# What logical operator is equivalent to “is like”

Currently I'm learning some logic and while surfing the net I found this post that states:

So I was wondering if I could prove that the statement is a Tautology. But first I wasn't sure how to interpret the "is like". What I ended up doing is treating it as biconditional. So let P = Samurai with a sword then, the expression would be $\neg{P} \iff (P \wedge \neg{P})$. Using a truth table I was able to confirm this to be a tautology:

$$\begin{array}{c|c|c|c|c|} P & \neg{P} & P \wedge \neg{P} & \neg{P} \implies (P \wedge \neg{P}) & (P \wedge \neg{P}) \implies \neg{P} \\ \hline \text{T} & \text{F} & \text{F} & \text{T} & \text{T} \\ \hline \text{F} & \text{T} & \text{F} & \text{T} & \text{T} \\ \hline \end{array}$$

Is my reasoning correct? Should I treate the "is like" as biconditional

Update: I just realized that I was wrong and the statement is not a tautology. The table should be:

$$\begin{array}{c|c|c|c|c|} P & \neg{P} & P \wedge \neg{P} & \neg{P} \implies (P \wedge \neg{P}) & (P \wedge \neg{P}) \implies \neg{P} \\ \hline \text{T} & \text{F} & \text{F} & \text{T} & \text{T} \\ \hline \text{F} & \text{T} & \text{F} & \text{F} & \text{T} \\ \hline \end{array}$$

But I'm still not sure if my interpretation of the statement is correct.

• Whether or not treating $\iff$ as the biconditional is another matter, but your truth table is wrong: the bottom value in the fourth column should be $F$. – Mees de Vries Dec 30 '16 at 20:09
• $\neg P \Leftrightarrow (P \wedge \neg P)$ is not a tautology. – Fabio Somenzi Dec 30 '16 at 20:09
• Yep you are right I edited the question – Kunashu Dec 30 '16 at 20:11
• I don't think it is a good idea to analyze 'is like' as a truth-functional operator at all. Also note that your $P$ isn't even a statement! – Bram28 Dec 30 '16 at 20:20
• You are right!!! A statement would be something like: P =The samurai has a sword. With that said it seems like the phrase can't be analyzed without rephrasing the whole thing, which would end up removing "is like"! – Kunashu Dec 30 '16 at 20:27

Let $S(x): \text{x is a samurai}$, P(x): $\text{x has a sword}$.

My suggestion is:

$$\forall x, S(x) \implies (\neg P(x) \iff P(x) \land \neg P(x))$$

This is a tautology iff there are no samurais or if every samurai has got a sword (which even makes some sense).

• I like your suggestion. How would it look like if there were no samurai? – Kunashu Dec 30 '16 at 20:47
• I think i got it how it would look like if there were no samurai: $\forall{x},S(x) = \text{False}$ and $\neg{P(x)} \iff P(x) \wedge \neg{P(x)} = \text{False}$ then $\text{False} \implies \text{False}$ which would make it true. – Kunashu Dec 30 '16 at 20:53
• Actually $\neg P(x) \iff P(x) \land \neg P(x)$ could still be true, while assuming the inexistence of samurai (this would mean everybody, samurai or not, had swords), because $False \implies True$ is still true. – J. C. Dec 30 '16 at 20:58
• Being a tautology is an even stronger property than being valid for a predicate-calculus sentence. $\forall x ~.~ S(x) \Rightarrow (\neg P(x) \Leftrightarrow (P(x) \wedge \neg P(x)))$ is not even valid. I concur with those who call for rephrasing the statement. – Fabio Somenzi Dec 30 '16 at 21:00
• @JoãoC. Yes, it's clear that the scope of the implication is everything that follows it. As you point out, in a world in which someone is a samurai and someone lacks a sword the sentence is false. Hence, it is not valid. That's all I wanted to point out. How the sentence in the OP should be translated is a different matter, which hinges on how it is to be understood. – Fabio Somenzi Dec 30 '16 at 21:11