(a) By the definition of piece arrow- P\downarrow Q=P↓Q= ~(P\lor Q)(P∨Q) P\downarrow QP↓Q =~(P\lor P)(P∨P) We have derived that P\downarrow PP↓P is logically equivalent with ~P ~P=P\downarrow PP=P↓P (b)(P\downarrow Q)\downarrow (P\downarrow Q)(P↓Q)↓(P↓Q) =(~(P\lor Q))\downarrowP∨Q))↓ (~(P\lor Q)(P∨Q) =(P\lor Q)\land (P\lor Q)\\ =P\lor Q=(P∨Q)∧(P∨Q) =P∨Q (c)(P\downarrow P)\downarrow (Q\downarrow Q)(P↓P)↓(Q↓Q) =(~(P\lor P))\downarrowP∨P))↓ (~(Q\lor Q))Q∨Q)) =(P\lor P)\land (Q\lor Q)\\ =P\land Q=(P∨P)∧(Q∨Q) =P∧Q d) P → Q\equiv \neg P \lor QP→Q≡¬P∨Q \neg P\equiv P\downarrow P¬P≡P↓P P \lor Q \equiv \neg(P\downarrow Q)P∨Q≡¬(P↓Q) \neg P \lor Q\equiv \neg(\neg P\downarrow Q)\equiv \neg((P\downarrow P)\downarrow Q)\equiv ((P\downarrow P)\downarrow Q)\downarrow ((P\downarrow P)\downarrow Q)¬P∨Q≡¬(¬P↓Q)≡¬((P↓P)↓Q)≡((P↓P)↓Q)↓((P↓P)↓Q) P → Q\equiv ((P\downarrow P)\downarrow Q)\downarrow ((P\downarrow P)\downarrow Q)P→Q≡((P↓P)↓Q)↓((P↓P)↓Q) e) P ↔ Q\equiv (P → Q) \land (Q → P)P↔Q≡(P→Q)∧(Q→P) P ↔ Q\equiv ((P\downarrow P)\downarrow Q)\downarrow ((P\downarrow P)\downarrow Q)\land((Q\downarrow Q)\downarrow P)\downarrow ((Q\downarrow Q)\downarrow P)\equivP↔Q≡((P↓P)↓Q)↓((P↓P)↓Q)∧((Q↓Q)↓P)↓((Q↓Q)↓P)≡ [((P\downarrow P)\downarrow Q)\downarrow ((P\downarrow P)\downarrow Q)\downarrow ((P\downarrow P)\downarrow Q)\downarrow ((P\downarrow P)\downarrow Q)]\downarrow[((P↓P)↓Q)↓((P↓P)↓Q)↓((P↓P)↓Q)↓((P↓P)↓Q)]↓ [((Q\downarrow Q)\downarrow P)\downarrow ((Q\downarrow Q)\downarrow P)\downarrow ((Q\downarrow Q)\downarrow P)\downarrow ((Q\downarrow Q)\downarrow P)][((Q↓Q)↓P)↓((Q↓Q)↓P)↓((Q↓Q)↓P)↓((Q↓Q)↓P)]

Solution:

injective function Definition:

A function f: A → B is said to be a one - one function or injective mapping if different elements of A have different f images in B. A function f is injective if and only if whenever f(x) = f(y), x = y. Example: f(x) = x + 9 from the set of real number R to R is an injective function. When x = 3,then :f(x) = 12,when f(y) = 8,the value of y can only be 3,so x = y.

(ii) surjective function Definition: If the function f:A→B is such that each element in B (co - domain) is the ‘f’ image of at least one element in A , then we say that f is a function of A ‘onto’ B .Thus f: A→B is surjective if, for all b ∈ B, there are some a ∈ A such that f(a) = b.

Example: The function f(x) = 2x from the set of natural numbers N to the set of non negative even numbers is a surjective function.

(iii) bijective function Definition: A function f (from set A to B) is bijective if, for every y in B, there is exactly one x in A such that f(x) = y. Alternatively, f is bijective if it is a one - to - one correspondence between those sets, in other words, both injective and surjective.

Example: If f(x) = x2,from the set of positive real numbers to positive real numbers is both injective and surjective. Thus, it is a bijective function.

$f:N\rightarrow N \\f(x) = x^2$

$\begin{aligned} &\begin{array}{l} x_{1}, x_{2} \in N \\ f\left(x_{1}\right)=f\left(x_{2}\right) \Rightarrow x_{1}^{2}=x_{2}^{2} \end{array} \\ &\Rightarrow x_{1}^{2}-x_{2}^{2}=0 \\ &\Rightarrow\left(x_{1}+x_{2}\right)\left(x_{1}-x_{2}\right)=0 \\ &\Rightarrow x_{1}=x_{2}\left\{\begin{array}{c} x_{1}+x_{2} \neq 0 \\ \text { as } x_{1}, x_{2} \in N \end{array}\right\} \end{aligned}$

hence f is injective, for some elements like, 2,3 etc has no preimage in N such that f(x)=2 hence not surjective.