver and Steel assuming a strong large cardinal axiom imply that $A_\infty$ is free; it is open whether the freeness of $A_\infty$ can be proved without the large cardinal assumption, or even in Peano arithmetic. The main result of this paper is the equivalence of this problem with the existence of a certain algebra of increasing functions on natural numbers, called an {\it embedding algebra}. Using this and results of the first author, we conclude that the freeness of $A_\infty$ is unprovable in primitive recursive arithmetic.
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