# First order logic - proving an argument is invalid

The question asks to construct a derivation of the argument if it's valid, and otherwise to provide an arithmetical interpretation which shows it's invalid. Am I doing this correctly?

Argument: "All good critics like every poet mentioned in the lecture. No good critic likes Edgar Guest, although Edgar Guest is a poet. Therefore, Edgar Guest was not mentioned in the lecture."

Let $$C(x)\leftrightarrow x\ is\ a\ good\ critic$$

$$L(x,y) \leftrightarrow x\ likes\ y$$

$$P(x) \leftrightarrow x\ is\ a\ poet$$

$$M(x) \leftrightarrow x\ was\ mentioned\ in\ the\ lecture$$

$$g = Edgar\ Guest$$

I symbolise the premises of the argument as:

Premise 1: $$\forall x(C(x)\rightarrow \forall y(M(y)\wedge P(y) \rightarrow L(x,y)))$$

Premise 2: $$P(g)\wedge\forall x(C(x) \rightarrow \neg L(x,g))$$

The conclusion is $$\neg M(g)$$

In my first attempt, I tried to prove that the argument is true by assuming the negation of the desired conclusion, $$M(g)$$, and finding a contradiction. However, it only led me to conclude from the premises that $$\forall x(M(g) \rightarrow \neg C(x))$$

So I tried to find an interpretation which makes the argument invalid and found the following:

Domain: $$\mathbb{Z}^+$$

$$C(x) \leftrightarrow x < 1$$

$$P(x) \leftrightarrow x\ is\ prime$$

$$M(x) \leftrightarrow x=2$$

$$L(x,y) \leftrightarrow x

$$g=2$$

Essentially I chose an interpretation of $$C(x)$$ so that implications in which $$C(x)$$ is an antecedent are always true because $$C(x)$$ is always false. This allows the first premise to be true. Note that this also makes the second term of the second premise true. I then choose an interpretation of $$P(x)$$ and $$M(x)$$ so that $$P(g)$$ ("2 is prime") is true while $$\neg M(g)$$ ("2$$\neq$$2") is false. Is this solution fine?

## 1 Answer

Indeed, in your first attempt you only managed to show that $$\forall x (M(g)\to\neg C(x))$$, which is equivalent to $$\forall x(C(x)\to\neg M(g))$$ (by contraposition). This indicates already that to conclude that $$g$$ was not mentioned in the lecture, the universe must contain at least one $$x$$ that is a good critic. In other words, "there is still the possibility that $$g$$ was mentioned and that is in case no one is a good critic".

So, you came up with a universe/model which has no "good critics" at all, to prove that the initial argument is not valid.

Well done!

• thanks for the feedback! :) – Unknowledgeable Mar 17 at 10:34