Issues debugging infix grammar in pyparsing
I wrote this script in order to parse statements with a syntax similar to prolog, treating connectives as operators with precedence:
import pyparsing as pyp
alphabet = "abcdefghijklmnopqrstuvwxyz"
alphabet = alphabet + alphabet.upper()
symbol = pyp.Word(alphabet)
predicate = symbol + "(" + pyp.ZeroOrMore(symbol + ",") + symbol + ")"
parenthetic = pyp.Forward()
pyp_formula = pyp.infixNotation((predicate | parenthetic),
[
(pyp.oneOf('~'), 1, pyp.opAssoc.RIGHT),
(pyp.oneOf(','), 2, pyp.opAssoc.LEFT),
(pyp.oneOf(';'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('::'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('->'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf(':-'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('--'), 1, pyp.opAssoc.LEFT),
(pyp.oneOf('.'), 1, pyp.opAssoc.LEFT),
])
parenthetic << "(" + pyp_formula + ")"
When I run
parse = pyp_formula.parseString('d(A, D), e(D) :- f(A), g(D); h(D).')
parse_list = formula_parse.asList()
print(parse_list)
I get that "d(A, D), e(D)" has not even been split into two predicates and "f(A), g(D); h(D)" is treated as a single terminal within a non-terminal.
[[[['d', '(', 'A', ',', 'D', ')', ',', 'e', '(', 'D', ')'],
':-', [['f', '(', 'A', ')', ',', 'g', '(', 'D', ')'],
';', 'h', '(', 'D', ')']], '.']]
I've tried several alternatives but I don't seem to be able to get the right parse.
Any help is welcome!
python pyparsing
add a comment |
I wrote this script in order to parse statements with a syntax similar to prolog, treating connectives as operators with precedence:
import pyparsing as pyp
alphabet = "abcdefghijklmnopqrstuvwxyz"
alphabet = alphabet + alphabet.upper()
symbol = pyp.Word(alphabet)
predicate = symbol + "(" + pyp.ZeroOrMore(symbol + ",") + symbol + ")"
parenthetic = pyp.Forward()
pyp_formula = pyp.infixNotation((predicate | parenthetic),
[
(pyp.oneOf('~'), 1, pyp.opAssoc.RIGHT),
(pyp.oneOf(','), 2, pyp.opAssoc.LEFT),
(pyp.oneOf(';'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('::'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('->'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf(':-'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('--'), 1, pyp.opAssoc.LEFT),
(pyp.oneOf('.'), 1, pyp.opAssoc.LEFT),
])
parenthetic << "(" + pyp_formula + ")"
When I run
parse = pyp_formula.parseString('d(A, D), e(D) :- f(A), g(D); h(D).')
parse_list = formula_parse.asList()
print(parse_list)
I get that "d(A, D), e(D)" has not even been split into two predicates and "f(A), g(D); h(D)" is treated as a single terminal within a non-terminal.
[[[['d', '(', 'A', ',', 'D', ')', ',', 'e', '(', 'D', ')'],
':-', [['f', '(', 'A', ')', ',', 'g', '(', 'D', ')'],
';', 'h', '(', 'D', ')']], '.']]
I've tried several alternatives but I don't seem to be able to get the right parse.
Any help is welcome!
python pyparsing
add a comment |
I wrote this script in order to parse statements with a syntax similar to prolog, treating connectives as operators with precedence:
import pyparsing as pyp
alphabet = "abcdefghijklmnopqrstuvwxyz"
alphabet = alphabet + alphabet.upper()
symbol = pyp.Word(alphabet)
predicate = symbol + "(" + pyp.ZeroOrMore(symbol + ",") + symbol + ")"
parenthetic = pyp.Forward()
pyp_formula = pyp.infixNotation((predicate | parenthetic),
[
(pyp.oneOf('~'), 1, pyp.opAssoc.RIGHT),
(pyp.oneOf(','), 2, pyp.opAssoc.LEFT),
(pyp.oneOf(';'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('::'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('->'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf(':-'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('--'), 1, pyp.opAssoc.LEFT),
(pyp.oneOf('.'), 1, pyp.opAssoc.LEFT),
])
parenthetic << "(" + pyp_formula + ")"
When I run
parse = pyp_formula.parseString('d(A, D), e(D) :- f(A), g(D); h(D).')
parse_list = formula_parse.asList()
print(parse_list)
I get that "d(A, D), e(D)" has not even been split into two predicates and "f(A), g(D); h(D)" is treated as a single terminal within a non-terminal.
[[[['d', '(', 'A', ',', 'D', ')', ',', 'e', '(', 'D', ')'],
':-', [['f', '(', 'A', ')', ',', 'g', '(', 'D', ')'],
';', 'h', '(', 'D', ')']], '.']]
I've tried several alternatives but I don't seem to be able to get the right parse.
Any help is welcome!
python pyparsing
I wrote this script in order to parse statements with a syntax similar to prolog, treating connectives as operators with precedence:
import pyparsing as pyp
alphabet = "abcdefghijklmnopqrstuvwxyz"
alphabet = alphabet + alphabet.upper()
symbol = pyp.Word(alphabet)
predicate = symbol + "(" + pyp.ZeroOrMore(symbol + ",") + symbol + ")"
parenthetic = pyp.Forward()
pyp_formula = pyp.infixNotation((predicate | parenthetic),
[
(pyp.oneOf('~'), 1, pyp.opAssoc.RIGHT),
(pyp.oneOf(','), 2, pyp.opAssoc.LEFT),
(pyp.oneOf(';'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('::'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('->'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf(':-'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('--'), 1, pyp.opAssoc.LEFT),
(pyp.oneOf('.'), 1, pyp.opAssoc.LEFT),
])
parenthetic << "(" + pyp_formula + ")"
When I run
parse = pyp_formula.parseString('d(A, D), e(D) :- f(A), g(D); h(D).')
parse_list = formula_parse.asList()
print(parse_list)
I get that "d(A, D), e(D)" has not even been split into two predicates and "f(A), g(D); h(D)" is treated as a single terminal within a non-terminal.
[[[['d', '(', 'A', ',', 'D', ')', ',', 'e', '(', 'D', ')'],
':-', [['f', '(', 'A', ')', ',', 'g', '(', 'D', ')'],
';', 'h', '(', 'D', ')']], '.']]
I've tried several alternatives but I don't seem to be able to get the right parse.
Any help is welcome!
python pyparsing
python pyparsing
asked Nov 15 '18 at 21:49
KipirpoKipirpo
103
103
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
There is no need to define parenthetic
, infixNotation
does that for you. And if you pyp.Group
your predicate
expression, then the symbol and symbol args will get grouped for you. pyparsing also has the delimitedList
helper as a short cut for expr, expr, expr
: use pyp.delimitedList(expr)
. The default delimiter is ',', but you can define other delimiters. Lastly, with a list of operators this long, it is likely that you will have performance issues with any non-trivial parse - add a call to ParserElement.enablePackrat()
to turn on an internal parsing cache.
Here is how your code looks with these changes:
import pyparsing as pyp
pyp.ParserElement.enablePackrat()
alphabet = "abcdefghijklmnopqrstuvwxyz"
alphabet = alphabet + alphabet.upper()
symbol = pyp.Word(alphabet)
predicate = pyp.Group(symbol + "(" + pyp.delimitedList(symbol) + ")")
pyp_formula = pyp.infixNotation(predicate,
[
(pyp.oneOf('~'), 1, pyp.opAssoc.RIGHT),
(pyp.oneOf(','), 2, pyp.opAssoc.LEFT),
(pyp.oneOf(';'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('::'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('->'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf(':-'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('--'), 1, pyp.opAssoc.LEFT),
(pyp.oneOf('.'), 1, pyp.opAssoc.LEFT),
])
pyp_formula.runTests(
'''
d(A, D) -> h(D).
d(A, D), e(D) :- f(A), g(D); h(D).
'''
)
And the results:
d(A, D) -> h(D).
[[[['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']], '.']]
[0]:
[[['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']], '.']
[0]:
[['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']]
[0]:
['d', '(', 'A', 'D', ')']
[1]:
->
[2]:
['h', '(', 'D', ')']
[1]:
.
d(A, D), e(D) :- f(A), g(D); h(D).
[[[[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]], '.']]
[0]:
[[[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]], '.']
[0]:
[[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]]
[0]:
[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']]
[0]:
['d', '(', 'A', 'D', ')']
[1]:
,
[2]:
['e', '(', 'D', ')']
[1]:
:-
[2]:
[[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]
[0]:
[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']]
[0]:
['f', '(', 'A', ')']
[1]:
,
[2]:
['g', '(', 'D', ')']
[1]:
;
[2]:
['h', '(', 'D', ')']
[1]:
.
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
There is no need to define parenthetic
, infixNotation
does that for you. And if you pyp.Group
your predicate
expression, then the symbol and symbol args will get grouped for you. pyparsing also has the delimitedList
helper as a short cut for expr, expr, expr
: use pyp.delimitedList(expr)
. The default delimiter is ',', but you can define other delimiters. Lastly, with a list of operators this long, it is likely that you will have performance issues with any non-trivial parse - add a call to ParserElement.enablePackrat()
to turn on an internal parsing cache.
Here is how your code looks with these changes:
import pyparsing as pyp
pyp.ParserElement.enablePackrat()
alphabet = "abcdefghijklmnopqrstuvwxyz"
alphabet = alphabet + alphabet.upper()
symbol = pyp.Word(alphabet)
predicate = pyp.Group(symbol + "(" + pyp.delimitedList(symbol) + ")")
pyp_formula = pyp.infixNotation(predicate,
[
(pyp.oneOf('~'), 1, pyp.opAssoc.RIGHT),
(pyp.oneOf(','), 2, pyp.opAssoc.LEFT),
(pyp.oneOf(';'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('::'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('->'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf(':-'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('--'), 1, pyp.opAssoc.LEFT),
(pyp.oneOf('.'), 1, pyp.opAssoc.LEFT),
])
pyp_formula.runTests(
'''
d(A, D) -> h(D).
d(A, D), e(D) :- f(A), g(D); h(D).
'''
)
And the results:
d(A, D) -> h(D).
[[[['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']], '.']]
[0]:
[[['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']], '.']
[0]:
[['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']]
[0]:
['d', '(', 'A', 'D', ')']
[1]:
->
[2]:
['h', '(', 'D', ')']
[1]:
.
d(A, D), e(D) :- f(A), g(D); h(D).
[[[[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]], '.']]
[0]:
[[[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]], '.']
[0]:
[[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]]
[0]:
[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']]
[0]:
['d', '(', 'A', 'D', ')']
[1]:
,
[2]:
['e', '(', 'D', ')']
[1]:
:-
[2]:
[[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]
[0]:
[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']]
[0]:
['f', '(', 'A', ')']
[1]:
,
[2]:
['g', '(', 'D', ')']
[1]:
;
[2]:
['h', '(', 'D', ')']
[1]:
.
add a comment |
There is no need to define parenthetic
, infixNotation
does that for you. And if you pyp.Group
your predicate
expression, then the symbol and symbol args will get grouped for you. pyparsing also has the delimitedList
helper as a short cut for expr, expr, expr
: use pyp.delimitedList(expr)
. The default delimiter is ',', but you can define other delimiters. Lastly, with a list of operators this long, it is likely that you will have performance issues with any non-trivial parse - add a call to ParserElement.enablePackrat()
to turn on an internal parsing cache.
Here is how your code looks with these changes:
import pyparsing as pyp
pyp.ParserElement.enablePackrat()
alphabet = "abcdefghijklmnopqrstuvwxyz"
alphabet = alphabet + alphabet.upper()
symbol = pyp.Word(alphabet)
predicate = pyp.Group(symbol + "(" + pyp.delimitedList(symbol) + ")")
pyp_formula = pyp.infixNotation(predicate,
[
(pyp.oneOf('~'), 1, pyp.opAssoc.RIGHT),
(pyp.oneOf(','), 2, pyp.opAssoc.LEFT),
(pyp.oneOf(';'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('::'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('->'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf(':-'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('--'), 1, pyp.opAssoc.LEFT),
(pyp.oneOf('.'), 1, pyp.opAssoc.LEFT),
])
pyp_formula.runTests(
'''
d(A, D) -> h(D).
d(A, D), e(D) :- f(A), g(D); h(D).
'''
)
And the results:
d(A, D) -> h(D).
[[[['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']], '.']]
[0]:
[[['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']], '.']
[0]:
[['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']]
[0]:
['d', '(', 'A', 'D', ')']
[1]:
->
[2]:
['h', '(', 'D', ')']
[1]:
.
d(A, D), e(D) :- f(A), g(D); h(D).
[[[[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]], '.']]
[0]:
[[[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]], '.']
[0]:
[[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]]
[0]:
[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']]
[0]:
['d', '(', 'A', 'D', ')']
[1]:
,
[2]:
['e', '(', 'D', ')']
[1]:
:-
[2]:
[[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]
[0]:
[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']]
[0]:
['f', '(', 'A', ')']
[1]:
,
[2]:
['g', '(', 'D', ')']
[1]:
;
[2]:
['h', '(', 'D', ')']
[1]:
.
add a comment |
There is no need to define parenthetic
, infixNotation
does that for you. And if you pyp.Group
your predicate
expression, then the symbol and symbol args will get grouped for you. pyparsing also has the delimitedList
helper as a short cut for expr, expr, expr
: use pyp.delimitedList(expr)
. The default delimiter is ',', but you can define other delimiters. Lastly, with a list of operators this long, it is likely that you will have performance issues with any non-trivial parse - add a call to ParserElement.enablePackrat()
to turn on an internal parsing cache.
Here is how your code looks with these changes:
import pyparsing as pyp
pyp.ParserElement.enablePackrat()
alphabet = "abcdefghijklmnopqrstuvwxyz"
alphabet = alphabet + alphabet.upper()
symbol = pyp.Word(alphabet)
predicate = pyp.Group(symbol + "(" + pyp.delimitedList(symbol) + ")")
pyp_formula = pyp.infixNotation(predicate,
[
(pyp.oneOf('~'), 1, pyp.opAssoc.RIGHT),
(pyp.oneOf(','), 2, pyp.opAssoc.LEFT),
(pyp.oneOf(';'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('::'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('->'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf(':-'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('--'), 1, pyp.opAssoc.LEFT),
(pyp.oneOf('.'), 1, pyp.opAssoc.LEFT),
])
pyp_formula.runTests(
'''
d(A, D) -> h(D).
d(A, D), e(D) :- f(A), g(D); h(D).
'''
)
And the results:
d(A, D) -> h(D).
[[[['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']], '.']]
[0]:
[[['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']], '.']
[0]:
[['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']]
[0]:
['d', '(', 'A', 'D', ')']
[1]:
->
[2]:
['h', '(', 'D', ')']
[1]:
.
d(A, D), e(D) :- f(A), g(D); h(D).
[[[[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]], '.']]
[0]:
[[[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]], '.']
[0]:
[[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]]
[0]:
[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']]
[0]:
['d', '(', 'A', 'D', ')']
[1]:
,
[2]:
['e', '(', 'D', ')']
[1]:
:-
[2]:
[[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]
[0]:
[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']]
[0]:
['f', '(', 'A', ')']
[1]:
,
[2]:
['g', '(', 'D', ')']
[1]:
;
[2]:
['h', '(', 'D', ')']
[1]:
.
There is no need to define parenthetic
, infixNotation
does that for you. And if you pyp.Group
your predicate
expression, then the symbol and symbol args will get grouped for you. pyparsing also has the delimitedList
helper as a short cut for expr, expr, expr
: use pyp.delimitedList(expr)
. The default delimiter is ',', but you can define other delimiters. Lastly, with a list of operators this long, it is likely that you will have performance issues with any non-trivial parse - add a call to ParserElement.enablePackrat()
to turn on an internal parsing cache.
Here is how your code looks with these changes:
import pyparsing as pyp
pyp.ParserElement.enablePackrat()
alphabet = "abcdefghijklmnopqrstuvwxyz"
alphabet = alphabet + alphabet.upper()
symbol = pyp.Word(alphabet)
predicate = pyp.Group(symbol + "(" + pyp.delimitedList(symbol) + ")")
pyp_formula = pyp.infixNotation(predicate,
[
(pyp.oneOf('~'), 1, pyp.opAssoc.RIGHT),
(pyp.oneOf(','), 2, pyp.opAssoc.LEFT),
(pyp.oneOf(';'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('::'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('->'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf(':-'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('--'), 1, pyp.opAssoc.LEFT),
(pyp.oneOf('.'), 1, pyp.opAssoc.LEFT),
])
pyp_formula.runTests(
'''
d(A, D) -> h(D).
d(A, D), e(D) :- f(A), g(D); h(D).
'''
)
And the results:
d(A, D) -> h(D).
[[[['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']], '.']]
[0]:
[[['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']], '.']
[0]:
[['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']]
[0]:
['d', '(', 'A', 'D', ')']
[1]:
->
[2]:
['h', '(', 'D', ')']
[1]:
.
d(A, D), e(D) :- f(A), g(D); h(D).
[[[[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]], '.']]
[0]:
[[[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]], '.']
[0]:
[[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]]
[0]:
[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']]
[0]:
['d', '(', 'A', 'D', ')']
[1]:
,
[2]:
['e', '(', 'D', ')']
[1]:
:-
[2]:
[[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]
[0]:
[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']]
[0]:
['f', '(', 'A', ')']
[1]:
,
[2]:
['g', '(', 'D', ')']
[1]:
;
[2]:
['h', '(', 'D', ')']
[1]:
.
answered Nov 16 '18 at 4:01
PaulMcGPaulMcG
46.7k969111
46.7k969111
add a comment |
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