All the results given in the paper hold true. In the proof of Theorem 1, change steps IV, V, and VI to
IV. for every $a,b,c \in T$ , add $(\langle a\rangle,\langle b \rangle,\langle c \rangle,\$) \rightarrow (\langle 0a \rangle,\langle 0b \rangle,\langle 0c \rangle,\S)$ to P ;
V. for every $a,b,c,d \in T$ , add $(Y, \langle 0a \rangle, Y, \langle 0b \rangle, Y, \langle 0c \rangle, \S) \rightarrow \# , \langle 0a \rangle, X, \langle 0b \rangle, Y, \langle 0c \rangle, \S)$ , $(\langle 0a\rangle, \langle 0b \rangle, \langle 0c \rangle, \S) \rightarrow (\langle 4a \rangle, \langle 1b \rangle, \langle 2c \rangle, \S)$ , $(\langle 4a \rangle, X, \langle 1b \rangle, Y, \langle 2c \rangle, \S) \rightarrow (\langle 4a \rangle, \# , \langle 1b \rangle, X, \langle 2c \rangle, \S)$ , $(\langle 4a \rangle, \langle 1b \rangle, \langle 2c \rangle, \langle d \rangle, \S \rightarrow (a, \langle 4b \rangle, \langle 1c \rangle, \langle 2d \rangle, \S)$ , $(\langle 4a \rangle, \langle 1b \rangle, \langle 2c \rangle, \S) \rightarrow (a, \langle 1b \rangle, \langle 3c \rangle, \S)$ , $(\langle 1a \rangle, X, \langle 3b \rangle, Y, \S) \rightarrow (\langle 1a \rangle, \# , \langle 3b \rangle, \# , \S)$ to P ;
VI. for every $a, b \in T$ , add $(\langle 1a \rangle, X, \langle 3b \rangle, \S) \rightarrow (a, \# , b, \# )$ to P .
