对称二元谓词的基本一阶逻辑推理失败

Question

超级基础的问题。我试图表达两个二元谓词（父项和子项）之间的对称关系。但是，通过以下陈述，我的决议证明者可以让我证明任何事情。转换后的 CNF 形式对我来说很有意义，通过解析证明也是如此，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 证明器。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

Answer 1

这是 ResolutionProver 的快速修复。

导致prover不健全的问题是当有多个互补字面量时，它没有正确执行解析规则。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}。两者都将被丢弃为重言式。当前的 NLTK 实现会产生 {C D}.

引入这可能是因为子句在 NLTK 中表示为列表，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} 和 {-A -A} 时，此规则确实会正确生成空子句，但通常此规则不正确。

看来，如果我们让子句不重复相同的文字，我们可以通过一些更改恢复健全。

首先定义一个删除相同文字的函数。

下面是这种函数的简单实现

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() 以确保输入无重复。

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 class 中存储文字的容器替换为基于内置 Python 哈希集的容器，但这似乎需要一个对证明者实施进行更彻底的返工，并引入一些性能测试基础设施。

对称二元谓词的基本一阶逻辑推理失败

Basic first order logic inference fails for symmetric binary predicate

nlp

artificial-intelligence

nltk

theorem-proving

first-order-logic