To prove  \(\neg(p \rightarrow q) \equiv p \wedge \neg q\)
Truth table for \(\neg(p \rightarrow q)\)

Truth table for \(p \wedge \neg q\)

The entries in the column \(\neg(p \rightarrow q) \text { and } p \wedge \neg q\) are identical and they are equivalent.

Truth table for \(\neg(p \wedge q)\)

Truth table for \(\neg p \vee \neg q\)

P	Q	\(\neg p\)	\(\neg q\)	\(\neg p \vee \neg q\)
T	T	F	F	F
T	F	F	T	T
F	T	T	F	T
F	F	T	T	T

From the table, it is clear that the column of p➝(¬q V ~r) and ~pV(~q V r) are identical
∴  p➝(¬q V ~r) ≡ ~pV(~q V r) 
Hence proved.

Prove that p⟶(q⟶r) = (p ∧ q)⟶r without using truth table. From example we know that p⟶ q = ~p V q
Consider LHS = p⟶(q⟶r)
= p ⟶ (~q V r) [Implication Law]
= ~p V (~q V r) [Implication Law]
= (~p V ~q) V r [associative property]
= ~(p ∧ q) V r [using Demorgan's law]
= (p ∧ q) ⟶ r [Implication Law]
Hence proved.

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

~(p V q) V (~p ∧ q) and ~p

The entries in column (5) and column (7) are identical.
∴ ~(p V q) V (~p ∧ q) and ~p are logically equivalent.

Given statement is p ➝(q ➝ p)
=p ➝ (~q v p)
[∴ By example p ➝ q =p V q]
= (~p) V (~q V p) [Again by example 12.17]
= (~p V ~q) V p [Associative law)]
= (~p V p) V ~q [Commutative law)]
= II V ~q [~p V p is also tautology]
= II [II V ~q] is also a tautology.

given A = {Q\{1}}
A is defined on A by x*y = x+y-xy
Let x,y ≠ 1
∴ x*y = x + y - xy
Now to prove that x + y - xy ≠ 1
Let us assume that x + y - xy = 1
x+y-xy-1 = 0
(x-1)-y(x-1) = 0
(x-1)(1-y) = 0
x =1 or y = 1 which is a false [∵x, y ≠ 1]
∴ is a binary operation on A.
Commutative property:
Let x,y ∈A ⇒ x, y≠1
∴x*y = x+y-xy
and y*x = y+x-yx
⇒x+y = y*x∀x, y∈A
A has commutative property under *
Associative property:
Let x,y,z ∊A ⇒x,y,z≠1
Consider (x*y)*z = (x+y-xy)*z
= x +y~xy +z- (x +y-xy)z
= x +y-xy+ z-xz- yz + xyz
= x +y+z-xy-yz-zx +xyz ...(1)
= x +y+z - yz - x (y + z - yz)
= x +y +z-yz-zy-xz +xyz ..(2)
From (1) & (2), (x*y)*z = x*(y*z)
A has associative property under *.

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.

∴ ((p⟶ q) ∧ (q ⟶ r)) ⟶ (p ⟶ r)
The given statement is a tautology.

Since this is neither a tautology not a contradiction
( p ⟶ q) ↔️ (~p ⟶ q) is a contingency.

(( p V q)∧ ~p)) ➝ q

The statement (( p V q)∧ ~p) ➝ q is a tautology.

The statement (p ∧ q) ∧ ~(p ∨ q)

Given \(a*b=\frac { a+b }{ 2 } \), where a.b ∈Q Let a,b ∈Q
An element e has to found out such that 
a*e = e*a = a
Let a = 5, Then 5*e = 5
\(\Rightarrow \frac { 5+e }{ 2 } =\)5 ⇒ 5 + e = 10
Let a = \(\frac{2}{3}\). Then \(\frac{2}{3}\)*e = \(\frac{2}{3}\)
\(\Rightarrow \frac { \frac { 2 }{ 3 } +e }{ 2 } =\frac { 2 }{ 3 } \)
\(\Rightarrow \frac { 2 }{ 3 } +e=\frac { 4 }{ 3 } \)
\(\Rightarrow e=\frac { 4 }{ 3 } -\frac { 2 }{ 3 } =\frac { 2 }{ 3 } \)
It is seen that for the binary operation * defined on Q, identity element e is not unique. Hence identity element not defined for the binary operation * on Q. 
The identity does not exist. Hence inverse also does not exist for the operation * on Q.

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.

The table for the operation x₁₁ is as follows.

Following the same kind of procedure as explained in the previous example, a brief outline of the process of verification of the properties of ×₁₁ on A is given below.
(i) Since each box has an unique element of A, ×₁₁ is a binary operation on A.
(ii) The entries are symmetrical about the main diagonal. Hence ×₁₁ has commutative property.
(iii) As usual, the associative property can be seen to be true.
(iv) The entries of both the row and column headed by the element 1 are identical. Hence 1 is the identity element.
(v) Since the identity 1 exists in each row and each column, the existence of inverse property is assured for ×₁₁. The inverse of 1 is 1, that of 3 is 4, that of 4 is 3, 5 is 9, and, that of 9 is 5.

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.

(i) The output m+ n - mn is clearly an integer and hence∗ is a binary operation on Z.
(ii) m*n = m+ n − mn = n + m − nm = n*m, ∀m,n∈Z. So ∗ has commutative property.
(iii) Consider (m*n)*p = (m+ n −m n)* p= (m+ n −mn) + p − (m+ n −m n) p
= m+ n + p −mn −m p − n p + m n p ... (1)
Similarly m*(n*p) = m*(n + p − n p) = m+ (n + p − n p) −m (n + p − n p)
= m+ n + p − n p −m n −mp + m n p ... (2)
From (1) and (2), we see that m*(n*p) = (m*n)*p. Hence * has associative property.
(iv) An integer e is to be found such that 
m*e = e*m = m, ∀m∈Z ⇒ m + e - m e = m
⇒e(1-m) = 0 ⇒ e = 0 or m = 1. But m is an arbitrary integer and hence need not be equal to 1. So the only possibility is e = 0. Also m*0 = 0*m = m, ∀m∈Z. Hence 0 is the identity element and hence the existence of identity is assured.
(v) An element m'∈ Z is to be found such that m*m' = m' * m = e = 0, ∀m∈Z.
m*m' = 0 ⇒ m+m'-m m'= 0⇒ m m' = 0 ⇒ \(\frac{m}{m-1}\). when m = 1, m' is not defined.
When m = 2, m' is an integer. But except m = 2, m′ need not be an integer for all values of m. Hence inverse does not exist in Z.

(i) It is true that a*b = a^b ∈N; ∀a,b∈N. So * is a binary operation on N.
(ii) a*b = a^b and b*a = b^a. Put, a = 2 and b = 3. Then a*b = 23 = 8 but b*a = 3² = 9 So a*b need not be equal to b*a. Hence * does not have commutative property.
(iii) Next consider a*(b*c) = a*(b^c) = a^_(bc₎ Take a = 2, b = 3 and c = 4. 
Then a*(b*c) = 2*(*4) = 2^₃4 = 2⁸¹
But (a*b)*c = (a^b)*c = (a^b)^c = a(bc) = a^(bc) = 2¹²
Hence a*(b*c) ≠ (a*b)*c. So * does not have associative property on N.

Consider the set Z_o of all odd integers Z_o = {2k+1 : k∈Z} = {.., -5,-3,-1,1,3,5,...}. + is not a binary operation on Z_o because when x = 2m+1, y = 2n +1, x + y = 2(m+ n) + 2 is even for all m and n. For instance, consider the two odd numbers 3, 7 ∈Z_o. Their sum 3 + 7 = 10 is an even number. In general, if x, y∈Z₀, then (x+y)∉Z_o. Other properties need not be checked as it is not a binary operation.

Consider the set of all even integers Z_e = {2k | k ∈ Z} = {...,−6, −4, −2, 0, 2, 4, 6,...}.
Let us verify the properties satisfied by + on Z_e.
(i) The sum of any two even integers is also an even integer.
Because x, y∈Z_e, ⇒ x = 2m and y = 2n , m,n∈Z.
So (x + y) = 2m + 2n = 2(m+n)∈Z_e. Hence + is a binary operation on Z_e.
(ii) ∀ x, y∈Z_e, (x + y) = 2(m + n) = 2(m + n) = 2(n + m) = (2n + 2m) = (y + x).
So + has commutative property
(iii) Similarly it can be seen that ∀x, y, z∈Z_e, (x + y) + z = x + ( y + z).
Hence the associative property is true.
(iv) Now take x = 2k , then 2k + e = e + 2k = 2k ⇒ e = 0.
Thus ∀ x ∈ Z_e, ヨ0∈Z_e, ⋺x+0 = 0+x = x.
So, 0 is the identity element.
(v) Taking x = 2k and x′ as its inverse, we have 2k+x' = 0 = x'+2k ⇒ x' = −2k. i.e., x' = −x.
Thus ∀x ∈ Z_e, ヨ-x∈Z_e ⋺x + (−x) = (−x) + x = 0
Hence -x is the inverse of x ∈Z_e.

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?).

(i) m + n∈Z, ∀m, n∈Z. Hence + is a binary operation on Z.
(ii) Also m + n = n + m,∀m, n∈Z. So the commutative property is satisfied
(iii) ∀m, n, p∈Z, m+ (n + p) = (m+ n) + p. Hence the associative property is satisfied.
(iv) m + e = e + m = m ⇒ e = 0. Thus ヨ 0∈Z⋺(m+ 0) = (0 + m) = m. Hence the existence of identity is assured.
(v) m + m' = m'+ m = 0 ⇒ m' = −m. Thus ∀∈Z,ョ−m∈Z ⋺ m+ (−m) = (−m) + m = 0. Hence, the existence of inverse property is also assured. Thus we see that the usual addition + on Z satisfies all the above five properties.

Given p : Jupiter is a planet and
q : India is an island.
(i) ¬p : Jupiter is not a planet.
(ii) p ∧ ¬q : Jupiter is a planet and India is not an island.
(iii) ¬p ∨ q : Jupiter is not a planet or India is an island.
(iv) p➝ ¬q : If Jupiter is a planet then India is not an island.
(v) p↔q : Jupiter is a planet if and only if India is an island.