therightwomanwf

Answered

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)

$C(x)=x$ is in calculus class. $D(y)=y$ is in discrete class. $S(x,y)=x$ is smarter than y

$\neg \mathrm{\exists}x(C(x)\to \mathrm{\forall}y(D(y)\wedge S(x,y)))$

But this is the solution given in the Velleman's book:

$\neg \mathrm{\exists}x[C(x)\wedge \mathrm{\forall}y(D(y)\to S(x,y))]$

I cannot understand how that answer is correct. Can someone explain the thing I'm missing there ?

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)

$C(x)=x$ is in calculus class. $D(y)=y$ is in discrete class. $S(x,y)=x$ is smarter than y

$\neg \mathrm{\exists}x(C(x)\to \mathrm{\forall}y(D(y)\wedge S(x,y)))$

But this is the solution given in the Velleman's book:

$\neg \mathrm{\exists}x[C(x)\wedge \mathrm{\forall}y(D(y)\to S(x,y))]$

I cannot understand how that answer is correct. Can someone explain the thing I'm missing there ?

Answer & Explanation

Oliver Shepherd

Expert

2022-07-14Added 24 answers

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.

$\mathrm{\neg}\mathrm{\exists}x{\textstyle (}C(x)\wedge \mathrm{\forall}y(D(x)\to S(x,y)){\textstyle )}$

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.

$\mathrm{\neg}\mathrm{\exists}x{\textstyle (}C(x)\wedge \mathrm{\forall}y(D(x)\to S(x,y)){\textstyle )}$

uplakanimkk

Expert

2022-07-15Added 6 answers

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 $\mathrm{\neg}S(x,y)$

3) $\mathrm{\forall}x\in C\mathrm{\exists}y\in D\left[\mathrm{\neg}S(x,y)\right]$

4) $\mathrm{\forall}x[x\in C\Rightarrow \mathrm{\exists}y[y\in D\wedge \mathrm{\neg}S(x,y)]]$

5) $\mathrm{\neg}\mathrm{\exists}x\mathrm{\neg}[x\in C\Rightarrow \mathrm{\exists}y[y\in D\wedge \mathrm{\neg}S(x,y)]]$

6) $\mathrm{\neg}\mathrm{\exists}x\mathrm{\neg}[x\notin C\vee \mathrm{\exists}y[y\in D\wedge \mathrm{\neg}S(x,y)]]$

7) $\mathrm{\neg}\mathrm{\exists}x[x\in C\wedge \mathrm{\neg}\mathrm{\exists}y[y\in D\wedge \mathrm{\neg}S(x,y)]]$

8) $\mathrm{\neg}\mathrm{\exists}x[x\in C\wedge \mathrm{\forall}y[y\notin D\vee S(x,y)]]$

9) $\mathrm{\neg}\mathrm{\exists}x[x\in C\wedge \mathrm{\forall}y[y\in D\Rightarrow S(x,y)]]$

There is quite some redundancy here, but I hope this give you understanding about the correctness of the answer.

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 $\mathrm{\neg}S(x,y)$

3) $\mathrm{\forall}x\in C\mathrm{\exists}y\in D\left[\mathrm{\neg}S(x,y)\right]$

4) $\mathrm{\forall}x[x\in C\Rightarrow \mathrm{\exists}y[y\in D\wedge \mathrm{\neg}S(x,y)]]$

5) $\mathrm{\neg}\mathrm{\exists}x\mathrm{\neg}[x\in C\Rightarrow \mathrm{\exists}y[y\in D\wedge \mathrm{\neg}S(x,y)]]$

6) $\mathrm{\neg}\mathrm{\exists}x\mathrm{\neg}[x\notin C\vee \mathrm{\exists}y[y\in D\wedge \mathrm{\neg}S(x,y)]]$

7) $\mathrm{\neg}\mathrm{\exists}x[x\in C\wedge \mathrm{\neg}\mathrm{\exists}y[y\in D\wedge \mathrm{\neg}S(x,y)]]$

8) $\mathrm{\neg}\mathrm{\exists}x[x\in C\wedge \mathrm{\forall}y[y\notin D\vee S(x,y)]]$

9) $\mathrm{\neg}\mathrm{\exists}x[x\in C\wedge \mathrm{\forall}y[y\in D\Rightarrow S(x,y)]]$

There is quite some redundancy here, but I hope this give you understanding about the correctness of the answer.

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