Question

Predicate Logic Question:

'There is a student in your class who has sent everyone else an email."

Domain {x,y = classmates}, M(x,y): x has emailed y

Here is the right answer.
\[existsx forally( x \neq y \rightarrow M(x,y)\]

Here is mine.
\[existsx forally( x \neq y \bigwedge M(x,y)\]

can someone explain to me why my answer is wrong?

Answer 1

sorry, missed a closing parenthesis

Answer 2

lol, i knew there was something missing

Answer 3

Interesting question, might have to do with the truth value of the function and it's negation.
\[\Large \exists x\forall y: (x \neq y \rightarrow M(x,y)) \]
is always true, even if the premise is wrong, but it doesn't seem like a sufficient argument to me yet.

Answer 4

It seemed to me like using a conjunction was more clear. can you see some obvious difference between using a conjunction instead of the conditional? I know that the two forms shouldn't be equivalent, but I'm having a hard time understanding why

Answer 5

Like you said, if x and y are equal, the conditional would still be true. But that isn't the case for the conjunction.

Answer 6

I'm often using conditionals when I should be using conjunctions and conjunctions when I should be using conditionals, so I must be missing something from my understanding..

Answer 7

Maybe if you type it out:

first one would read like
"if x is not equal to y then x has sent an email to y"

second one would read as:
"x is not equal to y and x has sent an email to y"

As it comes to my understanding, this problem requires a conditional because we first must guarantee something for it to be true. The first statement is a tautology, it has a lefthand side that can be evaluated for its truth value and a right hand side.

The second statement is one statement in general, it's a conjunction and it's entire phrase can only have the truth value right or wrong.

I think the best way that I can wrap my head around conditionals is to read it as:
\[p \rightarrow q \]
and assume that p is given, if p is given then q must be true as well, if p is false, then the implication is still true, so the crucial point is to check if p then q.

For a conjuction however:
\[p \wedge q \]
there is no such statement possible, we can't assume p, nor q, we could apply various truth values for either side and from there deduce if the statement is right or wrong.

Answer 8

One thing I thought of: Since x and y are from the same domain, and since we're considering all y, doesn't that set up a contradiction is we use the conjunction? What I mean is, since x and y are from the same domain, if we cycle through the entire domain, the qualifier would eventually fail

Answer 9

because x and y can't be not equal for the entire domain, since they're both from the same domain

Answer 10

so if we use the conditional instead, that would be valid because we're not necessarily looping through the entire domain?

Answer 11

yes that might be a reason as well, if you consider a computer algorithm that runs through the entire domain. A conjunction should always combine statements (enhance) or enlarge them, I believe what would work as well is this statement (as for relation or 2-tuples)
\[\Large \exists x \forall y: ((x,y)\wedge x\neq y \rightarrow M(x,y)) \] However in this statement we should clarify whether there is only one domain or a codomain as well. Maybe that adds up to it's trickiness :-)

Answer 12

hmmm.. I'm going to have to think about this a lot more I guess. I find this subject to be really difficult.

Answer 13

thanks for the help

Answer 14

@Spacelimbus Thanks for the help. I understand the problem now.

Answer 15

@sinusoidal, you're very welcome. Do you mind sharing which explanation gave you the best insight? I am always curious for other peoples perspective, as I am just a student myself :-)