\[ \newcommand{\tr}{\Rightarrow} \newcommand{\trs}{\tr^{\!\ast}} \newcommand{\rlnm}[1]{\mathsf{(#1)}} \newcommand{\rred}[1]{\xrightarrow{#1}} \newcommand{\rreds}[1]{\mathrel{\xrightarrow{#1}\!\!^*}} \newcommand{\cl}{\mathsf{Cl}} \newcommand{\pow}{\mathcal{P}} \newcommand{\matches}{\mathrel{\mathsf{matches}}} \newcommand{\kw}[1]{\mathsf{#1}} \newcommand{\andop}{\mathrel{\&\!\&}} \newcommand{\orop}{\parallel} \newcommand{\ff}{\mathsf{false}} \newcommand{\tt}{\mathsf{true}} \newcommand{\abra}[1]{\langle #1 \rangle} \newcommand{\bnfnt}[1]{\abra{\small \textsf{#1}}} \newcommand{\llbracket}{[\![} \newcommand{\rrbracket}{]\!]} \newcommand{\first}{\mathsf{First}} \newcommand{\nullable}{\mathsf{Nullable}} \newcommand{\follow}{\mathsf{Follow}} \newcommand{\tm}[1]{\mathsf{#1}} \newcommand{\nt}[1]{\mathit{#1}} \newcommand{\Coloneqq}{::=} \newcommand{\abs}[1]{|#1|} \]

The precedence of an operator is a convention used when inserting implicit parentheses (building abstract syntax trees). An operator $\otimes$ is said to be of higher precedence than an operator $\oplus$ just if every chain $u \otimes v \oplus w$ is considered syntactically identical to (that is, having the same abstract syntax as) $(u \otimes v) \oplus w$; and $u \oplus v \otimes w$ is considered syntactically identical to $u \oplus (v \otimes w)$.