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## Here's the Solution to this Question

(a) By the definition of piece arrow-

$P\downarrow Q=$ ~$(P\lor Q)$

$P\downarrow Q$ =~$(P\lor P)$

We have derived that $P\downarrow P$ is logically equivalent with ~P

~$P=P\downarrow P$

(b)$(P\downarrow Q)\downarrow (P\downarrow Q)$ =(~($P\lor Q))\downarrow$ (~$(P\lor Q)$

$=(P\lor Q)\land (P\lor Q)\\ =P\lor Q$

(c)$(P\downarrow P)\downarrow (Q\downarrow Q)$ =(~($P\lor P))\downarrow$ (~($Q\lor Q))$

$=(P\lor P)\land (Q\lor Q)\\ =P\land Q$

d)

$P → Q\equiv \neg P \lor Q$

$\neg P\equiv P\downarrow P$

$P \lor Q \equiv \neg(P\downarrow Q)$

$\neg P \lor Q\equiv \neg(\neg P\downarrow Q)\equiv \neg((P\downarrow P)\downarrow Q)\equiv ((P\downarrow P)\downarrow Q)\downarrow ((P\downarrow P)\downarrow Q)$

$P → Q\equiv ((P\downarrow P)\downarrow Q)\downarrow ((P\downarrow P)\downarrow Q)$

e)

$P ↔ Q\equiv (P → Q) \land (Q → P)$

$P ↔ Q\equiv ((P\downarrow P)\downarrow Q)\downarrow ((P\downarrow P)\downarrow Q)\land((Q\downarrow Q)\downarrow P)\downarrow ((Q\downarrow Q)\downarrow P)\equiv$

$[((P\downarrow P)\downarrow Q)\downarrow ((P\downarrow P)\downarrow Q)\downarrow ((P\downarrow P)\downarrow Q)\downarrow ((P\downarrow P)\downarrow Q)]\downarrow$

$[((Q\downarrow Q)\downarrow P)\downarrow ((Q\downarrow Q)\downarrow P)\downarrow ((Q\downarrow Q)\downarrow P)\downarrow ((Q\downarrow Q)\downarrow P)]$