Finished

[Arachnophobian]

I really meant it when I said logic was my weakness. GitHub is not much better.

/assign
@giacomodecolle
@avsculley
@JaronJCheung
@Finn1928

[Finn1928]

On question 3, I believe we need to define (a)-(e) in terms of P(x,y) and F(x). So, for example, in 3(a), I don't think you can use B(x,y), since that's what you're trying to define. Also, make sure you say x is "not" female (~F(x)). The answer I give for 3(a) is ∃x∃y∃z(P(z,y)∧P(z,x)∧¬Fx∧x≠y∧x≠z∧z≠y), which says that z and y share a parent, x is not female, and none of them are identical to each other.

For question 4, I'm still not entirely sure how to answer it fully. But Karl pointed out to me inverse roles in the textbook (p. 37), which can help us talk about one person being a child of someone else without having a child_of role to work with. It basically takes the parent_of relation and inverses it. I also make use of number restrictions (p. 39) For 4(a), I get B = M∩∃parent_of−.(≥2∃parent_of), but I do not know if this is correct. And I'm stuck on some of the other definitions in this section. But hopefully these suggestions can help you on your answers. It should also help out with question 5 (which I also don't have fully figured out, yet).

Question 7 looks like it can be done without using any of the description logic we've been learning about. Think about creating a possible world where one of the sentences is true and the other is false. There's probably multiple potential answers to this question. The answer I give draws a universe with 2 objects. One of the objects has an R relation pointing at itself, and another pointing to the other object. The other object only has one R relation that is pointing at itself. I think this is a scenario that makes the first sentence false and the second true.

For question 9, I couldn't figure out how to use the natural deduction proof tool either. If you know how to do deductive logic proofs, I just brute-forced this question that way. If not, I also suggest googling "proofs for DeMorgan's Theorems". You'll likely find people making proofs that you can copy.

Once you answer 9, I think one of the most significant differences for question 10 is that the tree proof starts off by assuming that what you're trying to show is false and attempting to draw out a contradiction given this. By contrast, the natural deduction can employ indirect derivation like this, but it does not need to start with this.

[JaronJCheung]

Dear @Arachnophobian,

Thank you for inviting me into the process of reviewing your work! I have reviewed your work and below are some suggestions, observations, and some sharing of material.

Q1: Very minor typo--for (a) and (c), you misspelled tautology. (I did this too for one of mine lol.)

Q2: You are right to say for (b) and (c) that it is a tautology, but it is not correct to say that there is no CNF form. A tautology takes the form of P ∨ ~P, so you can reduce (b) and (c) to the CNF form of "A ∨ ~A".

Also, you misspelled tautology for (b) and (c) also.

For (a), I think you can reduce it further. This is what I got via the distribution rule.

(a) (A→B)→C = (~A ∨ B) → C = ~(~A ∨ B) ∨ C = (~~A ∧ ~B) ∨ C = (A ∧ ~B) ∨ C = (A ∨ C) ∧ (~B ∨ C)

Q6: Although there are no specified subjects, you can explain the relationship between each quantified thing. Below is how I went about it. Please feel free to glean (or not) whatever is helpful!

(a) ∃x∀y and ∀x∃y

∃x∀y = "There exists an x such that for every y" or possibly "Something is identical with everything." This can be interpreted as a statement saying that there is at least one value of x for which the statement "for every y" is true. In other words, it asserts the existence of x in relation to all possible values of y.

∀x∃y = "For all x, there exists a y" or possibly "Everything is identical with something" This can be interpreted as a statement saying that for every possible value of x, there is at least one corresponding value of y. In other words, it asserts the existence of y in relation to all possible values of x.

(b) ∃x∀y∃z and ∀x∃y∀z

∃x∀y∃z = "There exists an x such that for every y, there exists a z." This can be interpreted as a statement saying that for some value of x, we can find a corresponding value of z for every possible value of y. It states the existence of x and z in relation to all possible values of y.

∀x∃y∀z = "For all x, there exists a y such that for all z." This can be interpreted as a statement saying that for all possible values of x, we can find a corresponding value of y that holds true for all values of z. In other words, it asserts the existence of y in relation to all possible values of x and z.

(c) ∀x∃y∀z∃w and ∃x∀y∃z∀w

∀x∃y∀z∃w = "For all x, there exists a y such that for all z, there exists a w." This can be interpreted as a statement saying that for all possible values of x, we can find a corresponding value of y, and for all possible values of z, we can find a corresponding value of w. In other words, it asserts the existence of y and w in relation to all possible values of x and z.

∃x∀y∃z∀w = "There exists an x such that for all y, there exists a z for all w." This can be interpeted as a statement saying that for some value x, we can find a corresponding value y, and for every y, there is some value z that holds for all of w.

Q9: I figured out how to use the natural deduction calculator found here: http://teachinglogic.liglab.fr/DN/. I changed up the mathematical symbols to logical symbols. Feel free to see and use below.

(a) ∀x∀y(¬(Px ∧ Qx) → (¬Px ∨ ¬Qx))

¬(Px ∧ Qx) → (¬Px ∨ ¬Qx)

assume ¬(Px ∧ Qx).
assume ¬(¬Px ∨ ¬Qx).
assume ¬Px.
¬Px ∨ ¬Qx.
F.
therefore ¬¬Px.
assume Qx.
Px.
assume ¬(¬Px ∨ ¬Qx).
Px ∧ Qx.
F.
therefore ¬¬(¬Px ∨ ¬Qx).
F.
therefore ¬Qx.
¬Px ∨ ¬Qx.
F.
therefore ¬¬(¬Px ∨ ¬Qx).
¬Px ∨ ¬Qx.
therefore ¬(Px ∧ Qx) → ¬Px ∨ ¬Qx.
(b) ∀x∀y(¬(Px ∨ Qx) → (¬Px ∧ ¬Qx))

¬(Px ∨ Qx) → (¬Px ∧ ¬Qx)

assume ¬(Px ∨ Qx).
assume Px.
Px ∨ Qx.
F.
therefore ¬Px.
assume Qx.
Px ∨ Qx.
F.
therefore ¬Qx.
¬Px ∧ ¬Qx.
therefore ¬(Px ∨ Qx) → ¬Px ∧ ¬Qx.
(c) ∀x∀y((¬Px ∨ ¬Qx) → ¬(Px ∧ Qx))

¬Px ∨ ¬Qx → ¬(Px ∧ Qx)

assume ¬Px ∨ ¬Qx.
assume Px ∧ Qx.
Px.
Qx.
assume ¬Px.
F.
therefore ¬Px → F.
assume ¬Qx.
F.
therefore ¬Qx → F.
F.
therefore ¬(Px ∧ Qx).
therefore ¬Px ∨ ¬Qx → ¬(Px ∧ Qx).
(d) ∀x∀y((¬Px ∨ ¬Qx) → ¬(Px ∧ Qx))

¬Px ∨ ¬Qx → ¬(Px ∧ Qx)

assume ¬Px ∨ ¬Qx.
assume Px ∧ Qx.
Px.
Qx.
assume ¬Px.
F.
therefore ¬Px → F.
assume ¬Qx.
F.
therefore ¬Qx → F.
F.
therefore ¬(Px ∧ Qx).
therefore ¬Px + ¬Qx → ¬(Px ∧ Qx).

Besides what Finn and I have addressed, it looks good!

Please let me know if you have any questions.

Best,
Jaron