Given \(A=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) ,B=\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) and\quad C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 0 & 0 \\ 0 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 0 \\ 0 & 0 \\ 0 & 1 \end{matrix} \right) \vee \left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
\((A\wedge B)\vee C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)

Given \(A=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) ,B=\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) and\quad C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
\(\left( \begin{matrix} 1 & 1 \\ 1 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 1 \\ 1 & 1 \\ 0 & 1 \end{matrix} \right) \wedge \left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) =\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) \)

Given \(A=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) ,B=\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) and\quad C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) \vee \left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 1\vee 0 & 0\vee 1 \\ 0\vee 1 & 1\vee 0 \\ 1\vee 1 & 0\vee 0 \end{matrix}\begin{matrix} 1\vee 0 & 0\vee 1 \\ 0\vee 1 & 1\vee 0 \\ 0\vee 0 & 1\vee 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 1 & 1 \\ 1 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 1 \\ 1 & 1 \\ 0 & 1 \end{matrix} \right) \) [∵ a∨b=max(a,b)]

√b is not defined for negative values, b which also ∈ R.
Hence, a√b is not defined for all a, b ∈R
* is not a binary operation on R

Given a*b = a.|b| on R
a,b ∈ R \(\Rightarrow\) a.|b| ∈R as a ∈ R and |b|∈R.
Hence * is a binary operation on R

In this problem a ∗ b is in the quotient form. Since the division by 0 is undefined, the denominator b -1 must be nonzero.
It is clear that b −1 = 0 if b = 1. As 1∈Q, ∗ is not a binary operation on the whole of Q. However it can be found that by omitting 1 from Q, the output a ∗b exists in Q\{1}. Hence ∗ is a binary operation on Q\{1}.

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

~(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 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 *.

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.

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.

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) 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 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) 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.

Let (S, *) be an algebraic structure and a ∈ S.  Assume that the inverse of a exists in S. It is to be proved that the inverse of a is unique. The existence of inverse in S ensures the existence of the identity element e in S. 
Let a ∈ S. It is to be proved that the inverse a (if exists) is unique.
Suppose that a has two inverses, say a₁, a₂
Treating a₁ as an inverse of a gives \(a * a_{1}=a_{1} * a=e\)  ........(1)
Next treating a₂ as the inverse of a gives \(a * a_{2}=a_{2} * a=e\)  ........(2)
\(a_{1}=a_{1} * e=a_{1} *\left(a * a_{2}\right)=\left(a_{1} * a\right) * a_{2}=e * a_{2}=a_{2}(\text { by }(1) \text { and }(2))\)
So, a₁= a₂. Hence the inverse of a is unique which completes the proof.

Let (S, *) be an algebraic structure. Assume that the identity element of S exists in S .
It is to be proved that the identity element is unique. Suppose that e₁ and e₂ be any two identity elements of S .
First treat e₁ as the identity and e₂ as an arbitrary element of S.
Then by the existence of identity property \(e_{2} * e_{1}=e_{1} * e_{2}=e_{2}\)  ..........(1)
Interchanging the role of e₁ and e₂  \(e_{2}, e_{1} * e_{2}=e_{2} * e_{1}=e_{1}\)  ..........(2)
From (1) and (2), e₁ = e₂. Hence the identity element is unique which completes the proof.

The entries in the columns corresponding q ➝ p and ~p ➝ ~q are identical and hence they are equivalent.
q ➝ p ≡ ~p ➝ ~q
Hence proved

Truth Table for (~p ⟶ r) ∧ ( p ↔️ q)

​​​​​​Truth Table for ~(p ∧ ~q)

Truth Table for ​​​​​​~p ∧ ~q