Prove that the set {→, ¬} is functionally complete

Question

I am not sure how to do this question. I have looked at some of the other similar questions but to no avail

I know that for a set of operators to be functionally complete, the set can be used to express all possible truth tables by combining members of the set into a Boolean expression

Do you know of another functionally complete set? Can you build equivalents to all the operations there with $\to$ and $\lnot$? — Arthur, Oct 15 '15 at 11:10

Stefan Hante · Accepted Answer · 2015-10-15T11:38:57.537

5

The implication $\to$ is defined by $$ a\to b \equiv \neg a \vee b. $$

This means, that $$ \neg a \to b \equiv a \vee b$$ and thus, you can express logical or using $\to$ and $\neg$. Furthermore you know de Morgans rules and you have $$ \neg(\neg a \vee \neg b) \equiv \neg\neg a\wedge \neg\neg b = a\wedge b.$$

Thus you can express all logical operators using $\to$ and $\neg$. The set of these is known to be functionally complete.

edited Oct 15 '15 at 11:38

answered Oct 15 '15 at 11:11

Stefan Hante

2,726

what about the functional completeness of {∨, ¬}? Would this be proven in the same manner? – bawse Oct 15 '15 at 11:19
Absolutely, yes. – Stefan Hante Oct 15 '15 at 11:25
Would you be able to give me a hint on how to start the proof for that? – bawse Oct 15 '15 at 11:26
You can very easily get $\to$ using its definition and for $\wedge$ you can do the same trick as I did in my answer. – Stefan Hante Oct 15 '15 at 11:31
Why would you need to get $\to$? The set I am referring to is {∨, ¬} – bawse Oct 15 '15 at 11:32
In my answer we constructed $\vee$ and $\wedge$ from $\neg$ and $\to$. If you start with $\vee$ and $\neg$, you have to construct $\to$ and $\wedge$. – Stefan Hante Oct 15 '15 at 11:40
What if I want to prove it in terms of the functionally complete set {∨, & , ¬}? – bawse Oct 15 '15 at 11:42
Also, I was looking at the operator $\to$ and was wondering if a set consisting of that operator only is functionally complete too? – bawse Oct 16 '15 at 02:40

score 2 · Answer 2 · answered Oct 16 '15 at 08:20

2

No, $\to$ alone is not complete because you cannot get to negation, for example.

answered Oct 16 '15 at 08:20

Stefan Hante

2,726

Prove that the set {→, ¬} is functionally complete

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