Conditional lexer rules in Antlr4 -

given grammar:

grammar colontest;  main        : statement* eof; statement   : num_literal expression semicolon; expression  : primary (mult_op primary)*; primary     : word+;  num_literal : [0-9]+; semicolon   : ';'; mult_op     : '*' | '/'; // | ':'; word        : ('a'..'z' | 'a'..'z')+; ws          : [ \t\r\n]+ -> skip; colon       : ':' -> skip; 

and input:

1 : statement; 2 : splitted statement   : 2 lines; 3 : * b / c; 4 : * b : c; 

in 4'th line second colon skipped because of "colon" lexer rule. need colon, because part of language (lets should part of mult_op keyword too). how achieve this?

edit 1:

after deleting colon : ':' -> skip; , inserting:

statement   : num_literal ':' expression (':' expression)* semicolon; 

the tree looks this:

the desired tree should this:

edit 2:

how this? implicit tokens defined - works.

grammar multiline;  main        : statement* eof; statement   : num_literal ':' expression semicolon; expression  : primary ((mult_op|':') primary)*; primary     : word+;  num_literal : [0-9]+; semicolon   : ';'; mult_op     : '*' | '/'; word        : ('a'..'z' | 'a'..'z')+; ws2         : [\r\n]+ [ \t]+ ':' -> skip; // removes not needed colons ws          : [ \t\r\n]+ -> skip; 

can improve code in other way?

tokens such : should not skipped. then, alter statement rule be:

statement : num_literal expression (':' expression)* semicolon;