Tomassi: Logic

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Reply Sat 4 Mar, 2017 04:13 pm
Hi everyone, does anyone know what the answers to these questions would be? I am really struggling with this particular assignment, so would appreciate any help or explanations. I have to prove that the following statements are valid in proven logic. Thank you Smile
¬ = not → = if/ then
¬ ¬(P & Q) : ¬ ¬ (Q & P) (6)
¬ P → ¬Q: Q → P (6)
: (P →Q) → (¬Q →¬P) (5) Principle of transposition
Q → R : (¬Q →¬P) →(P →R) (9)
(P & Q) →¬R : R →(P →¬Q) (11)
P: [(¬(Q → R) →¬P)] →[( ¬R →¬Q)] (9)
P, ¬Q: ¬ (P →Q) (6)
P, ¬P : Q (8)
: ¬P → (P →Q) (10) Law of Dun Scotus
P → ¬P : ¬P (11)
alethes sophia
Reply Tue 4 Jul, 2017 01:48 am
I take it that " : " is to stand for " ∴ "?

I'll just explain using euler/venn diagrams, using basic enclosed shapes. Basically, just imagine areas inside of closed shapes (usually circles, but sometimes boxes) to be "possibly true" areas. I can't help with writing out a proof. But if you can understand the statement and visualize it as a euler/venn diagram, then it shouldn't be too hard to write out a proof by simply making factual observations about the diagram. My mind works better with shapes; it's faster that way, I think.

(1) ¬ ¬(P & Q) : ¬ ¬ (Q & P)
I think this is just saying that the order of P and Q doesn't matter, because the relationship that binds them is an "&" instead of an "or". The diagram will be of two areas P and Q, which have a partial overlapping area called "P & Q".

(2) ¬ P → ¬Q: Q → P
I think this just means that Q is an area totally inside of P, so that whenever Q is actually true, P is also always actually true.

(3) (P →Q) → (¬Q →¬P)
This is the reverse situation. P is an area totally inside of Q, so that whenever Q is not actually true, then P cannot be actually true either (since P is an area totally within Q).

(4) Q → R : (¬Q →¬P) →(P →R)
This situation is going to be 3 totally nested areas, one inside of the other, with P being the smallest inner area that is totally inside of Q, and Q being a medium sized area that is totally inside of R.

(5) (P & Q) →¬R : R →(P →¬Q)
I think the diagram will be two mutually exclusive, but adjacent areas called R and Q, with P being a smaller area that overlaps both R and Q. So imagine a square with a smaller circle nested inside of the square. The area inside of the smaller circle is to be called P. Then divide up the square into two parts, so that the division also cuts through P. Call one part of the square R, and the other part of the square Q. Where P and Q overlap, the area is not going to be any kind of R. But whenever we're talking about any area of R, it will always be the case that if it's also P, then it will be P that doesn't overlap with Q.

(6) P: [(¬(Q → R) →¬P)] →[( ¬R →¬Q)]
I'm going to skip this one. Sorry.

(7) P, ¬Q: ¬ (P →Q)
This is just saying that P and Q are totally mutually exclusive areas. So the diagram would be something like two enclosed areas P and Q, such that P and Q are not overlapping in any way.

(8) P, ¬P : Q
Ah, this one is fun. Tricky, but easy enough. Well, it looks like a large area called Q that is divided into two non-overlapping sub-sections called P and not-P. So basically imagine a circle called Q, and divide that into two non-overlapping parts. Label one part P and the other one gets the unoriginal label of "not-P". Since both P and not-P are "kinds" of Q, it's possible for both P and not-P to be actually true so long as we are claiming that the whole of Q is actually true. Yes, it's paradoxical. That's what makes it fun -- and tricky.

(9) : ¬P → (P →Q)
Is this one missing something? Maybe I'm getting tired. It seems to be a repeat situation of the above diagram in (8). Except that the conditional is simply stating that whenever not-P is actually true, it's still true that the part of Q that is P is still totally inside of Q.

(10) P → ¬P : ¬P
This is the same diagram as in (8) and (9). Except the larger area once called Q is not given a proper label here. I don't like this one. It's deliberately tricky, and it's just kind of a sophisticated cleverness and not very useful -- not to mention a redundancy of (8).

Okay. Well that's my help. But I wonder if it's helpful at all.

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