Consistency can be defined in either semantic or syntactic terms.
From a syntactical point of view, considering a proof system and the corresponding relation of derivability ($\vdash$), a set of sentences $\Sigma$ is consistent if it does not contain a contradiction.
Thus, $\Sigma$ is inconsistent if $\Sigma \vdash P \land \lnot P$, for some formula $P$.
In some treatments of logic, the logical constant $\bot$ is used, representing a proposition that is always false, i.e. a contradiction.
Thus, inconsistency of $\Sigma$ can be equivalently formulated as : $\Sigma \vdash \bot$.
From a semantical point of view, a set $\Sigma$ of sentences is consistent if and only if it has a model, i.e., there exists an interpretation under which all formulas in the set are true.
In this case, we say also that the set is satisfiable.
To say that a proof system is sound means that :
only valid formulas can be derived with it. In symbols : $\text { if } \vdash \varphi, \text { then } \vDash \varphi$.
More generally : $\text { if } \Gamma \vdash \varphi, \text { then } \Gamma \vDash \varphi$.
Thus, soundness implies consistency, because $\bot$ is not valid.
(Semantic) completeness of a proof system is the "twin" property of soundness. It means that :
all valid formulas of the calculus are provable. In symbols : $\text { if } \vDash \varphi, \text { then } \vdash \varphi$.
So-called Strong completeness formalize the concept of logical consequence.
A proof system is strongly complete iff for every set of formulas $Γ$, any formula that semantically follows from $Γ$ is derivable from $Γ$. That is: $\text { if } \Gamma \vDash \varphi, \text { then } \Gamma \vdash \varphi$.
In classical logic, where Ex falso holds, an inconsistent set of sentences is obviously unsound but trivially complete : being inconsistent, it proves every formula, and thus also the valid ones.
Strictly related to completeness is the Model Existence Theorem :
If a set $\Sigma$ of sentences is consistent, then $\Sigma$ is satisfiable (i.e. it has a model).
From Model Existence Theorem, strong completeness follows :
(i) $\text {if } \Gamma \nvdash \varphi, \text { then } \Gamma \cup \{ \lnot \varphi \} \text { is consistent}$.
Thus,
(ii) $\Gamma \cup \{ \lnot \varphi \} \text { has a model}$.
This means that
(iii) $\Gamma \nvDash \varphi$.