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Is it possible to represent the english sentence with numerical value in First order Logic. For example if the sentence is: Nobody has more than one mother.

I am wandering who can i show the quantitative sentence in FOL. I am studying AI, and i came across this problem, it not my homework..

Thanks in advance

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5 Answers 5

$$\forall x\forall y\forall z((M(x,z) \land M(y,z))\longrightarrow (x=y))$$

Here $M(u,v)$ is the predicate that says $u$ is a mother of $v$.

Added: We describe the above sentence using more words. It says that for any (person) $z$, if $x$ is a mother of $z$ and $y$ is a mother of $z$, then $x$ and $y$ are the same.

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Will you please elaborate the answer. I am not sure if this satisfies "Nobody has more than one mother"... Correct me if I am wrong, What I can extract from your answer (∀x∀y∀z((M(x,z)∧M(y,z))⟶(x=y))) is Everybody has same mother.. I dont know, it looks so confusing. May be it would be ∀x∀y∃z((M(x,z)∧M(y,z))⟶(x=y)) –  Vivek Bhusal Mar 25 at 6:17
    
More or less in words, it says for any $z$, if $x$ is a mother of $z$, and $y$ is also a mother of $z$, then $x$ and $y$ are the same. (It does not say everybody has a mother, we were not asked to say that. It just says nobody has more than one mother.) Your version says there is somebody who has no more than one mother. –  André Nicolas Mar 25 at 6:22
    
Actually, because of the order of quantifiers in Vivek's version, it says even less than "there is somebody who has no more than one mother." It's awkward to express it directly in English, but it's equivalent to "No two distinct people are both everybody's mother." –  Andreas Blass Mar 25 at 20:33

Let $\alpha$ be a variable, where $\alpha$ = a child. Let $\beta$ = a mother of the child $\alpha$. Proof: $\forall$ ($\alpha$ $\in$ $\Re^{3}$) $\exists$ ($\beta$) such that $f(\alpha, \beta)$ is a one-to-one function. That is as good as I have. Hope this helps.

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If, for example, you want to assert there are exactly three things that have property $P$, you can do so by asserting:

  • There exists $x$ that has property $P$
  • There exists $y$ that has property $P$
  • There exists $z$ that has property $P$
  • $x \neq y$
  • $x \neq z$
  • $y \neq z$
  • For every $w$ with property $P$: $w = x$ or $w = y$ or $w = z$.
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Sorry, I would have liked to add a comment to your question but as I don't have 50 credits I'm giving an answer which is bound be be false. You can break down the question into parts and then find the conclusion like so, (Going from the hypothesis "Nobody has more than one mother") i) Everyone has a mother ii) A mother may have more than one child iii) A mother may not have everyone as a child iv) One person may not have everyone's mother as mother v) Hence a person can only have one mother

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You can use the 'at most one' quantifier $\;\langle !x :: P(x) \rangle \;$, which reads 'there is at most one $\;x\;$ for which $\;P(x)\;$ holds'. The answer is then $$ \langle \forall z :: \langle ! x :: x\text{ is mother of }z \rangle \rangle \tag{0} $$ ("Every $\;z\;$ has at most one mother $\;x\;$.")

To get rid of this unfamiliar quantifier, $\;\langle !x :: P(x) \rangle \;$ can be viewed as an abbreviation for either $$ \langle \forall x,y :: P(x) \land P(y) \;\Rightarrow\; x = y \rangle \tag{a} $$ or equivalently (less traditional but more brief) $$ \langle \exists y :: \langle \forall x :: P(x) \;\Rightarrow\; x = y \rangle \rangle \tag{b} $$

Expanding that abbreviation gives us an answer of either $$ \langle \forall z,x,y :: x\text{ is mother of }z \land y\text{ is mother of }z \;\Rightarrow\; x = y \rangle \tag{0a} $$ or $$ \langle \forall z :: \langle \exists y :: \langle \forall x :: x\text{ is mother of }z \;\Rightarrow\; x = y \rangle \rangle \rangle \tag{0b} $$

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