Predicate Logic and Inference

Question

Assume that given three predicates are presented below:

$H(x)$: $x$ is a horse

$A(x)$: $x$ is an animal

$T(x,y)$: $x$ is a tail of $y$

Then, translate the following inference into an inference using predicate logic expressions and prove whether inference is valid or not (for instance, using natural deduction):

Horses are animals.

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Horses' tails are tails of animals.

My thoughts: I am quite good at translating predicate logic expressions, but here I struggled to come up with formula for Horses' tails. My initial idea was to consider similar sentence such as "w is a tail of a horse" to form required inference, but it was not successful. Would be welcomed to hear your ideas about this task.

Answer 1

As correctly suggested in lemontree's answer, "Horses' tails are tails of animals" can be formalized as $\forall x \forall y \big((H(y) \land T(x,y)) \to A(y) \big)$ or more precisely, $\forall x \forall y \big((H(y) \land T(x,y)) \to (A(y) \land T(x,y))\big)$.

Of course, the argument

$\frac{\text{Horses are animals}}{\text{Horses' tails are tails of animals}} \quad \text{i.e.} \quad \frac{\forall y (H(y) \to A(y))}{\forall x \forall y \big((H(y) \land T(x,y)) \to (A(y) \land T(x,y))\big)}$

is valid. First, I give you an informal proof of that.

We want to prove that $\forall x \forall y \big((H(y) \land T(x,y)) \to (A(y) \land T(x,y))\big)$, under the hypothesis $\forall y (H(y) \to A(y) )$. So, let us fix arbitrary individuals $x$ and $y$ and let us suppose that $H(y) \land T(x,y)$, we have to show that $A(y) \land T(x,y)$. Since by hypothesis $\forall y (H(y) \to A(y) )$, hence $H(y) \to A(y)$ holds for the particular $y$ we have chosen. Moreover, we are supposing that $H(y) \land T(x,y)$ and in particular $H(y)$ holds. By modus ponens, from $H(y) \to A(y)$ and $H(y)$ it follows that $A(y)$. Also, since we are supposing that $H(y) \land T(x,y)$, in particular $T(x,y)$ holds. So, $A(y) \land T(x,y)$. Therefore, we have proved that, for arbitrary $x$ and $y$, if $H(y) \land T(x,y)$ then $A(y) \land T(x,y)$. Thus, $\forall x \forall y \big((H(y) \land T(x,y)) \to (A(y) \land T(x,y))\big)$ holds, under the hypothesis $\forall y (H(y) \to A(y))$.

You can formalize this proof in natural deduction as follows:

$$ \dfrac {\dfrac {\dfrac {\dfrac{\dfrac{\forall y (H(y) \to A(y))}{H(y) \to A(y)}\forall_\text{elim} \qquad \dfrac{[H(y) \land T(x,y)]^*}{H(y)}\land_\text{elim}\!\!\!\!\!\!\!\!\!\!\!}{A(y)}\to_\text{elim} \quad \dfrac{[H(y) \land T(x,y)]^*}{T(x,y)}\land_\text{elim}} {A(y) \land T(x,y)}\land_\text{intro}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! } {(\!\!\!\!\!\!\!\!H(y) \land T(x,y)) \to (A(y) \land T(x,y))} \to_\text{intro}^*\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! } {\dfrac {\forall y \big((H(y) \land T(x,y)) \to (A(y) \land T(x,y))\big)} {\forall x \forall y \big((H(y) \land T(x,y)) \to (A(y) \land T(x,y))\big)} \forall_\text{intro} } \forall_\text{intro} $$

Answer 2

Hints:

"$x$ is a $P$'s tail" means that $x$ is a tail of $y$ and $y$ is a $P$.

"Horses' tails are tails of animals" means that for all tails $x$ and tail-bearers $y$, the tail being a horse's tail implies the tail being an animal's tail (where for "being a $P$'s tail" insert the above definition).

With the appropriate formalization of this paraphrase, it is possible to find a formal proof of the inference.