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first time here!

I have been trying to wrap my head around this assignment from my book:

"Certain of the logical connectives can be defined in terms of others. For example, $(p\to q)$ can be defined as an abbreviation for $(\lnot p \lor q)$, since the two statements are logically equivalent. Hence, all formulas containing the connective $\to$ could be replaced by formulas containing $\lor$ and $\lnot$

  • Define $\to$ in terms of $\land$ and $\lnot$.

  • Show how the five connectives could be reduced to just $\land$ and $\lnot$.

From what I understood, the conditional is equivalent to the contrapositive and the converse is equivalent to the inverse.

None of these formulas have a conjunction and negation in them. I find it difficult to understand what they want me to do. Are they asking me to say how it is not possible to do so? Or is it possible?

Can someone guide me into the process of this? I feel like I'm missing something here and maybe you could give me that push I need.

Thank you!

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  • $\begingroup$ Hint: Look at the truth tables and build the solution from there. $\endgroup$ – Luke Feb 12 '20 at 15:21
  • $\begingroup$ Hint: When does an implication become false? $\endgroup$ – lemontree Feb 12 '20 at 15:21
  • $\begingroup$ Just use DeMorgan: \begin{align*}p\to q &\equiv \neg p \lor q \\ &\equiv \neg (p \land \neg q)\end{align*}. $\endgroup$ – Adrian Keister Feb 12 '20 at 15:22
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$$p\to q \equiv \lnot p \lor q $$ $$ \equiv \lnot p \lor \lnot(\lnot q)\tag{Double Negation}$$ $$\equiv \lnot (p \land \lnot q)\tag {Demorgan's.}$$

Literally, $p\to q$ holds whenever it is not the case that $p$ is true and $q$ is false.

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