SMT公式的ANTLR语法

时间:2015-05-06 06:34:26

标签: parsing antlr grammar antlr4 smt

我正在尝试为SMT公式制作语法,这是我到目前为止所做的

grammar Z3input;

startRule : formulaList? EOF;

LEFT_PAREN : '(';
RIGHT_PAREN : ')';
COMMA : ',';
SEMICOLON : ';';

PLUS : '+';
MINUS : '-';
TIMES : '*';
DIVIDE : '/';

DIGIT : [0-9];
INTEGER : '0' | [1-9] DIGIT*;
FLOAT : DIGIT+ '.' DIGIT+;
NUMERICAL_LITERAL : FLOAT | INTEGER;
BOOLEAN_LITERAL : 'True' | 'False';
LITERAL : MINUS? NUMERICAL_LITERAL | BOOLEAN_LITERAL;

COMPARISON_OPERATOR : '>' | '<' | '>=' | '<=' | '!=' | '==';
WHITESPACE: [ \t\n\r]+ -> skip;
IDENTIFIER : [a-uw-zB-DF-Z]+ ([a-zA-Z0-9]? [a-uw-zB-DF-Z])*; // omits 'v', 'A', 'E' and cannot end in those characters

IMPLIES : '->' | '-->' | 'implies';
AND : '&' | 'and' | '^';
OR : 'or' | 'v' | '|';
NOT : '~' | '!' | 'not';
QUANTIFIER : 'A' | 'E' | 'forall' | 'exists';

formulaList : formula ( SEMICOLON formula )*; 
argumentList : expression ( COMMA expression )*; 

formula : formulaConjunction 
        | LEFT_PAREN formula RIGHT_PAREN OR LEFT_PAREN formulaConjunction RIGHT_PAREN
        | formula IMPLIES LEFT_PAREN formulaConjunction RIGHT_PAREN;

formulaConjunction : formulaNegation | formulaConjunction AND         formulaNegation;
formulaNegation : formulaAtom | NOT formulaNegation;
formulaAtom : BOOLEAN_LITERAL 
        | IDENTIFIER ( LEFT_PAREN argumentList? RIGHT_PAREN )?
        | QUANTIFIER '.' LEFT_PAREN formulaAtom RIGHT_PAREN
        | compareExpn;

expression : boolConjunction | expression OR boolConjunction;
boolConjunction : boolNegation | boolConjunction AND boolNegation;
boolNegation : compareExpn | NOT boolNegation;

compareExpn : arithExpn COMPARISON_OPERATOR arithExpn;
arithExpn : term | arithExpn PLUS term | arithExpn MINUS term;

term : factor | term TIMES factor | term DIVIDE factor;
factor : primary | MINUS factor;

primary : LITERAL 
        | IDENTIFIER ( LEFT_PAREN argumentList? RIGHT_PAREN )? 
        | LEFT_PAREN expression RIGHT_PAREN;

SMT公式是具有函数符号的一阶逻辑的公式(可以使用多个参数调用的标识符),变量,布尔文字的比较(即&#39; True&#39;或者&#39; False& #39;)或数字文字或函数调用或变量,算术运算符&#39; +&#39;,&#39; *&#39;,&#39; - &#39;和&#39; ; /&#39 ;.基本上这些公式是一些签名的一阶逻辑,为了我的目的,我选择这个签名作为理性理论。

我可以对'True ^ True'之类的内容进行正确的解释,但更复杂的事情,甚至包括'True | True',似乎总是会产生某些内容

... mismatched input '|' expecting {<EOF>, ';', IMPLIES, AND}

所以我想帮助纠正语法。为了记录,我宁愿保持语法运行时独立。

1 个答案:

答案 0 :(得分:1)

您的formula规则似乎导致此问题:LEFT_PAREN formula RIGHT_PAREN OR LEFT_PAREN formulaConjunction RIGHT_PAREN

这就是说语言只接受(FORMULA)|(CONJUNCTIVE)形式的公式。

而是为每个运算符指定优先级规则,并为每个优先级使用非终结符。例如,您的语法可能如下所示:

formula            : (QUANTIFIER IDENTIFIER '.')? formulaImplication;
formulaImplication : formulaConjunction (IMPLIES formula)?;
formulaConjunction : formulaDisjunction (AND formulaConjunction)?;
formulaDisjunction : formulaNegation (OR formulaDisjunction)?;
formulaNegation    : formulaAtom | NOT formulaNegation;
formulaAtom        : BOOLEAN_LITERAL | IDENTIFIER ( LEFT_PAREN argumentList? RIGHT_PAREN )? | LEFT_PAREN formula RIGHT_PAREN | compareExpn;

expression : boolConjunction | expression OR boolConjunction;
boolConjunction : boolNegation | boolConjunction AND boolNegation;
boolNegation : compareExpn | NOT boolNegation;

compareExpn : arithExpn COMPARISON_OPERATOR arithExpn;
arithExpn : term | arithExpn PLUS term | arithExpn MINUS term;

term : factor ((TIMES | DIVIDE) term)?;
factor : primary | MINUS factor;
primary : LITERAL | IDENTIFIER ( LEFT_PAREN argumentList? RIGHT_PAREN )? | LEFT_PAREN expression RIGHT_PAREN;