# Is this theorem about soundness or completeness?

$\def\True{\top}\def\False{\bot}$ In Kaye's math logic, $X$ is a set of propositional letters, and $BT(X)$ is the set of Boolean terms over $X$. There is a theorem about its valuation on the binary Boolean algebra $\{\True, \False \}$:

1. Why is it named "completeness theorem"? I think it is not about "completeness" but about "soundness", because completeness is from $\Sigma \vDash \True$ to $\Sigma \vdash \True$, while soundness is from $\Sigma \vdash \True$ to $\Sigma \vDash \True$:

2. In particular, I am not sure if "$\Sigma_0$ is consistent" means $\Sigma_0 \vdash \True$.

Thanks.

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Regarding consistency [see also your previous post], the answer is : NO. See page 65 : "we will say that $Σ$ is inconsistent if $Σ \vdash ⊥$, and consistent otherwise." Thus (properties of negation) : $\Sigma$ is consistent iff it does not prove "the false", i.e. iff $\Sigma \nvdash \bot$. –  Mauro ALLEGRANZA Jul 25 at 6:46
Regarding Completeness Th, see page 89 : "Theorem 7.13 (Completeness Theorem, second form) : Let $X$ be a set, and suppose that $Σ ⊆ BT(X), τ ∈ BT(X)$ with $Σ \vDash τ$. Then $Σ \vdash τ$, i.e. there is a formal derivation of $τ$ from $Σ$", which is exactly what you expected. It is proved as a corollary of the first form of the Completeness Theorem. –  Mauro ALLEGRANZA Jul 25 at 8:10

That trivially implies that if some sentences $\Gamma, \neg\phi$ can't all be true together, then $\Gamma, \neg\phi$ aren't consistent.
In other words, if $\Gamma \vDash \phi$ [= no valuation makes all $\Gamma$ true and $\neg\phi$ true as well] then $\Gamma \vdash \phi$ [for if $\Gamma, \neg\phi$ are inconsistent, then by reductio $\Gamma \vdash \phi$].