Functional completeness of ¬ and ∧

Asked Jan 23 '21 at 07:23

Active Jan 23 '21 at 07:23

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I need to show that ¬ and ∧ are a functionally complete set of logical operators, but I'm completely stumped on how to approach proving that I can replace p → q and p ↔ q with these two logical operators.

I'm a complete beginner at this, so I appreciate any help I can get

asked Jan 23 '21 at 07:23

amy

2

Once you get $\lor$, it’s easy: $p\to q$ is equivalent to $\neg p\lor q$, and $p\leftrightarrow q$ is equivalent to $(p\to q)\land(q\to p)$. – Brian M. Scott Jan 23 '21 at 07:25
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I got stuck at ¬p ∨ q and wasn't sure where to go from there, so I assumed I was doing something wrong. Where do I go from there? – amy Jan 23 '21 at 07:32
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Have you figured out how to get $p\lor q$? – Brian M. Scott Jan 23 '21 at 07:33
No, I don't know how to get rid of the negative on p – amy Jan 23 '21 at 07:37
That’s not what I asked. Forget about $\neg p\lor q$ for a moment; how do you get $p\lor q$? – Brian M. Scott Jan 23 '21 at 07:40
By doing ¬p⇒q right? – amy Jan 23 '21 at 07:42
No: you don’t yet have implication. How do you get $p\lor q$ using only the connectives $\neg$ and $\land$? That’s the first step to getting everything. – Brian M. Scott Jan 23 '21 at 07:44
I have to shut down for the night, but I’ll leave a hint: start with the De Morgan law that $\neg(p\lor q)$ is equivalent to $\neg p\land\neg q$, and tinker with it a bit. – Brian M. Scott Jan 23 '21 at 07:52

Functional completeness of ¬ and ∧

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