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.

~(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 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.

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)