The entries in column (4) and (6) are identical  ~(p↔️q) ≡ p↔️~q

The entries in column (3) and column (4) are not identical.

The entries in the columns corresponding to p ↔ q and ( p ⟶ q) ∧ (q ⟶ p) are identical and hence they are equivalent

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

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

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)]

Given A = {a, b, c, d} and * is defined as follows.

1) Commulative Property:
\(a * b=c\)  but  \(b \times a=d \Rightarrow a * b \neq b * a\)
\(a * c=b\) but \(c * a=c \Rightarrow a * c \neq c * a\) 
\(\therefore\) \(\text { * }\) is not commutative on A..
2) Associative Property: 
\((a+b) * c=c * c=a\) 
\(a *(b * c)=a * b=c\) 
\(\therefore(a * b) * c \neq a *(b * c)\) 
\(\text { * }\) is not associative on A.

(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.

(i) If 6 + 2 = 5, then the milk is white.
Let p: 6 + 2 = 5 (F)
q: Milk is white (T)  
p ➝ q is having the truth value T
(ii) China is in Europe or \(\sqrt3\) is an integer.
p: China is in Europe (F)
q: \(\sqrt3\) is an integer (F)
p v q is having the truth value (F).
(iii) It is not true time 5 + 5 = 9 or Earth is a planet.
Let P: 5 + 5 = 9 is not true (T)
q: Earth is a planet (T
~p ∨ q is having the truth value T
(iv) 11 is a prime number and all the sides of a rectangle are equal.
p:11 is a prime number (T)
q: Allthe sides of arectangle areequal (F)
p _^ q is having the truth value F

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.

A = {a+\(\sqrt5\) b:a,b ∈ z}
Let C = a+\(\sqrt5\) b
B = c+\(\sqrt5\)d∈A
where a, b, c, d ∈ Z
[∵ ac + 5bd∈Z and ad+bc ∈Z]
∴ B = (a+\(\sqrt5\)b).(c+\(\sqrt5\)d)
= ac+\(\sqrt5\)ad+cb\(\sqrt5\) + 5bd
= (ac+5bd)+\(\sqrt5\)(ad+bc)∈A
∴ C.B ∈A∀ a, b, c, d∈Z
∴ Usual multiplicaition is a binary operation on.

Given a*b = a + b + ab -7, ∀ a,b ∈R
If a ∈R, b∈R then ab ∈ R
(a*b) = a +b+ ab - 7 ∈R
For example, let 1, 2 ∈ R
(1*2) = 1+2+(1)(2)-7
= 2 ∈ R
* a binary operation on R

[Here a = 3, b = \(\frac{-7}{15}\)]
\(=3-\frac { 7 }{ 15 } -\frac { 21 }{ 15 } -7\)
\(\therefore 3*\left( \frac { -7 }{ 15 } \right) =\frac { -88 }{ 15 } \)

Given m*n = mⁿ + n^m ∀m, n∈Z
Let us take - 2, 2∈Z
Consider m = -3, n = 2
(m * n) = (-2 * 2) = (-2)² + (2)^-2
\(=4+\frac { 1 }{ 4 } =\frac { 17 }{ 4 } \notin Z\)
∴ * is not a binary operation on Z

The entries in the columns corresponding to p → q and ㄱp ν q are identical and hence they are equivalent.

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