Class C中有十个Instance，Class C有二个属性 hasA,hasB
Instance名称分别为 ins_1、ins_2、ins_3到ins_10，

hasA属性中已经有设定ins_1、ins_2、ins_3三个实例，
我要使用SWRL推理出hasB的实列(为hasA没有的的实例)

C(?i) ^ “这边该如何写”→ hasB(do something)

ps. HasB属性中就是要放hasA所没有的实例。

Thanks.

hasA已有三个instance，推理结果hasB要有七个instance

谢谢回答。

http://www.w3.org/Submission/SWRL/

C(?x)  ∧  C(?y)  ∧  C(?z)  ∧  hasA(?x, ?y)  ∧  hasName(?y, ?n1)  ∧  hasName(?z, ?n2)  ∧  swrlb:notEqual(?n1, ?n2)  →  hasB(?x, ?z)

这样写会有问题，错误如下
Exception importing SWRL rules and OWL knowledge: edu.stanford.smi.protegex.owl.swrl.bridge.jess.exceptions.SWRLJessBridgeException: Jess reported an error in routine invokeSWRLBuiltIn
 while executing (invokeSWRLBuiltIn Rule-1 swrlb:notEqual ?n1 ?n2)
 while executing rule LHS (Node2)
 while executing rule LHS (Node2)
 while executing rule LHS (Node2)
 while executing rule LHS (Node2)
 while executing rule LHS (Node2)
 while executing rule LHS (Node2)
 while executing rule LHS (MTELN)
 while executing rule LHS (TECT).

Message: Error invoking built-in 'swrlb:notEqual' from Jess: Exception thrown by built-in 'swrlb:notEqual' in rule 'Rule-1': Expecting ordered type for argument 0 for built-in 'swrlb:notEqual'.

请问该如何解决此问题，感谢您。

Thanks.