Since the entries in the last column are F, (p ∧ ~p) ∧ (~q ∧ P) is a contradiction

It can be obtained by using examples 12.15 and 12.16 that
p↔q ≡ (¬ p ∨ q)  ∧ (¬q ∨ p) ... (1)
≡ (¬p∨q)  ∧ ( p∨ ¬q) (by Commutative Law) ... (2)
≡ (¬p ∧ ( p ∨ ¬q)) ∨ (q ∧ ( p ∨ ¬q)) (by Distributive Law)
≡ (¬p ∧ p) ∨ (¬p ∧ ¬q) ∨ (q ∧ p) ∨ (q ∧ ¬q) (by Distributive Law)
≡ F ∨ (¬p ∧ ¬q) ∨ (q ∧ p) ∨ F; (by Complement Law)
≡ (¬p ∧ ¬q) ∨ (q ∧ p) ; (by Identity Law)
≡ ( p ∧ q) ∨ (¬p ∧ ¬q) ; (by Commutative Law)
Finally (1) becomes p ↔️ q ≡ ( p ∧ q) v (ㄱp ∧ ㄱq)

Also the above result can be proved without using truth tables. This proof will be provided after studying the logical equivalence

i) Though - is not binary on N; it is binary on Z. To check the validity of any more properties satisfied by – on Z, it is better to check them for some particular simple values.
ii) Take m = 4 , n = 5 and (m− n) = (4 − 5) = −1and (n −m) = (5 − 4) = 1.
Hence (m− n) ≠ (n −m). So the operation - is not commutative on Z.
iii) In order to check the associative property, let us put m = 4, n = 5 and p = 7 in both (m- n) - p and m- (n - p).
(m−n)− p = (4−5)−7 = (−1−7) = −8 …(1)
m−(n− p) = 4−(5−7) = (4+2) = 6  …(2)
From (1) and (2), it follows that (m - n) - p m - (n - p).
Hence – is not associative on Z.
iv) Identity does not exist (why?).
v) Inverse does not exist (why?).

∴p v (q Λ r) is a contingency

(i) 4 + 7 = 12
it s a proposition as its truth value is F
(ii) What are you doing?
It is a question and not a proposition 
(iii) 3ⁿ ≤ 81, n ∈ N
It is a proposition as it is true when
n = 1, 2, 3, 4
(iv) Peacock is our national bird. It is a proposition as its truth value is T.
(v) How tall this mountain is!
This is an exclamation, not a proposition.

Let p: 19 is a prime number.
q: All the angles of a triangle are equal be two simple statements.
(i) 19 is not a prime number and all the angles of a triangle are equal.
~p ∧ q
(ii) 19 is a prime number or all the angles of a triangle are not equal.
p ∧ ~q
(iii) 19 is a prime number and all the angles of a triangle are equal.
p ∧ q
(iv) 19 is not a prime number.
~p.

Then A∨ B =\(\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\vee \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 0\vee 1 & 1\vee 1 \\ 1\vee 0 & 1\vee 1 \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\)
\(A\wedge B=\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\wedge \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 0\wedge 1 & 1\wedge 1 \\ 1\wedge 0 & 1\wedge 1 \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}\)

p v ~q

p ⇔ q is F

Let a = 2, b = -5
∴ a * b = a^b
⇒ 2 * -5 = 2^-5 = \(\frac{1}{2^5}\) ∉ z
* is not a binary operation on z.