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We know : $P \rightarrow Q $ $\equiv $ $P'+ Q \equiv \overline{P \overline{Q}} $

And $P ↓ Q \equiv \overline{P} \,\,\overline{Q}$

So, $\overline{P} ↓ Q \equiv P \overline{Q}$

$P \rightarrow Q $ $\equiv$ $((P↓P)  ↓ Q ) ↓ ((P↓P)  ↓ Q) $
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