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超级基础的问题。我试图表达两个二元谓词（父母和孩子）之间的对称关系。但是，通过以下陈述，我的解决方案证明者允许我证明任何事情。转换后的 CNF 形式对我来说很有意义，解析证明也是如此，但这应该是一个明显的错误案例。我错过了什么？

forall x,y (is-parent-of(x,y) <-> is-child-of(y,x))

我正在使用 nltk python 库和 ResolutionProver 证明者。这是nltk代码：

from nltk.sem import Expression as exp
from nltk.inference import ResolutionProver as prover

s = exp.fromstring('all x.(all y.(parentof(y, x) <-> childof(x, y)))')
q = exp.fromstring('foo(Bar)')
print prover().prove(q, [s], verbose=True)

输出：

[1] {-foo(Bar)}                             A 
[2] {-parentof(z9,z10), childof(z10,z9)}    A 
[3] {parentof(z11,z12), -childof(z12,z11)}  A 
[4] {}                                      (2, 3) 

True
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1 回答 1

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这是 ResolutionProver 的快速修复。

导致证明者不健全的问题是，当存在多个互补文字时，它没有正确实现解析规则。例如，给定子句{A B C}和{-A -B D}二进制解析将产生子句{A -A C D}和{B -B C D}. 两者都将作为重言式被丢弃。当前的 NLTK 实现将生成{C D}.

这可能是因为子句在 NLTK 中表示为列表而引入的，因此相同的文字可能在一个子句中出现不止一次。当应用于子句{A A}and时，此规则确实会正确生成一个空子句{-A -A}，但通常此规则是不正确的。

似乎如果我们保持从句不重复相同的文字，我们可以通过一些改变来恢复健全性。

首先定义一个删除相同文字的函数。

这是此类功能的天真实现

import nltk.inference.resolution as res

def _simplify(clause):
    """
    Remove duplicate literals from a clause
    """
    duplicates=[]
    for i,c in enumerate(clause):
       if i in duplicates:
          continue
       for j,d in enumerate(clause[i+1:],start=i+1):
          if j in duplicates:
             continue
          if c == d:
               duplicates.append(j)
    result=[]
    for i,c in enumerate(clause):
        if not i in duplicates:
           result.append(clause[i])
    return res.Clause(result)

现在我们可以将这个函数插入到nltk.inference.resolution模块的一些函数中。

def _iterate_first_fix(first, second, bindings, used, skipped, finalize_method, debug):
    """
    This method facilitates movement through the terms of 'self'
    """
    debug.line('unify(%s,%s) %s'%(first, second, bindings))

    if not len(first) or not len(second): #if no more recursions can be performed
        return finalize_method(first, second, bindings, used, skipped, debug)
    else:
        #explore this 'self' atom
        result = res._iterate_second(first, second, bindings, used, skipped, finalize_method, debug+1)

        #skip this possible 'self' atom
        newskipped = (skipped[0]+[first[0]], skipped[1])
        result += res._iterate_first(first[1:], second, bindings, used, newskipped, finalize_method, debug+1)
        try:
            newbindings, newused, unused = res._unify_terms(first[0], second[0], bindings, used)
            #Unification found, so progress with this line of unification
            #put skipped and unused terms back into play for later unification.
            newfirst = first[1:] + skipped[0] + unused[0]
            newsecond = second[1:] + skipped[1] + unused[1]

            # We return immediately when `_unify_term()` is successful
            result += _simplify(finalize_method(newfirst,newsecond,newbindings,newused,([],[]),debug))
        except res.BindingException:
            pass
    return result
res._iterate_first=_iterate_first_fix

同样更新 res._iterate_second

def _iterate_second_fix(first, second, bindings, used, skipped, finalize_method, debug):
    """
    This method facilitates movement through the terms of 'other'
    """
    debug.line('unify(%s,%s) %s'%(first, second, bindings))

    if not len(first) or not len(second): #if no more recursions can be performed
        return finalize_method(first, second, bindings, used, skipped, debug)
    else:
        #skip this possible pairing and move to the next
        newskipped = (skipped[0], skipped[1]+[second[0]])
        result = res._iterate_second(first, second[1:], bindings, used, newskipped, finalize_method, debug+1)

        try:
            newbindings, newused, unused = res._unify_terms(first[0], second[0], bindings, used)
            #Unification found, so progress with this line of unification
            #put skipped and unused terms back into play for later unification.
            newfirst = first[1:] + skipped[0] + unused[0]
            newsecond = second[1:] + skipped[1] + unused[1]

            # We return immediately when `_unify_term()` is successful
            result += _simplify(finalize_method(newfirst,newsecond,newbindings,newused,([],[]),debug))
        except res.BindingException:
            #the atoms could not be unified,
            pass

    return result
res._iterate_second=_iterate_second_fix

最后，将我们的函数插入clausify()以确保输入是无重复的。

def clausify_simplify(expression):
    """
    Skolemize, clausify, and standardize the variables apart.
    """
    clause_list = []
    for clause in res._clausify(res.skolemize(expression)):
        for free in clause.free():
            if res.is_indvar(free.name):
                newvar = res.VariableExpression(res.unique_variable())
                clause = clause.replace(free, newvar)
        clause_list.append(_simplify(clause))
    return clause_list
res.clausify=clausify_simplify

应用这些更改后，证明者应该运行标准测试并正确处理parentof/childof关系。

print res.ResolutionProver().prove(q, [s], verbose=True)

输出：

[1] {-foo(Bar)}                                  A 
[2] {-parentof(z144,z143), childof(z143,z144)}   A 
[3] {parentof(z146,z145), -childof(z145,z146)}   A 
[4] {childof(z145,z146), -childof(z145,z146)}    (2, 3) Tautology
[5] {-parentof(z146,z145), parentof(z146,z145)}  (2, 3) Tautology
[6] {childof(z145,z146), -childof(z145,z146)}    (2, 3) Tautology

False

更新：实现正确性并不是故事的结局。一种更有效的解决方案是将用于在Clause类中存储文字的容器替换为基于内置 Python 基于哈希集的容器，但这似乎需要对证明者实现进行更彻底的返工并引入一些性能测试基础设施也是。

于 2017-04-21T14:44:40.103 回答