Suppose there is a statement -

Some students in this class has visited Mexico.

Solution given is: considering the universe of discourse for the variable $x$ consists of all the people. Our solution becomes $\exists{x}[S(x)\land{M(X)}]$ where the predicate $S(x)$ tells $x$ is a student in the class and $M(x)$ tells $x$ has visited Mexico.

The solution also tells $\exists{x}[S(x)\implies{M(X)}]$ is wrong because this expression is true when there is not someone in the class. So the statement can't be described in this way.

My questions is that only - What's wrong in giving true when there is not someone in the class (or true even when the someone is not student) ? Why exactly can't the statement be expressed using implication?


Rephrasing that second version back into English:

There exists a person $x$, such that if $x$ is a student, then $x$ has gone to Mexico.

If there exists anyone who's not a student, this ends up being true. "If false then [anything]" is always true, so when "$x$ is a student" is false, "if $x$ is a student, then $x$ has gone to Mexico" is true—whether or not $x$ has gone to Mexico.

What you want is:

There exists a person $x$, such that $x$ is a student, and $x$ has gone to Mexico.

That is indeed what the first given solution means.

  • $\begingroup$ Thanks but I know about vicious proofs. What I'm asking is what's wrong in giving true when that someone is not student? $\endgroup$ – Mr. Sigma. Dec 16 '18 at 7:32
  • $\begingroup$ @Mr. Sigma: that's irrelevant. What matters is that the statement is false when no student in the class has visited Mexico. $\endgroup$ – reinierpost Dec 16 '18 at 12:30
  • $\begingroup$ @Mr.Sigma. The statement "Some students in the class have visited Mexico" should, intuitively, be false when no students have visited Mexico. But in this version, if there exists any person who is not a student, the statement returns true. That's wrong. $\endgroup$ – Draconis Dec 16 '18 at 16:51

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