I'm self studying How to Prove book and have been working out the following problem...

therightwomanwf

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
¬ x ( C ( x ) y ( D ( y ) S ( x , y ) ) )
But this is the solution given in the Velleman's book:
¬ x [ C ( x ) y ( D ( y ) S ( x , y ) ) ]
I cannot understand how that answer is correct. Can someone explain the thing I'm missing there ?

Answer & Explanation

Oliver Shepherd

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.
¬ x ( C ( x ) y ( D ( x ) S ( x , y ) ) )
uplakanimkk

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 ¬ S ( x , y )
3) x C y D [ ¬ S ( x , y ) ]
4) x [ x C y [ y D ¬ S ( x , y ) ] ]
5) ¬ x ¬ [ x C y [ y D ¬ S ( x , y ) ] ]
6) ¬ x ¬ [ x C y [ y D ¬ S ( x , y ) ] ]
7) ¬ x [ x C ¬ y [ y D ¬ S ( x , y ) ] ]
8) ¬ x [ x C y [ y D S ( x , y ) ] ]
9) ¬ x [ x C y [ y D 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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