AAμ:A×AA\mu:A\times A\to A
μが結合法則を満たす:    x,y,zA;μ(x,μ(y,z))=μ(μ(x,y),z)\mu\text{が結合法則を満たす}:\iff\forall x,y,z\in A; \mu(x,\mu(y,z))=\mu(\mu(x,y),z)
使x,y,zA;xμ(yμz)=(xμy)μz\forall x,y,z\in A; x\mu(y\mu z)=(x\mu y)\mu z


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