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I am having a bit of trouble with predicate logic involving family members.

This is a question from an assignment I am doing.

Let the domain be a group of three dogs, Tiger, Ashes and Smokey. Consider the following premises:

• ∀x∀y(Puppy(x) ∧ ¬Puppy(y) → Parent(y, x))

• ∀x∀y(Parent(x, y) → ¬Puppy(x))

• ∀x∀y(Parent(x, y) → Puppy(y))

• Parent(Smokey, Ashes)

• ¬Puppy(Tiger)

Use predicate logic natural deduction to figure out the relationship between Tiger and Ashes. Hint: if you have two (or more) universal quantifiers in a row, you can treat them as one quantifier for a pair of elements, and instantiate them together in universal modus ponens. Also, use the rule that from A and B you can derive A ∧ B in one step.

It seems that Tiger is the parent of Ashes. I don't know how to prove that though (outside of just looking at it and making a deduction)

asdfjkdsg
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2 Answers2

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It seems that Tiger is the parent of Ashes. I don't know how to prove that though (outside of just looking at it and making a deduction)

That is basically what you have to do. Just state what parts you are looking at when using logical deduction, and which rule use use of the two rules of inference you are allowed.

  • Parent(Smokey, Ashes) and ∀x∀y(Parent(x, y) → Puppy(y)) entails that Puppy(Ashes), via universal modus ponens.

  • Puppy(Ashes), and ¬Puppy(Tiger) entails what conclusion via which rule.

  • This conclusion, and which premise entails Parent(Tiger, Ashes) via which rule.

Also

  • Parent(Smokey, Ashes) and ∀x∀y(Parent(x, y) → ¬Puppy(x)) entails that ¬Puppy(Smokey), via universal modus ponens.

Note: if you can also use universal modus tolens you can also similarly make quite a few more conclusions.

Graham Kemp
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  • Thank you, this really helped. I appreciate it. Unfortunately I am unable to give you + due to my noob rating. – asdfjkdsg Oct 24 '17 at 20:23
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$\def\fitch#1#2{\begin{array}{|l}#1 \\ \hline #2\end{array}}$

$\fitch{ 1. \forall x \forall y \ ( (Puppy(x) \land \neg Puppy(y))\rightarrow Parent(y,x))\\ 2. \forall x \forall y \ (Parent(x,y) \rightarrow \neg Puppy(x))\\ 3. \forall x \forall y \ (Parent(x,y) \rightarrow Puppy(y))\\ 4. Parent(smokey, ashes)\\ 5. \neg Puppy(tiger)}{ 6. Parent(smokey, ashes)\rightarrow Puppy(ashes) \quad \forall \ Elim \ 3\\ 7. Puppy(ashes) \quad \rightarrow \ Elim \ 4,6\\ 8. Puppy(ashes) \land \neg Puppy(tiger) \quad \land \ Intro \ 5,7\\ 9. (Puppy(ashes) \land \neg Puppy(tiger)) \rightarrow Parent(tiger,ashes) \quad \forall \ Elim \ 1\\ 10. Parent(tiger,ashes) \quad \rightarrow \ Elim \ 8,9 }$

Bram28
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