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Find a compound proposition logically equivalent to $p \rightarrow q$ using only the logical operator $\downarrow$?
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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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