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I need to find a proof for my goal , given some assumptions. The assumptions and the goal were translated from english sentences.

My assumptions(KB)

John ,Mary,Helen,George are the only members of club1:

-Member(Club1,John) ∧ Member(Club1,Mary) ∧ -Member(Club1,Helen) ∧ Member(Club1,George)

John is Mary's husband.

-Married(John,Mary)

George is Helen's brother.

-Siblings(Goerge,Helen)

The husband or the wife of each person is also a part of the same club

- (∀x)(∀y)(∀s)(Married(x,y) & Member(s,x) → Member(s,y))

My Goal(Φ)

Helen is not married.

- (∀x)(¬(Married(x,Helen)).

With common logic we can prove that helen (being ethical and not incest) , is not married. But the above assumptions , can't prove my goal with propotitional logic. I need to add my own assumptions , so that I can prove my goal. What assumptions should I add to my KB to prove φ.

I tried the following ones , but they are not enough:

(∀x)(∀y)(Siblings(x,y) -> -Married(x,y)).
(∀x)(∀y)(∀z)( Married(x,y) -> -Married(x,z) & -Married(y,z)).

The above assumptions , tells that if two atoms are siblings then they can't be marry each other and if they can't marry more than one person. I am using prover9 , to also prove my goal , but I can't find what else does it need , to prove it.

EDIT 1 So , far , what I have done enter image description here

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    Another rule is missing: not member if different of Helen, ... – 2017-01-12
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    @MauroALLEGRANZA The OP tried to say that one cannot marry more than one person with the last sentence but didn't do a very good job ... – 2017-01-12

2 Answers 2

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You also need:

$\forall x (Member(Club1,x) \rightarrow (x = John \lor x = Mary \lor x = Helen \lor x = George))$

Since these four are the only members of the club, i.e. there are no others!

Also, let's make sure they are all different people:

$John \not = Mary$

$John \not = Helen$

$John \not = George$

$Mary \not= Helen$

$Mary \not = George$

$Helen \not = George$

Also, I think the puzzle assumes no gay marriage, so:

$\forall x \forall y ((Male(x) \land Married(x,y) \rightarrow Female(y))$

$\forall x \forall y ((Female(x) \land Married(x,y) \rightarrow Male(y))$

And so you'll also want to add:

$Male(John)$

$Female(Mary)$

$Female(Helen)$

$Male(George)$

$\forall x (Male(x) \leftrightarrow \neg Female(x))$

Finally, your last sentence where you try to say that someone cannot be married to more than one person isn't correct; it should be:

$\forall x \forall y \forall z (Married(x,y) \color{red}{ \land z \not = y}) \rightarrow \neg Married(x,z))$

and to use that, you'll probably also need:

$\forall x \forall y (Married(x,y) \rightarrow Married(y,x))$

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    Still can't find a solultion... :/ – 2017-01-12
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    Oh! You should add that they are all different people, since in logic, just because you have different constant symbols does not mean they denote different objects. I'll add that to my post. – 2017-01-12
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    Still no answer , should I post a picture of what i have done so far ? – 2017-01-12
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    @SocialProgrammer yeah, go ahead and show what you have. – 2017-01-12
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    here you go brother – 2017-01-12
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    @SocialProgrammer you misspelled George as 'Goerge' in the Siblings predicate – 2017-01-12
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    Still, exhausted prover , still exhausting me ! D: – 2017-01-12
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    @SocialProgrammer you also need to add that the four people are members of the club; my statement about the club only said that there are no pther members than those four, but that doesn't make those four actual members of the club. So, either add back the four member claims you had, or just make the conditional a biconsitinal in my claim. – 2017-01-12
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    That's 'biconditional', not 'biconsitinal' ... Talk about misspelling :) – 2017-01-12
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    Yeahhhhhh, we made it fellow brother , posting my question soon , please check it out , an tell me if it seems true ! – 2017-01-12
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    @SocialProgrammer Woo!!! – 2017-01-12
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Finally , the answer : enter image description here

and the proof: enter image description here

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    @Bram28 Answer is up ! – 2017-01-12
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    Looks good! looks like the prover is using some kind of resolution. – 2017-01-12
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    Yeah. Anyways , thanks a lot bram. Can i add you as a friend ? – 2017-01-12
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    I don't think this site has a friends feature, but if you ever have another problem you know how to reach me! – 2017-01-12
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    Ok , I am new here , goodnight sir – 2017-01-12
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    @Bram28 Prover9 does use resolution. – 2017-01-14