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53.
p  q  p ∨\mathop{¬}q  p ∧ q  (p ∨\mathop{¬}q) ∧ (p ∧ q) 
1  1  1  1  1 
1  0  1  0  0 
0  1  0  0  0 
0  0  1  0  0 
54.
p  q  \mathop{¬}p ∧\mathop{¬}q  \mathop{¬}p ∨ q  (\mathop{¬}p ∧\mathop{¬}q) ∨ (\mathop{¬}p ∨ q) 
1  1  0  1  1 
1  0  0  0  0 
0  1  0  1  1 
0  0  1  1  1 
55.
p  q  r  \mathop{¬}p ∧ q  (\mathop{¬}p ∧ q) ∧ r 
1  1  1  0  0 
1  1  0  0  0 
1  0  1  0  0 
1  0  0  0  0 
0  1  1  1  1 
0  1  0  1  0 
0  0  1  0  0 
0  0  0  0  0 
(Alternate from homeworks: If the area code is 216, then it is Tom Shaffer’s. This is another reasonable interpretation. As is: If you are Tom Shaffer, then your area code is 216.)
The clause p ∧ 1 simplifies, leading to p → 1. From the truth table, we see that the value of p does not matter and the result always is true. So the statement is true.
58.
p  q  \mathop{¬}q →\mathop{¬}p  (\mathop{¬}q →\mathop{¬}p) →\mathop{¬}q 
1  1  1  0 
1  0  0  1 
0  1  1  0 
0  0  1  1 
60.
p  q  p ∧ q  p ∨ q  (p ∧ q) → (p ∨ q) 
1  1  1  1  1 
1  0  0  1  0 
0  1  0  1  0 
0  0  0  0  1 
Quick note on Problem 67. Authentic Persian rugs and other rugs from that region must have flaws in the design. That is a religious requirement. And you can tell if it’s handmade by checking the seams near the edges. If the seams are too small, the rug was machinemade.
55. Literally translated, \mathop{∀}k \text{Items}\kern 1.66702pt \mathop{∀}s \text{Stores} : \mathop{¬}(k\text{ is available in }s). We can pull out the negation to see that this is the same as \mathop{¬}(\mathop{∃}k \text{Items}\kern 1.66702pt \mathop{∃}s \text{Stores} : k\text{ is available in }s. So no item is available in any store. This is not the correct statement. The advertisement means to state that some items may not be available in all stores. We do not have the proper symbols to translate may.
56. The direct translation is \mathop{∀}p \text{People} : \mathop{¬}(p\text{ has time to maintain his/her car properly}). We can pull the negation out to see that \mathop{¬}(\mathop{∃}p : p\text{ has time to maintain his/her car properly}), and the original statement says that no one maintains his/her car properly. The intent likely was that some people do not have time to maintain their cars properly, or \mathop{∃}p : \mathop{¬}(p\text{ has time to maintain his/her car properly}).
The first statement is about every student, \mathop{∀}s : \mathop{¬}(s\text{ passed the test})
The second negates the entire thing, so there exists some student who did not pass, because \mathrel{⊧}\mathop{¬}\mathop{∀}s : s\text{ passed the test} ≡\mathop{∃}s : s\text{ did not pass the test}.
Note that the latter statement is still true if no student passed. So you cannot infer that anyone passed in either statement.
Hint: Quantifiers do not necessarily exclude each other.
Both statements are true but they have different quantifiers.
The original statement (for some real number) is true if you pick one number (say zero) and demonstrate its truth ({0}^{2} = 0 ≥ 0).
Your friend’s statement applies to all real numbers. To demonstrate its validity, the statement must be proven to be true regardless of the value of x.
There is a number p for all numbers q such that the difference between p and q is 2.
Symbolically,
\mathop{∀}q\mathop{∃}p : p  q = 2.

Note the order of the quantifiers. So the negation is
\mathop{¬}(\mathop{∀}q\mathop{∃}p : p  q = 2) ≡\mathop{∃}q\mathop{∀}p : p  q≠2.

And in English:
For some number q, for all numbers p, the difference between p and q is not 2.
Quite often the phrasing is less bizarre if you push the negation only partway through. Here, negating just the initial \mathop{∀}q gives the phrase:
For some number q there is no number p such that the difference between p and q is 2.
The initial statement is true, and its negation is false.
For all sets A, for all sets B, there is a set C such that A ∩ B = C and C is not ∅.
Translating:
\mathop{∀}A\mathop{∀}B\mathop{∃}C : A ∩ B = C ∧ C≠∅.

Pushing the negation through the quantifiers and applying De Morgan’s laws,
\mathop{¬}(\mathop{∀}A\mathop{∀}B\mathop{∃}C : A ∩ B = C ∧ C≠∅) ≡\mathop{∃}A\mathop{∃}B\mathop{∀}C : A ∩ B≠C ∨ C = ∅.

Back into English:
There is a set A for which there is a set B such that for all C, A ∩ B≠C or C = ∅.
Note that the statement
\mathop{∀}A\mathop{∀}B\mathop{∃}C : A ∩ B = C ∧ C≠∅

is false. If two sets A and B are disjoint, then A ∩ B = ∅. Thus the negation is true.
Derive a logic expression from the following truth table. Attempt to simplify it remembering the distributive property, De Morgan’s laws, and that \mathrel{⊧}z ∨\mathop{¬}z ≡ 1 and \mathrel{⊧}z ∧\mathop{¬}z ≡ 0.
p  q  r  f(p,q,r) 
1  1  1  1 
1  1  0  0 
1  0  1  1 
1  0  0  1 
0  1  1  0 
0  1  0  0 
0  0  1  0 
0  0  0  0 
There are fewer true values, so we start by listing individual conditions where the function is true. These are p ∧ q ∧ r, p ∧\mathop{¬}q ∧ r, and p ∧\mathop{¬}q ∧\mathop{¬}r. Joining these with or,
\mathrel{⊧}f(p,q,r) ≡ (p ∧ q ∧ r) ∨ (p ∧\mathop{¬}q ∧ r) ∨ (p ∧\mathop{¬}q ∧\mathop{¬}r).

Using the distributive property, we can pull out a p. Then
\mathrel{⊧}f(p,q,r) ≡ p ∧ ((q ∧ r) ∨ (\mathop{¬}q ∧ r) ∨ (\mathop{¬}q ∧\mathop{¬}r)).

Now we can pull out either \mathop{¬}q or r to try simplifying inside the parenthesis. With the former:
\mathrel{⊧}f(p,q,r) ≡ p ∧ ((q ∧ r) ∨ (\mathop{¬}q ∧ (r ∨\mathop{¬}r))).

Because \mathrel{⊧}r ∨\mathop{¬}r ≡ 1,
\mathrel{⊧}f(p,q,r) ≡ p ∧ ((q ∧ r) ∨\mathop{¬}q).

Now redistribute \mathop{¬}q and simplify to see
So a simplified form is \mathrel{⊧}f(p,q,r) ≡ p ∧ (r ∨\mathop{¬}q). Another equivalent form is \mathrel{⊧}f(p,q,r) ≡ p ∧ (r → q).