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

"A samurai without a sword is like a samurai with the sword but without the sword"

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.

  • $\begingroup$ 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 $F$. $\endgroup$ Dec 30, 2016 at 20:09
  • $\begingroup$ $\neg P \Leftrightarrow (P \wedge \neg P)$ is not a tautology. $\endgroup$ Dec 30, 2016 at 20:09
  • $\begingroup$ Yep you are right I edited the question $\endgroup$
    – Kunashu
    Dec 30, 2016 at 20:11
  • $\begingroup$ 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! $\endgroup$
    – Bram28
    Dec 30, 2016 at 20:20
  • $\begingroup$ 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"! $\endgroup$
    – Kunashu
    Dec 30, 2016 at 20:27

1 Answer 1


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).

  • $\begingroup$ I like your suggestion. How would it look like if there were no samurai? $\endgroup$
    – Kunashu
    Dec 30, 2016 at 20:47
  • $\begingroup$ 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. $\endgroup$
    – Kunashu
    Dec 30, 2016 at 20:53
  • $\begingroup$ 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. $\endgroup$
    – J. C.
    Dec 30, 2016 at 20:58
  • 1
    $\begingroup$ 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. $\endgroup$ Dec 30, 2016 at 21:00
  • 1
    $\begingroup$ @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. $\endgroup$ Dec 30, 2016 at 21:11

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