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Which of the following are functionally complete sets of sentential connectives? (if any?) {∧,∨} {∧,¬} {→,¬}

I know that {¬,∨,∧} {¬,∨} {¬,∧} {¬,→} {f,→} functionally complete sets of sentential connectives. Does the order of connectives within the set matter? I.e. Since {¬,∧} and {¬,→} are functionally complete sets, does this then mean that {∧,¬} {→,¬} also are? If not, why not?

BrianO
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JB28
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    Do you mean, "which of these sets are functionally complete sets of sentential connectives?" Because "sentential connective" is usually used to describe the individual connectves, e.g. "$\wedge$ is a sentential connective." – Noah Schweber Oct 28 '15 at 22:53
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    The order in which elements in any set are listed doesn't matter. ${a,b} = {b,a}$ and so on. – BrianO Oct 28 '15 at 22:55
  • Your question isn't exactly a duplicate of this question, but you may get some mileage from the answer there. You already know that the last two sets of connectives are functionally complete! So the only remaining part of the puzzle is ${\wedge, \vee}$. – BrianO Oct 28 '15 at 22:58

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You already know that the last two sets of connectives are functionally complete! $\{\neg, \to\} = \{\to, \neg\}$, for example.

The only remaining part of the puzzle is {∧,∨}. This set of connectives is not functionally complete. To show this, it's enough to show that some truth function can't be expressed using only these two connectives.

First, a few conventions and definitions. Assume that the symbols of the language $\mathscr{L}$ are the connectives, parentheses and propositional variables $p_0, p_1, \dots, p_n, \dots$, and that formulas are built up in the usual way. The variables of a formula $\varphi$ are all the $p_i$ occurring in $\varphi$.

Let $\mathsf{TV} = \{0,1\}$ be the set of truth values. Given an integer $n$, call the members of $\mathsf{TV}^n$ *valuations: a valuation $v \in \mathsf{TV}^n$ can be thought of as assigning the value $v_i$ to the $i$th variable $p_i$. $n$ is large enough for a formula $\varphi$ if for every variable $p_i$ occurring in $\varphi$, $i < n$. A valuation $v$ is long enough for $\varphi$ if its length is large enough.

The value of a formula under a valuation that's long enough for it is defined by induction on the construction of formulas:

  • $Val(p_i, v) = v_i$
  • $Val((\varphi \wedge \psi), v) = Val(\varphi, v) \cdot Val(\psi, v)$
  • $Val((\varphi \vee \psi), v) = \max(Val(\varphi, v), Val(\psi, v))$

Finally, an $n$-ary truth function is a function $\mathsf{TV}^n \to \mathsf{TV}$, i.e. a mapping from $n$-ary valuations to truth values. Every formula $\varphi$ induces truth functions $f_{\varphi} \colon \mathsf{TV}^n \to \mathsf{TV}$ for every large enough $n$:

  • $f_{p_i}(v) = v_i$
  • $f_{(\varphi \wedge \psi)}(v) = f_{\varphi}(v) \cdot f_{\psi}(v)$
  • $f_{(\varphi \vee \psi)}(v) = \max(f_{\varphi}(v), f_{\psi}(v))$

To show that $\{\wedge, \vee\}$ is not functionally complete, show that:

For any formula $\varphi$ of $\mathscr{L}$, if $v = (1)_{i<n}$ is a constantly-$1$ valuation that's long enough for $\varphi$, then $f_{\varphi}(v) = 1$.

The proof is by induction on the construction of formulas, and follows easily from the inductive definition of $f_{\varphi}$. From this, the result is immediate:

If $g \colon \mathsf{TV}^n \to \mathsf{TV}$ is a truth function taking the value $0$ on the constantly-$1$ valuation $(1)_{i<n}$, then there is no formula $\varphi$ of $\mathscr{L}$ such that $g = f_{\varphi}$. That is, $g$ is not represented by any formula of $\mathscr{L}$.

BrianO
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