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If I know that the set of operators {∨, & , ¬} is functionally complete, how do I go about proving/disproving the functional completeness of the following set of operators?

a) $\{\vee,\neg\}$

b) $\{\to,\neg\}$

c) $\{\to\}$

I have looked at the answer here for (b) : Prove that the set {→, ¬} is functionally complete

Tian Vlasic
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user141834
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2 Answers2

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a)

$\{\vee, \neg\}$ is functionally complete. You only have to show that $\wedge$ (alias $\&$) is definable from $\{\vee, \neg\}$, because then you can express all the connectives of the complete set $\{\vee, \wedge, \neg\}$ (and the functions they induce). Like so: $$ p \wedge q \colon = \neg (\neg p \vee \neg q) $$

b)

$\{\to, \neg\}$ is functionally complete. By a), it suffices to show that $\vee$ is definable from $\{\to, \neg\}$. Thus: $$ p \vee q \colon = (\neg p \to q) $$

c)

$\{\to \}$ is not functionally complete. The proof is by induction on the complexity of propositional formulas in two variables. Let $2 = \{0,1\}$ be the set of truth values. We show that the always-false function $$F_{false} \colon (u,v) \mapsto 0 \colon 2 \times 2 \to 2$$ is not definable. Let $$ \begin{align} F_{\to} &\colon (u, v) \mapsto (\text{$1$ if $u\le v$ else $0$}) \colon 2 \times 2 \to 2 \\ \end{align} $$ $F_{\to}$ is just the truth table for implication "$\to$".

For any propositional formula $A = A(p,q)$ in variables $p, q$, we define the function $f_A \colon 2 \times 2 \to 2$ corresponding to A, inductively:

$$ \begin{align} f_p(u, v) &= u \tag{$pr_1$} \\ f_q(u, v) &= v \tag{$pr_1$} \\ f_{B\to C}(u,v) &= F_{\to}(f_B(u,v), f_C(u,v)) \tag{$Rule_{\to}$} \end{align} $$

Induction on formulas

Atomic formulas are $p$ and $q$. Clearly, neither $f_p$ nor $f_q$ is always false ($0$).

Now suppose $A$ is $B \to C$ with $B, C$ of lesser complexity (height, length). By induction hypothesis, neither $f_B$ nor $f_C$ is always false. Suppose $f_{B \to C}$ is always false: $$f_{B \to C}(u,v) = 0 \;\;\text{for all $u, v$.}$$ By ($Rule_{\to}$), this means: $$F_{\to}(f_B(u,v), f_C(u,v)) = 0 \;\;\text{for all $u, v$.}$$ By the definition of $F_{\to}$, this can only be so when $$ \begin{align} f_B(u,v) &= 1 \;\;\text{for all $u, v$, and} \tag{i} \\ f_C(u,v) &= 0 \;\;\text{for all $u, v$.} \tag{ii} \\ \end{align} $$ But by induction hypothesis, (ii) cannot be.

BrianO
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  • Is there any other way to prove the falsity of (iii) that does not involve induction? – user141834 Oct 16 '15 at 01:25
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    Hmm good question. I couldn't think of one, anyway, or I would have used it assuming it was simpler. And if there really is no other way to prove it, that's harder still to prove (lol). Whatever the approach, you have to show that some function can't be expressed by any formula built up from 2 variables and implication. Perhaps there's some other theorem we could appeal to that would get rid of the induction -- maybe something involving "parity" of the truth function; but I'd bet that the proof of that theorem would be a much heavier induction. Basically, when you want to prove something.. – BrianO Oct 16 '15 at 01:35
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    (cont) about all formulas built up in a standard way, the standard approach is by induction on the construction of the formulas. I'll search a bit, though, see what I can see. – BrianO Oct 16 '15 at 01:36
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    There is a theorem characterizing the functionally complete sets of connectives, but I doubted you could just cite it in your answer -- it's more high-powered. Here's what Emil Post proved, in the 1940s: https://en.wikipedia.org/wiki/Functional_completeness#Characterization_of_functional_completeness. This section (&probably the whole article) is also worth checking out: https://en.wikipedia.org/wiki/Truth_function#Functional_completeness – BrianO Oct 16 '15 at 01:46
  • I have come across this statement: – user141834 Oct 16 '15 at 02:42
  • https://en.wikipedia.org/wiki/Material_conditional If you look at the sentence just above the contents panel, you can see that they have expressed $p \to q$ in terms of &, OR and NOT. Doesn't that make it functionally complete? – user141834 Oct 16 '15 at 02:43
  • expressing $\to$ in terms of $\vee$: $\neg A \vee B$

    expressing $\to$ in terms of $\wedge$: $\neg(A \wedge \neg B)$

    Would this be correct?

     – user141834 Oct 16 '15 at 02:54
  • No: that's the "other direction". That just shows that you can express it using those other sets of connectives, which are known to be functionally complete :) Since e.g. "or" and "not" form a functionally complete sets, for example, of course you can express implication with them, because you can express every truth function with them. – BrianO Oct 16 '15 at 03:52
  • "But by induction hypothesis, (ii) cannot be." What are you referring to by (ii)? – user141834 Oct 16 '15 at 05:06
  • I numbered the two last lines (look in right margin). (ii) is the statement "$f_C(u,v) = 0$, all u,v". The induction hypothesis is that for any formula D that's smaller than $B\to C$, $f_D$ is not the always-false function. In particular, $C$ is a smaller formula than $B\to C$, but (ii) says that it IS always false -- or anyway, it would have to be if $f_{A\to B}$ were always false. Last sentence means: the induction hyp. rules out (ii), therefore $f_{A\to B}$ is not always false. // So, there's no way to represent the always-false function. You can represent always-true: $f_{p\to p}$ – BrianO Oct 16 '15 at 05:18
  • Also, I don't quite understand the meaning of: Let $2={0,1}$ be the set of truth values.

    Is there a particular reason why you used "2"?

     – user141834 Oct 16 '15 at 05:21
  • It's a convenient notation for a 2-element set, especially because in set theory it actually IS = {0,1} (the usual construction of the integers is: 0 = $\emptyset$, $n+1 = n \cup {n}$). If you prefer {F,T} or explicit {0,1}, go ahead and use it. 2 just looks neater. If you use {F,T} you'll have to rephrase the definition of $F_{\to}$ but no big deal (F < T). – BrianO Oct 16 '15 at 05:27
  • Would you show that ${ \lnot, \leftrightarrow }$ are not complete the same way? how would you do it? – boozi Mar 07 '22 at 09:23
  • @boozi Well, not really "the same" except in the broadest sense: that proof is also by structural induction on formulas, but the statement you prove inductively is quite different. There are several write-ups, search a little. (1) The question has been asked & answered here on SO: https://math.stackexchange.com/questions/556662/show-that-lnot-leftrightarrow-is-not-functional-complete. The answer by Bram28 (who answered this question too) is the best, but see my comment there on needed fixups. You may also find http://jdh.hamkins.org/the-hierarchy-of-logical-expressivity enlightening. – BrianO Mar 08 '22 at 20:45
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Here is a proof that $\{ \rightarrow \}$ is not complete:

We'll show by structural induction that for any expression $\phi(p,q)$ that is composed of any number of instances of variables $p$ and $q$, and any number of $\rightarrow$ connectives: $\phi(T,T)=T$.

Base: $\phi(p,q) = p$ and $\phi(p,q) = q$. In both cases we have $\phi(T,T)=T$

Step: Suppose $\phi(p,q) = \phi_1(p,q) \rightarrow \phi_2(p,q)$ and that (inductive hypothesis) $\phi_1(T,T)=T$ and $\phi_2(T,T)=T$. Then $\phi(T,T) = \phi_1(T,T) \rightarrow \phi_2(T,T) = T \rightarrow T = T$

We have proven that for no such expression $\phi(T,T)=F$, and hence $\{ \rightarrow \}$ is not complete.

Bram28
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