therightwomanwf

2022-07-13

I'm self studying How to Prove book and have been working out the following problem in which I have to analyze it to logical form:
Nobody in the calculus class is smarter than everybody in the discrete math class
Now, this is how, I started solving it:
¬(Somebody in the calculus class is smarter than everybody in the discrete math class) ¬(If x is in calculus class then x is smartert than everybody in the discrete maths class)
$C\left(x\right)=x$ is in calculus class. $D\left(y\right)=y$ is in discrete class. $S\left(x,y\right)=x$ is smarter than y
$¬\mathrm{\exists }x\left(C\left(x\right)\to \mathrm{\forall }y\left(D\left(y\right)\wedge S\left(x,y\right)\right)\right)$
But this is the solution given in the Velleman's book:
$¬\mathrm{\exists }x\left[C\left(x\right)\wedge \mathrm{\forall }y\left(D\left(y\right)\to S\left(x,y\right)\right)\right]$
I cannot understand how that answer is correct. Can someone explain the thing I'm missing there ?

Oliver Shepherd

Expert

Step 1
Your answer asserts that there does not exist anyone x, who, iF x is in Calculus, then (all students y are both in Discrete math and x is smarter than them.) This is clearly not what is conveyed in the original statement.
Step 2
What we need, essentially, is "There does not exist someone x who is enrolled in Calculus AND such that, for all students y, if y is enrolled in Discrete math, then x is smarter than y.
$\mathrm{¬}\mathrm{\exists }x\left(C\left(x\right)\wedge \mathrm{\forall }y\left(D\left(x\right)\to S\left(x,y\right)\right)\right)$

uplakanimkk

Expert

Step 1
Let C denote the set of members of calculus class and let D the set of members of discrete math class.
The following statements are equivalent (explore step by step) and the last one is the Velleman answer:
Step 2
1) Nobody in the calculus class is smarter than everybody in the discrete math class
2) For every person x in C there is a person y in D such that $\mathrm{¬}S\left(x,y\right)$
3) $\mathrm{\forall }x\in C\mathrm{\exists }y\in D\left[\mathrm{¬}S\left(x,y\right)\right]$
4) $\mathrm{\forall }x\left[x\in C⇒\mathrm{\exists }y\left[y\in D\wedge \mathrm{¬}S\left(x,y\right)\right]\right]$
5) $\mathrm{¬}\mathrm{\exists }x\mathrm{¬}\left[x\in C⇒\mathrm{\exists }y\left[y\in D\wedge \mathrm{¬}S\left(x,y\right)\right]\right]$
6) $\mathrm{¬}\mathrm{\exists }x\mathrm{¬}\left[x\notin C\vee \mathrm{\exists }y\left[y\in D\wedge \mathrm{¬}S\left(x,y\right)\right]\right]$
7) $\mathrm{¬}\mathrm{\exists }x\left[x\in C\wedge \mathrm{¬}\mathrm{\exists }y\left[y\in D\wedge \mathrm{¬}S\left(x,y\right)\right]\right]$
8) $\mathrm{¬}\mathrm{\exists }x\left[x\in C\wedge \mathrm{\forall }y\left[y\notin D\vee S\left(x,y\right)\right]\right]$
9) $\mathrm{¬}\mathrm{\exists }x\left[x\in C\wedge \mathrm{\forall }y\left[y\in D⇒S\left(x,y\right)\right]\right]$
There is quite some redundancy here, but I hope this give you understanding about the correctness of the answer.

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