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Using predicates $Px$ for "x is a person" and $Rx,y$ for "x is the father of y" and secondly taking reference to people as implicit and using only the two-place predicate Rx,y.

  1. No one has no father.
  2. No one has fathered no one.
  3. Some people are the fathers of someone’s father.
  4. Anyone who has fathered someone’s father has fathered someone.

My ideas:

  1. $\neg\exists x(Px \land \forall y(Py\rightarrow\neg Ry,x) \equiv \neg\exists x\forall y(\neg Ry,x)$

  2. $\neg\exists x(Px \land \forall y(Py\land\neg Rx,y) \equiv \neg\exists x\forall y(\neg Rx,y)$

  3. $\exists x(\exists y(Ry,x) \land \exists z(Rx,z))$

  4. $\forall y\exists z\exists x(Ry,z \land Rz,x \land Ry,x)$

Any help would be kindly appreciated, I'm not too sure what the solutions are to these formalisations. They are quite tricky! Thank you very much in advanced.

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    I disagree on 2. "No one has fathered no one" means "Everyone has fathered (i.e. has a son/daughter)" : $∀x∃y(Rx,y)$. – 2017-01-23
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    For 3. I suggest : $∃x∃y∃z[(Rx,y) ∧ (Ry,z)]$. – 2017-01-23
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    So I think we agree on 1. and 2. now. I've changed my question above to my new suggestions for 3. and 4.! Thanks! :) – 2017-01-24

2 Answers 2

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  1. $\forall x (Px \rightarrow \exists y (Py \land Ry,x))$ or $\neg \exists x (Px \land \forall y (Py \land \neg Ry,x))$ (without $Px$: $\forall x \exists y Ry,x$ or $\neg \exists x \forall y \neg Ry,x$)

  2. $\neg \exists x (Px \land \forall y (Py \rightarrow \neg Rx,y))$ or $\forall x (Px \rightarrow \exists y (Py \land Rx,y))$ (without $Px$: $\neg \exists x \forall y \neg Rx,y$ or $\forall x \exists y Rx,y$)

  3. $\exists x \exists y \exists z (Rx,y \land Ry,z)$

  4. $\forall x (\exists y \exists z (Rx,y \land Ry,z) \rightarrow \exists w Rx,w)$

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  1. is fine, though you could also do $\forall x (P(x) \rightarrow \exists y ( P(y) \land R(y,x)))$ and $\forall x \exists y R(y,x)$ (everyone has a father)

  2. is like 1, but with the role of $x$ and $y$ reversed. So: $\neg \exists x (P(x) \land \forall y (P(y) \rightarrow \neg R(x,y)))$ and $\neg \exists x \forall y \neg R(x,y)$. Alternatively: $\forall x (P(x) \rightarrow \exists y (P(y) \land R(x,y)))$ and (as Mauro suggests) $\forall x \exists y R(x,y)$

  3. Why do you have a negation in here?! Try again.

  4. Same here: why the negation?! Also, you will need four quantifiers: one for the first 'anyone', one for the persn they fathered (who itself is the father of someone, so that's a third), and a final one for the last 'someone'. Again, try this one again.

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    Thanks again @Bram28 ! Have realised that I'm going to need 3 quantifiers for no. 3 and 4 for no.4. Will try again and edit my inserts soon. – 2017-01-23
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    So I've got $\neg\exists x(Px \land \forall y(Py\land\neg Rx,y) \equiv \neg\exists x\forall y(\neg Rx,y)$ which is the same as what you suggest for no. 2! Will work on 3 and 4 later. Thanks! – 2017-01-23
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    Hi, so I realised that I typed out my answers for no. 3 and no. 4 incorrectly, hence why the random $\neg$ appears! – 2017-01-23
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    3. $\exists x(Px \land \forall y(Py \rightarrow\exists z (Rz,y,x)))$ – 2017-01-23
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    @Andy That universal isn't right: it says *some* people are the father of *some* one's father ... so you need a bunch of existentials there! Also, $Rz,y,x$ isn't right since $R$ only takes 2 arguments. In English, what are you trying to say about $x$, $y$, and $z$? – 2017-01-23
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    3. $\exists x(\exists y(Ry,x) \land \exists z(Rx,z))$ Would this be better? So someone y fathered someone x and someone x fathered someone z? – 2017-01-24
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    4. Looks a little like this perhaps: [Son - 'Someone'] $\rightarrow$ [Dad] $\rightarrow$ [Grandad - 'Anyone']? So I think this should look like: $\forall y\exists z\exists x(Ry,z \land Rz,x \land Ry,x)$ Which basically means: For all Grandads that fathered a Dad, the Dad fathered a Son and so the Grandad fathered the Son. – 2017-01-24