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如何为 FOL with Equality 添加运算符优先级到 lark 文法?

[英]How do I add operator precedence to a lark grammar for FOL with Equality?

How do I modify this grammar so it matches parenthesis that are further away?我如何修改此语法以匹配更远的括号?

?wff: compound_wff
?compound_wff: biconditional_wff
?biconditional_wff: conditional_wff (SPACE? BICONDITIONAL_SYMBOL SPACE? biconditional_wff)*
?conditional_wff: disjunctive_wff (SPACE? CONDITIONAL_SYMBOL SPACE? conditional_wff)*
?disjunctive_wff: conjunctive_wff (SPACE? DISJUNCTION_SYMBOL SPACE? disjunctive_wff)*
?conjunctive_wff: negated_wff (SPACE? CONJUNCTION_SYMBOL SPACE? conjunctive_wff)*
?negated_wff: (NEGATION_SYMBOL SPACE)* atomic_wff
?atomic_wff: predicate
           | term EQUAL_TO term
           | quantified_wff* LEFT_PARENTHESIS SPACE? wff SPACE? RIGHT_PARENTHESIS
?term: function
    | NAME
    | VARIABLE
?predicate: PREDICATE_NAME [LEFT_PARENTHESIS term (COMMA term)* RIGHT_PARENTHESIS]
?function: FUNCTION_NAME LEFT_PARENTHESIS term (COMMA SPACE? term)* RIGHT_PARENTHESIS
?quantified_wff: curly_quantifiers | quantifiers
?curly_quantifiers: quantifier_symbol LEFT_CURLY_BRACE VARIABLE (COMMA SPACE? VARIABLE)* RIGHT_CURLY_BRACE SPACE?
?quantifiers: quantifier_symbol SPACE? VARIABLE (COMMA SPACE? VARIABLE)* SPACE?

SPACE: /\s+/
COMMA: ","
EQUAL_TO: "="
LEFT_PARENTHESIS: "("
RIGHT_PARENTHESIS: ")"
LEFT_CURLY_BRACE: "{"
RIGHT_CURLY_BRACE: "}"
quantifier_symbol: UNIVERSAL_QUANTIFIER_SYMBOL | EXISTENTIAL_QUANTIFIER_SYMBOL
UNIVERSAL_QUANTIFIER_SYMBOL: "\\forall" | "∀"
EXISTENTIAL_QUANTIFIER_SYMBOL: "\\exists" | "∃"
NAME: /[a-t]/ | /[a-t]_[1-9]\d*/
VARIABLE: /[u-z]/ | /[u-z]_[1-9]\d*/
PREDICATE_NAME: /[A-HJ-Z]/ | /[A-HJ-Z]_[1-9]\d*/
FUNCTION_NAME: /[a-z]/ | /[a-z]_[1-9]\d*/
NEGATION_SYMBOL: "\\neg" | "\\lnot" | "¬"
CONJUNCTION_SYMBOL: "\\wedge" | "\\land" | "∧"
DISJUNCTION_SYMBOL: "\\vee" | "\\lor" | "∨"
CONDITIONAL_SYMBOL: "\\rightarrow" | "\\Rightarrow" | "\\Longrightarrow" | "\\implies" | "→" | "⇒"
BICONDITIONAL_SYMBOL: "\\leftrightarrow" | "\\iff" | "↔" | "⇔"

I'm trying to parse this using my grammar:我正在尝试使用我的语法来解析它:

方程

Which in LaTeX is:其中 LaTeX 是:

\exists{x} \forall{y} (P(f(x, y)) \vee \forall{z}(V(z) \iff \neg R(a) \wedge B(a)))

I followed thecalculator example and modified my original grammar to add operator precedence which resulted in this.我遵循了计算器示例并修改了我的原始语法以添加导致此结果的运算符优先级。 But it's no longer accepting the input string.但它不再接受输入字符串。

I'm getting this error:我收到此错误:

lark.exceptions.UnexpectedCharacters: No terminal matches '\' in the current parser context, at line 1 col 35

\exists{x} \forall{y} (P(f(x, y)) \vee \forall{z}(V(z) \iff \neg R(a) \wed
                                  ^
Expected one of: 
    * RIGHT_PARENTHESIS

Ideally I want to force requiring parenthesis wherever possible except in the case of in front of a negated atomic_wff without parenthesis.理想情况下,我想尽可能强制要求括号,除非在没有括号的否定atomic_wff前面。 This is to make sure only one parse tree is produced even on explicit ambiguity setting.这是为了确保即使在明确的歧义设置下也只生成一个解析树。 How do I resolve this issue?我该如何解决这个问题?

Edit 1编辑 1

I want to clarify that operator precedence for the same operators should be right associative.我想澄清的是,相同运算符的运算符优先级应该是右结合的。 So P(a) ∧ Q(a) ∧ R(a) will resolve as P(a) ∧ (Q(a) ∧ R(a))所以P(a) ∧ Q(a) ∧ R(a)将解析为P(a) ∧ (Q(a) ∧ R(a))

Edit 2编辑 2

I have made the lark grammar easier to debug using proper terminals.我已经使百灵鸟语法更容易使用适当的终端进行调试。 It now parses the long latex equation, but is still ambiguous for simpler inputs like P(a) ∧ Q(a) ∧ R(a) which produces two parse trees.它现在解析长 latex 方程,但对于更简单的输入仍然是模棱两可的,例如P(a) ∧ Q(a) ∧ R(a)会生成两个解析树。 I still don't know what I am doing wrong.我仍然不知道我做错了什么。 The grammar is doing right recursion and still failing to produce a single parse tree.语法正在执行正确的递归,但仍然无法生成单个解析树。

Edit 3编辑 3

My latest attempted solution works in every case except for giving negation a higher precedence.我最近尝试的解决方案在每种情况下都有效,除了给予否定更高的优先级。 Any idea?任何的想法?

?wff: compound_wff
compound_wff: biconditional_wff
biconditional_wff: conditional_wff (SPACE? BICONDITIONAL_SYMBOL SPACE? conditional_wff)*
conditional_wff: disjunctive_wff (SPACE? CONDITIONAL_SYMBOL SPACE? disjunctive_wff)*
disjunctive_wff: conjunctive_wff (SPACE? DISJUNCTION_SYMBOL SPACE? conjunctive_wff)*
conjunctive_wff: negated_wff (SPACE? CONJUNCTION_SYMBOL SPACE? negated_wff)*
negated_wff: (NEGATION_SYMBOL SPACE?)* atomic_wff
atomic_wff: predicate
          | term EQUAL_TO term
          | quantified_wff* LEFT_PARENTHESIS SPACE? wff SPACE? RIGHT_PARENTHESIS
term: function
    | NAME
    | VARIABLE
predicate: PREDICATE_NAME [LEFT_PARENTHESIS term (COMMA term)* RIGHT_PARENTHESIS]
function: FUNCTION_NAME LEFT_PARENTHESIS term (COMMA term)* RIGHT_PARENTHESIS
quantified_wff: curly_quantifiers | quantifiers
curly_quantifiers: quantifier_symbol LEFT_CURLY_BRACE VARIABLE (COMMA SPACE? VARIABLE)* RIGHT_CURLY_BRACE SPACE?
quantifiers: quantifier_symbol SPACE? VARIABLE (COMMA SPACE? VARIABLE)* SPACE?

SPACE: /\s+/
COMMA: /,\s*/
EQUAL_TO: /\s*=\s*/
LEFT_PARENTHESIS: "("
RIGHT_PARENTHESIS: ")"
LEFT_CURLY_BRACE: "{"
RIGHT_CURLY_BRACE: "}"
quantifier_symbol: UNIVERSAL_QUANTIFIER_SYMBOL | EXISTENTIAL_QUANTIFIER_SYMBOL
UNIVERSAL_QUANTIFIER_SYMBOL: "\\forall" | "∀"
EXISTENTIAL_QUANTIFIER_SYMBOL: "\\exists" | "∃"
NAME: /[a-t]/ | /[a-t]_[1-9]\d*/
VARIABLE: /[u-z]/ | /[u-z]_[1-9]\d*/
PREDICATE_NAME: /[A-HJ-Z]/ | /[A-HJ-Z]_[1-9]\d*/
FUNCTION_NAME: /[a-z]/ | /[a-z]_[1-9]\d*/
NEGATION_SYMBOL: "\\neg" | "\\lnot" | "¬"
CONJUNCTION_SYMBOL: "\\wedge" | "\\land" | "∧"
DISJUNCTION_SYMBOL: "\\vee" | "\\lor" | "∨"
CONDITIONAL_SYMBOL: "\\rightarrow" | "\\Rightarrow" | "\\Longrightarrow" | "\\implies" | "→" | "⇒"
BICONDITIONAL_SYMBOL: "\\leftrightarrow" | "\\iff" | "↔" | "⇔"

%ignore SPACE
%ignore COMMA
%ignore LEFT_CURLY_BRACE
%ignore RIGHT_CURLY_BRACE

So it fails to to produce a single parse tree for ¬P(a) ∧ Q(b) .因此它无法为¬P(a) ∧ Q(b)生成单个解析树。

The issue was each wff rule was not matching the antecedent rule in it's parts.问题是每个 wff 规则与其部分的先行规则不匹配。 So:所以:

?biconditional_wff: conditional_wff (SPACE? BICONDITIONAL_SYMBOL SPACE? biconditional_wff)*

Should be changed to:应改为:

?biconditional_wff: conditional_wff (SPACE? BICONDITIONAL_SYMBOL SPACE? conditional_wff)*

The solution to this issue is to have a grammar like this:这个问题的解决方案是使用这样的语法:

?wff: compound_wff
compound_wff: biconditional_wff
biconditional_wff: conditional_wff (SPACE? BICONDITIONAL_SYMBOL SPACE? conditional_wff)*
conditional_wff: disjunctive_wff (SPACE? CONDITIONAL_SYMBOL SPACE? disjunctive_wff)*
disjunctive_wff: conjunctive_wff (SPACE? DISJUNCTION_SYMBOL SPACE? conjunctive_wff)*
conjunctive_wff: negated_wff (SPACE? CONJUNCTION_SYMBOL SPACE? negated_wff)*
negated_wff: (NEGATION_SYMBOL SPACE?)* atomic_wff
atomic_wff: predicate
          | term EQUAL_TO term
          | quantified_wff* LEFT_PARENTHESIS SPACE? wff SPACE? RIGHT_PARENTHESIS
term: function
    | NAME
    | VARIABLE
predicate: PREDICATE_NAME [LEFT_PARENTHESIS term (COMMA term)* RIGHT_PARENTHESIS]
function: FUNCTION_NAME LEFT_PARENTHESIS term (COMMA term)* RIGHT_PARENTHESIS
quantified_wff: curly_quantifiers | quantifiers
curly_quantifiers: quantifier_symbol LEFT_CURLY_BRACE VARIABLE (COMMA SPACE? VARIABLE)* RIGHT_CURLY_BRACE SPACE?
quantifiers: quantifier_symbol SPACE? VARIABLE (COMMA SPACE? VARIABLE)* SPACE?

SPACE: /\s+/
COMMA: /,\s*/
EQUAL_TO: /\s*=\s*/
LEFT_PARENTHESIS: "("
RIGHT_PARENTHESIS: ")"
LEFT_CURLY_BRACE: "{"
RIGHT_CURLY_BRACE: "}"
quantifier_symbol: UNIVERSAL_QUANTIFIER_SYMBOL | EXISTENTIAL_QUANTIFIER_SYMBOL
UNIVERSAL_QUANTIFIER_SYMBOL: "\\forall" | "∀"
EXISTENTIAL_QUANTIFIER_SYMBOL: "\\exists" | "∃"
NAME: /[a-t]/ | /[a-t]_[1-9]\d*/
VARIABLE: /[u-z]/ | /[u-z]_[1-9]\d*/
PREDICATE_NAME: /[A-HJ-Z]/ | /[A-HJ-Z]_[1-9]\d*/
FUNCTION_NAME: /[a-z]/ | /[a-z]_[1-9]\d*/
NEGATION_SYMBOL: "\\neg" | "\\lnot" | "¬"
CONJUNCTION_SYMBOL: "\\wedge" | "\\land" | "∧"
DISJUNCTION_SYMBOL: "\\vee" | "\\lor" | "∨"
CONDITIONAL_SYMBOL: "\\rightarrow" | "\\Rightarrow" | "\\Longrightarrow" | "\\implies" | "→" | "⇒"
BICONDITIONAL_SYMBOL: "\\leftrightarrow" | "\\iff" | "↔" | "⇔"

Also remove the %ignore directives.同时删除%ignore指令。

%ignore SPACE
%ignore COMMA
%ignore LEFT_CURLY_BRACE
%ignore RIGHT_CURLY_BRACE

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