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After a little bit of time learning discrete mathematics, I attempted this

This is the question

Determine whether or not the following arguments are valid. If they are valid, then state the rules of inference used to prove validity. If they are invalid, outline precisely why they are invalid.

a. If it is summer, then it is humid outside. If Santa is delivering presents, then it is not the case that it is hot and humid outside. It is currently hot outside and Santa is delivering presents. Therefore, it must not be summer.

b. If it is Thursday, then I do not have to go to class. I do not go to class. Therefore, it must be Thursday.

c. Everyone taking Computer loves math. Everyone taking Statistics loves math. Therefore, everyone in Statistics is taking Computer.

This is my answer

1A)
    Q = it is summer
    P = it is humid 
    R = Santa is delivering presents  

    Invalid R →( ¬Q ∧ ¬P)
    It is currently hot and Santa is delivering presents 
    Therefore false 

 B)
    Q = it is Thursday 
    P = I have to go to class 
    R = I went to class

    If it is Thursday then I do not have to go to class
    Q → ¬P
    I did not go to class
    ¬R
    It is Thursday 
    Q

    1 Q → ¬P
    2 ¬R
    3 Q
    3 ¬R Modus Tollens


C) 
    Q = Taking computer 
    P= loves math
    R = Taking Statistics

Everyone taking Computer loves math
Q ∧ P
Everyone taking Stat 2507 loves math
R ∧ P
Therefore, everyone in statistics is taking comp
R → Q

1 Q ∧ P
2 R ∧ P
3 R → Q
4 Q 1 Simplification
5 P 1 Simplification

Any help is appreciated thanks in advance

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  • If it is summer, then it is humid outside.
  • If Santa is delivering presents, then it is not the case that it is hot and humid outside.
  • It is currently hot outside and Santa is delivering presents.
  • Therefore, it must not be summer.

Look at each of the statements.   You need a predicate for Hot (such as $S$) as well as Humid ($P$).   Note: you are not told that Summer ($Q$) implies it is Hot, just that it is Humid.   Don't presume it is so for where-ever this is.

$$\begin{split}Q \to~& P\\ R\to~& \neg(S\land P)\\& S \land R\\\hline \therefore\qquad&\neg Q \end{split}$$

That seems valid to me.   We can infer $S$ and $R$ from $S\wedge R$.   Then we infer $\neg (S\wedge P)$ from $R$ and $R\to\neg(S\wedge P)$.   We can infer $\neg P$ from $S$ and $\neg(S\land P)$ (well, this may require a sub proof).   Finally this lets us infer $\neg Q$ from $\neg P$ and $Q\to P$.

The names of the rules of inference are left to you.   That s what you do.


For $(b)$ you are missing a statement that claims "if you have to go to class, then you will go to class", although that obligation might be considered implicit by the mention of "have to".   Well, anyway, "do not have to go to class" and "do not go to class" are not equivalent.

Still, even with that, $Q\to\neg P, P\to R, \neg R$ will not allow you to infer $\neg Q$.   Modus Tollens will only let you infer $\neg P$ from $\neg R$ and $P\to R$.   You cannot validly infer $\neg Q$ from $\neg P$ and $Q\to\neg P$.


For (c) you are given implications, not conjunctions.   "Everyone who does this is doing that" not "Everyone is doing this and doing that."

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  • $\begingroup$ Here's a spelled out version of the first argument. $\endgroup$ – Derek Elkins May 30 '18 at 3:34
  • $\begingroup$ Ah, sequent calculus tree format, using left- and right- side introduction rules for the connectives? @DerekElkins A little cluttered up top, but clear enough. Yes, that is one way to do it. $\endgroup$ – Graham Kemp May 30 '18 at 4:12

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