(( p ∧ q) ∨ (¬r ∨¬s)) ∧ (¬ t ∧ v)) contains 6 variables p, q, r, s, t, and v. Hence the corresponding truth table will contain 2⁶ = 64 rows.

p ∨ ¬ t ( p ∨ ¬s) contains 3 variables p, s, and t. Hence the corresponding truth table will contain 2³ = 8 rows

(1) ¬p: It is not cold.
(2) p ∧ q: It is cold and raining.
(3) p ∨ q: It is cold or raining.
(4) q ∨ ¬p: It is raining or it is not cold
Observe that the statement formula ¬ p has only 1 variable p and its truth table has 2 = ( 2¹ ) rows. Each of the statement formulae p ∧ q and p ∨ q has two variables p and q. The truth table corresponding to each of them has 4 = (2² ) rows. In general, it follows that if a statement formula involves n variables, then its truth table will contain 2ⁿ rows.

Existence of identity:
We have to find an element a' ∈ A such that
a*a' = a'*a = e
⇒ a + e-ae = a
⇒ a + e-ae = a
⇒ e-ae = 0
⇒ e(1-a) = 0
⇒ e = \(\frac{0}{1-a}\) = 0∈A
∴ A has an identity under *.
Existence of inverse:
For every a ∈ A, there exist a' ∈ A such that a*a' = a' * a = e
⇒ a+a'-aa' = 0  [∵e=0]
⇒ a+a'(1-a) = 0
⇒ a'(1-a) = -a
⇒ a' = \(\frac{-a}{1-a}\)
To prove that \(\frac{-a}{1-a}\)≠ 1
Suppose \(\frac{-a}{1-a}\) = 1
⇒ -a = 1-a
⇒ -a + a = 1 ⇒ 0 ≠ 1
∴ Our assumption is wrong
⇒ \(\frac{-a}{1-a}\) ≠ 1
∴ A has inverse for every element x ∈ A

Given M = \(\left\{ \left( \begin{matrix} x & x \\ x & x \end{matrix} \right) :x\in R-\{ 0\} \right\} \) and ∗ be the matrix multiplication.
Let A = \(\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \)and 
B = \(\left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \)∈M
Where x, y ∈R-{0}.
\(A*B=\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \)
\(=\left( \begin{matrix} xy+xy & xy+xy \\ xy+xy & xy+xy \end{matrix} \right) \)
\(=\left( \begin{matrix} 2xy & xy \\ 2xy & 2xy \end{matrix} \right) \in M\\ \)
[∵ 2xy∈R-{0}]
∴ M is closed under M
Identity:
Since identity of 2\(\times\)2 matrices is I =\(\left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right) \)∉M 
∴ M has no identity under *.
Inverse:
Since it has no identity, it won't have inverse also.

Given M = \(\left\{ \left( \begin{matrix} x & x \\ x & x \end{matrix} \right) :x\in R-\{ 0\} \right\} \) and * be the matrix multipilication.
Let A \(=\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \) and B = \(\left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \in M\)
Where x, y ∈R-{0}.
\(A*B=\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \)
\(\\ =\left( \begin{matrix} 2xy & 2xy \\ 2xy & 2xy \end{matrix} \right) \in M\)
[∵ 2xy ∈R-{0}]
∴ M is closed under *.
Commutative property:
we know A*B =\(\left( \begin{matrix} 2xy & 2xy \\ 2xy & 2xy \end{matrix} \right) ..(1)\)
Let x,y∈R-{0}
Now B + A \(=\left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \)
\(=\left( \begin{matrix} xy+xy & xy+xy \\ xy+xy & xy+xy \end{matrix} \right) \)
\(=\left( \begin{matrix} 2xy & 2xy \\ 2xy & 2xy \end{matrix} \right) \\ \)
From (1) &(2), A*B = B*A
∴ *has commutative property on M
Associative property:
Let A = \(\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \)
B =\(\left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \) and
C = \(\left( \begin{matrix} z & z \\ z & z \end{matrix} \right) \)
for x, y, z ∈R-{0}
\((A*B)*C=\left( \begin{matrix} 2xy & 2xy \\ 2xy & 2xy \end{matrix} \right) *\left( \begin{matrix} z & z \\ z & z \end{matrix} \right) \)
\(=\left( \begin{matrix} 2xyz+2xyz & 2xyz+2xyz \\ 2xyz+2xyz & 2xyz+2xyz \end{matrix} \right) \)
\(=\left( \begin{matrix} 4xyz & 4xyz \\ 4xyz & 4xyz \end{matrix} \right) ...(1)\)
Now\(A*(B*C)=A*\left( \begin{matrix} 2yz & 2yz \\ 2yz & 2yz \end{matrix} \right) \)
\(=\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) *\left( \begin{matrix} 2yz & 2yz \\ 2yz & 2yz \end{matrix} \right) \)
\(=\left( \begin{matrix} 4xyz & 4xyz \\ 4xyz & 4xyz \end{matrix} \right) ...(2)\\ \)
From (1)&(2), (a*B)*C = A*B*C)
Since matrix multiplication is associative, this axiom holds good for M.

Given \(a*b=\frac { a+b }{ 2 } \forall \in Q\)
i) Closure property:
Let a, b ∈ Q
∴ a*b = \(\frac{a+b}{2}\)∈Q
[∵ addition and division are closed on Q]
* is closed on Q.
(ii) Commutative property:
Let a, b ∈ Q
Then a+b \(=\frac { a+b }{ 2 } =\frac { b+a }{ 2 } =b*a\)
∴ a*b = b*a ∀a,b∈Q
∴ * is commutative on Q.
(iii) Associative property :
Let a, b, c ∈ Q
a*(b*c) = (a*b)*c
Let a = 2, b = 3, c-5
∴ a*(b*c) = 2*(3*-5)
\(=*\left( \frac { 3-5 }{ 2 } \right) \)
\(=2*(-1)=\frac { 2+(-1) }{ 2 } \)
\(=\frac { 1 }{ 2 } \quad \quad ...(1)\)
Now (a*b)*c = (2*3)*(-5)
\(=\left( \frac { 2+3 }{ 2 } \right) *(-5)\)
\(=\frac { 5 }{ 2 } *-5=\frac { \frac { 5 }{ 2 } +(-5) }{ 2 } \)
\(=\frac { 5-10 }{ 4 } =\frac { -5 }{ 4 } ...(2)\)
From (1) & (2), a*(b*c) ≠ (a*b)*c
∴ * is not associative on Q.

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)

It is known that Z₅ = {[0], [1], [2], [3], [4]}. The table corresponding to addition modulo 5 is as follows: We take reminders {0,1,2,3,4} to represent the classes {[0], [1], [2], [3], [4]}.

(i) Since each box in the table is filled by exactly one element of Z₅, the output a +₅ b is unique and hence +₅ is a binary operation.
(ii) The entries are symmetrically placed with respect to the main diagonal. So +₅ has commutative property
(iii) The table cannot be used directly for the verification of the associative property. So it is to be verified as usual
For instance, (2+₅3)+₅ 4 = 0+₅ 4 = 4(mod 5)
and 2+₅(3+₅4) = 2 +₅ 2 = 4(mod5)
Hence (2+₅3)+₅4 = 2+₅(3+₅4)
Proceeding like this one can verify this for all possible triples and ultimately it can be shown that +₅ is associative
(iv) The row headed by 0 and the column headed by 0 are identical. Hence the identity element is 0.
(v) The existence of inverse is guaranteed provided the identity 0 exists in each row and each column. From Table, it is clear that this property is true in this case. The method of finding the inverse of any one of the elements of Z₅, say 2 is outlined below.
First find the position of the identity element 0 in the III row headed by 2. Move horizontally along the III row and after reaching 0, move vertically above 0 in the IV column, because 0 is in the III row and IV column. The element reached at the topmost position of IV column is 3. This element 3 is nothing but the inverse of 2, because, 2+₅ 5+ = 0 (mod5). In this way, the inverse of each and every element of Z₅ can be obtained. Note that the inverse of 0 is 0, that of 1 is 4, that of 2 is 3, that of 3 is 2, and, that of 4 is 1.

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