1) Need an explanation on why this argument is not valid:
p --> r
q --> r
¬(p or q)
-------------------
therefore: ¬r
2) How do I prove or disprove that the proposition "if it is not hot, then it is hot" is equivalent to "it is hot".
1) Need an explanation on why this argument is not valid:
p --> r
q --> r
¬(p or q)
-------------------
therefore: ¬r
2) How do I prove or disprove that the proposition "if it is not hot, then it is hot" is equivalent to "it is hot".
~(p or q) = ~p and ~q
If either p or q occurs, then r will occur.
But if p doesn't occur and q doesn't occur, that doesn't mean that r can't still occur.
You could prove it's not a valid argument by setting up a truth table and finding the case(s) when the premises are true but the conclusion is false.
Not sure how formal you need to be, I mean logic things it is usually a good idea to do a truth table if you formally need to prove them, but I will show you why the first statement is wrong and the second is true.
1)
not(p or q) means that neither p nor q holds. But if this is the case, then both the first two statements are vacuously true, so there is no way to assume r is not true.
If pigs fly, then it is tuesday. Well, pigs don't fly, so there is no way to know what day of the week it is, so the statement is said to be vacuously true. I assume you are aware of this definition if you are doing logic, but just to be complete.
2) I am assuming there are only two options it either is hot, or it is not.
To see these are logically equivalent, suppose it were not hot. Then number 1 tells you it is in fact hot, but it is not hot by assumption, so this statement is false. Similarly the statement "it is hot" is false because we assumed it was not hot.
Suppose it were hot. Then the first statement is vacuously true, and the second statement is clearly true.
I apologize for not being able to do the arrays properly, but this matrix will have to suffice. 0 represents false and 1 is true. p is the statement "it is hot".
$\displaystyle \begin{pmatrix}p & \sim p & p & (\sim p \Rightarrow p) & p\\
0 & 1 & 0 & 0\\
1 & 0 & 1 & 1 \end{pmatrix}$
As you can see the two statements have the same truth values so they are logically equivalent.
(1)just give a counterexample
(2)maybe you can use some algebraic ways to solve it,though very trivial.
let p be "it is hot"
then the condition is ¬p->p
but ¬p->p=¬¬p or p=p or p=p
note that p->q=¬p or q
or even you can use "axioms system" if you want it to be extremly strict.
I'm not good at setting up a truth table involving 3 variables. I do know that there should be eight combinations of truth values for p, q, and r like below:
p q r which propositions do I compare to over here on this side?
T T T
T T F
T F T
T F F
F T T
F T F
F F T
F F F
red is where p implies r goes wrong
blue is where q implies r goes wrong
green is where we have not p and not q going wrong ie, where either p or q is true
But this means that these ones that remain in black are possibilities still where the hypotheses are met. But like I said, it is inconclusive for the truth of r, it could be true or it could be false. you cannot assume (not r), given these hypotheses. because if p is false, q is false and r is true, you have a contradiction while still meeting the hypotheses. get it?