Question

我曾尝试使用prover9来证明一个对人类来说很明显的非常简单的陈述，但幸运的是，我无法使其发挥作用。我有以下情况：

% Three boys - Dan, Louise and Tom have t-shirts in three diffrent colors
% (white, yellow and green) and with three different patterns: (giraffe, camel and
% panda). Dan has the t-shirt with giraffe, Louise has the yelow one and Tom has
% not the white t-shirt. The boy with the yellow one has not the one with camel
% pattern. Task:
% Represent exercise with classical boolean statements and using 
% resolution algorithm answer the question: "who has the t-shirt with the camel pattern?"

formulas(sos).
%      (pattern(Dan, Giraffe) & pattern(Louise, Panda) & pattern(Tom, Camel))
%    | (pattern(Dan, Giraffe) & pattern(Louise, Camel) & pattern(Tom, Panda))
%    | (pattern(Dan, Panda) & pattern(Louise,Giraffe) & pattern(Tom, Camel))
%    | (pattern(Dan, Panda) & pattern(Louise, Camel) & pattern(Tom, Giraffe))
%    | (pattern(Dan, Camel) & pattern(Louise, Panda) & pattern(Tom, Giraffe))
%    | (pattern(Dan, Camel) & pattern(Louise, Giraffe) & pattern(Tom, Panda)).
    % This does not works, unfortunately

      (pattern(Dan, Giraffe) & pattern(Louise, Panda) & pattern(Tom, Camel))
    | (pattern(Dan, Giraffe) & pattern(Louise, Camel) & pattern(Tom, Panda)).
    % This works

      (color(Dan, White) & color(Louise, Yellow) & color(Tom, Green))
    | (color(Dan, White) & color(Louise, Green) & color(Tom, Yellow))
    | (color(Dan, Yellow) & color(Louise,White) & color(Tom, Green))
    | (color(Dan, Yellow) & color(Louise, Green) & color(Tom, White))
    | (color(Dan, Green) & color(Louise, Yellow) & color(Tom, White))
    | (color(Dan, Green) & color(Louise, White) & color(Tom, Yellow)).

    pattern(Dan, Giraffe).
    color(Louise, Yellow).

    -color(Tom,White).
    all x (color(x,Yellow) -> -pattern(x,Camel)).
end_of_list.

formulas(goals).
    pattern(Tom,Camel). % Our solution
    % pattern(Louise, Panda).
end_of_list.

和2.公式是写下所有可能性而没有约束 - 简单的排列3！（即使我们知道丹有长颈鹿，我们可以写下两种可能性）。它不应该修改问题添加额外的或语句不应该切断我们现有的证据，但它在我当前的解决方案。 3.声明（模式（Dan，Girrafe）事实上切断了不必要的可能性（没有程序找到正确的解决方案）。

我不知道，我是否使用了严重的prover9或只是忽略了我的问题中的某些东西（或者它在经典布尔语句中的表示）。我能做错什么？

Answer 1

添加一条声明，说明特定动物只能在一件T恤上：所有x（模式（Dan，x） - ＆gt;（ - 模式（Tom，x）＆amp; -pattern（Louise，x）））。

这解释了我们在解释时所做的事情：丹有长颈鹿而露易丝没有骆驼。由于露易丝不能拥有长颈鹿，路易丝必须拥有熊猫。

添加完成后，您的第一组语句和问题信息将导致证明。

事情与第二种约束形式（但不是第一种形式）相关的原因是选项较少。分辨率将语句更改为连接正常形式。给定的陈述的形式为（A＆amp; B＆amp; C）| （A＆amp; D＆amp; E）。应用分配法导致9个单独的陈述：A | A，A | D，A | E，B | A，B | D，B | E，C | A，C | D，C | E.这些中的每一个都只有两部分。一旦-pattern（Louise，Camel）派生解析可以将这些语句中的一些减少到单个原语并完成证明。

约束的第一个表述有更多的选择 - 并且将其改为合取的正常形式会导致诸如模式之类的陈述（Dan，Giraffe）|模式（丹，熊猫）|模式（路易丝，熊猫）|图案（路易丝，长颈鹿）。

% Saved by Prover9-Mace4 Version 0.5, December 2007.

set(ignore_option_dependencies). % GUI handles dependencies

if(Prover9). % Options for Prover9   assign(max_seconds, 60).
end_if.

if(Mace4).   % Options for Mace4   assign(max_seconds, 60). end_if.

formulas(assumptions).

% Three boys - Dan, Louise and Tom have t-shirts in three different colors 
% (white, yellow and green) and with three different patterns: (giraffe, camel and 
% panda). Dan has the t-shirt with giraffe, Louise has the yellow one and Tom has 
% not the white t-shirt. The boy with the yellow one has not the one with camel 
% pattern. Task: 
% Represent exercise with classical boolean statements and using  
% resolution algorithm answer the question: "who has the t-shirt with the camel pattern?" 

%formulas(sos).
     (pattern(Dan, Giraffe) & pattern(Louise, Panda) & pattern(Tom, Camel))
    | (pattern(Dan, Giraffe) & pattern(Louise, Camel) & pattern(Tom, Panda))
    | (pattern(Dan, Panda) & pattern(Louise,Giraffe) & pattern(Tom, Camel))
    | (pattern(Dan, Panda) & pattern(Louise, Camel) & pattern(Tom, Giraffe))
    | (pattern(Dan, Camel) & pattern(Louise, Panda) & pattern(Tom, Giraffe))
    | (pattern(Dan, Camel) & pattern(Louise, Giraffe) & pattern(Tom, Panda)).
    % The above now works

      (color(Dan, White) & color(Louise, Yellow) & color(Tom, Green))
    | (color(Dan, White) & color(Louise, Green) & color(Tom, Yellow))
    | (color(Dan, Yellow) & color(Louise,White) & color(Tom, Green))
    | (color(Dan, Yellow) & color(Louise, Green) & color(Tom, White))
    | (color(Dan, Green) & color(Louise, Yellow) & color(Tom, White))
    | (color(Dan, Green) & color(Louise, White) & color(Tom, Yellow)).

    pattern(Dan, Giraffe).
    color(Louise, Yellow).

    -color(Tom,White).
    all x (color(x,Yellow) -> -pattern(x,Camel)).

   % Dan has the giraffe and Louise does NOT have the camel; therefore Louise     
   % has the Panda (because Louise cannot also have the giraffe)    
   % A pattern on Dan's t-shirt cannot be on Tom's or Louise's t-shirt
    all x (pattern (Dan, x) -> (- pattern(Tom, x) & -pattern (Louise, x))).

end_of_list.

formulas(goals).

pattern(Tom, Camel).

end_of_list.

Prover9无法找到正确的解决方案

1 个答案: