Formulas consist of the operators &, |, ~, ^, ->, <->, corresponding to and, or, not, xor, if...then, if and only if. Operators can be applied to variables that consist of a leading letter and trailing underscores and alphanumerics. Parentheses may be used to explicitly show order of operation.
EXAMPLES:
Create boolean formulas and combine them with ifthen() method:
sage: import sage.logic.propcalc as propcalc
sage: f = propcalc.formula("a&((b|c)^a->c)<->b")
sage: g = propcalc.formula("boolean<->algebra")
sage: (f&~g).ifthen(f)
((a&((b|c)^a->c)<->b)&(~(boolean<->algebra)))->(a&((b|c)^a->c)<->b)
We can create a truth table from a formula:
sage: f.truthtable()
a b c value
False False False True
False False True True
False True False False
False True True False
True False False True
True False True False
True True False True
True True True True
sage: f.truthtable(end=3)
a b c value
False False False True
False False True True
False True False False
sage: f.truthtable(start=4)
a b c value
True False False True
True False True False
True True False True
True True True True
sage: propcalc.formula("a").truthtable()
a value
False False
True True
Now we can evaluate the formula for a given set of inputs:
sage: f.evaluate({'a':True, 'b':False, 'c':True})
False
sage: f.evaluate({'a':False, 'b':False, 'c':True})
True
And we can convert a boolean formula to conjunctive normal form:
sage: f.convert_cnf_table()
sage: f
(a|~b|c)&(a|~b|~c)&(~a|b|~c)
sage: f.convert_cnf_recur()
sage: f
(a|~b|c)&(a|~b|~c)&(~a|b|~c)
Or determine if an expression is satisfiable, a contradiction, or a tautology:
sage: f = propcalc.formula("a|b")
sage: f.is_satisfiable()
True
sage: f = f & ~f
sage: f.is_satisfiable()
False
sage: f.is_contradiction()
True
sage: f = f | ~f
sage: f.is_tautology()
True
The equality operator compares semantic equivalence:
sage: f = propcalc.formula("(a|b)&c")
sage: g = propcalc.formula("c&(b|a)")
sage: f == g
True
sage: g = propcalc.formula("a|b&c")
sage: f == g
False
It is an error to create a formula with bad syntax:
sage: propcalc.formula("")
Traceback (most recent call last):
...
SyntaxError: malformed statement
sage: propcalc.formula("a&b~(c|(d)")
Traceback (most recent call last):
...
SyntaxError: malformed statement
sage: propcalc.formula("a&&b")
Traceback (most recent call last):
...
SyntaxError: malformed statement
sage: propcalc.formula("a&b a")
Traceback (most recent call last):
...
SyntaxError: malformed statement
It is also an error to not abide by the naming conventions:
sage: propcalc.formula("~a&9b")
Traceback (most recent call last):
...
NameError: invalid variable name 9b: identifiers must begin with a letter and contain only alphanumerics and underscores
AUTHORS:
Initialize the data fields
INPUT:
OUTPUT:
None
EXAMPLES:
This example illustrates the creation of a statement.
sage: import sage.logic.propcalc as propcalc
sage: s = propcalc.formula("a&b|~(c|a)")
sage: s
a&b|~(c|a)
Combine two formulas with the given operator.
INPUT:
OUTPUT:
The result as an instance of BooleanFormula
EXAMPLES:
This example shows how to create a new formula from two others.
sage: import sage.logic.propcalc as propcalc
sage: s = propcalc.formula("a&b")
sage: f = propcalc.formula("c^d")
sage: s.add_statement(f, '|')
(a&b)|(c^d)
sage: s.add_statement(f, '->')
(a&b)->(c^d)
Convert boolean formula to conjunctive normal form.
INPUT:
OUTPUT:
An instance of BooleanFormula in conjunctive normal form
EXAMPLES:
This example illustrates how to convert a formula to cnf.
sage: import sage.logic.propcalc as propcalc
sage: s = propcalc.formula("a ^ b <-> c")
sage: s.convert_cnf()
sage: s
(a|b|~c)&(a|~b|c)&(~a|b|c)&(~a|~b|~c)
We now show that convert_cnf() and convert_cnf_table() are aliases.
sage: t = propcalc.formula("a ^ b <-> c")
sage: t.convert_cnf_table(); t
(a|b|~c)&(a|~b|c)&(~a|b|c)&(~a|~b|~c)
sage: t == s
True
Note
This method creates the cnf parse tree by examining the logic
table of the formula. Creating the table requires time
where
is the number of variables in the formula.
Convert boolean formula to conjunctive normal form.
INPUT:
OUTPUT:
An instance of BooleanFormula in conjunctive normal form
EXAMPLES:
This example hows how to convert a formula to conjunctive normal form.
sage: import sage.logic.propcalc as propcalc
sage: s = propcalc.formula("a^b<->c")
sage: s.convert_cnf_recur()
sage: s
(~a|a|c)&(~b|a|c)&(~a|b|c)&(~b|b|c)&(~c|a|b)&(~c|~a|~b)
Note
This function works by applying a set of rules that are guaranteed to convert the formula. Worst case the converted expression has an O(2^n) increase in size (and time as well), but if the formula is already in CNF (or close to) it is only O(n).
This function can require an exponential blow up in space from the original expression. This in turn can require large amounts of time. Unless a formula is already in (or close to) being in cnf convert_cnf() is typically preferred, but results can vary.
Convert boolean formula to conjunctive normal form.
INPUT:
OUTPUT:
An instance of BooleanFormula in conjunctive normal form
EXAMPLES:
This example illustrates how to convert a formula to cnf.
sage: import sage.logic.propcalc as propcalc
sage: s = propcalc.formula("a ^ b <-> c")
sage: s.convert_cnf()
sage: s
(a|b|~c)&(a|~b|c)&(~a|b|c)&(~a|~b|~c)
We now show that convert_cnf() and convert_cnf_table() are aliases.
sage: t = propcalc.formula("a ^ b <-> c")
sage: t.convert_cnf_table(); t
(a|b|~c)&(a|~b|c)&(~a|b|c)&(~a|~b|~c)
sage: t == s
True
Note
This method creates the cnf parse tree by examining the logic
table of the formula. Creating the table requires time
where
is the number of variables in the formula.
Convert the string representation of a formula to conjunctive normal form.
INPUT:
OUTPUT:
None
EXAMPLES:
We show how the converted formula is printed in conjunctive normal form.
sage: import sage.logic.propcalc as propcalc
sage: s = propcalc.formula("a^b<->c")
sage: s.convert_cnf_recur(); s #long time
(~a|a|c)&(~b|a|c)&(~a|b|c)&(~b|b|c)&(~c|a|b)&(~c|~a|~b)
Convert a parse tree to the tuple form used by bool_opt.
INPUT:
OUTPUT:
A 3-tuple
EXAMPLES:
This example illustrates the conversion of a formula into its corresponding tuple.
sage: import sage.logic.propcalc as propcalc, sage.logic.logicparser as logicparser
sage: s = propcalc.formula("a&(b|~c)")
sage: tree = ['&', 'a', ['|', 'b', ['~', 'c', None]]]
sage: logicparser.apply_func(tree, s.convert_opt)
('and', ('prop', 'a'), ('or', ('prop', 'b'), ('not', ('prop', 'c'))))
Note
This function only works on one branch of the parse tree. To apply the function to every branch of a parse tree, pass the function as an argument in apply_func() in logicparser.py.
Distribute ~ operators over & and | operators.
INPUT:
OUTPUT:
A new list
EXAMPLES:
This example illustrates the distribution of ‘~’ over ‘&’.
sage: import sage.logic.propcalc as propcalc, sage.logic.logicparser as logicparser
sage: s = propcalc.formula("~(a&b)")
sage: tree = ['~', ['&', 'a', 'b'], None]
sage: logicparser.apply_func(tree, s.dist_not) #long time
['|', ['~', 'a', None], ['~', 'b', None]]
Note
This function only operates on a single branch of a parse tree. To apply the function to an entire parse tree, pass the function as an argument to apply_func() in logicparser.py.
Distribute | over &.
INPUT:
OUTPUT:
A new list
EXAMPLES:
This example illustrates the distribution of ‘|’ over ‘&’.
sage: import sage.logic.propcalc as propcalc, sage.logic.logicparser as logicparser
sage: s = propcalc.formula("(a&b)|(a&c)")
sage: tree = ['|', ['&', 'a', 'b'], ['&', 'a', 'c']]
sage: logicparser.apply_func(tree, s.dist_ors) #long time
['&', ['&', ['|', 'a', 'a'], ['|', 'b', 'a']], ['&', ['|', 'a', 'c'], ['|', 'b', 'c']]]
Note
This function only operates on a single branch of a parse tree. To apply the function to an entire parse tree, pass the function as an argument to apply_func() in logicparser.py.
Determine if two formulas are semantically equivalent.
INPUT:
OUTPUT:
A boolean value to be determined as follows:
True - if the two formulas are logically equivalent
False - if the two formulas are not logically equivalent
EXAMPLES:
This example shows how to check for logical equivalence.
sage: import sage.logic.propcalc as propcalc
sage: f = propcalc.formula("(a|b)&c")
sage: g = propcalc.formula("c&(a|b)")
sage: f.equivalent(g)
True
sage: g = propcalc.formula("a|b&c")
sage: f.equivalent(g)
False
Evaluate a formula for the given input values.
INPUT:
OUTPUT:
The result of the evaluation as a boolean.
EXAMPLES:
This example illustrates the evaluation of a boolean formula.
sage: import sage.logic.propcalc as propcalc
sage: f = propcalc.formula("a&b|c")
sage: f.evaluate({'a':False, 'b':False, 'c':True})
True
sage: f.evaluate({'a':True, 'b':False, 'c':False})
False
Return a full syntax parse tree of the calling formula.
INPUT:
OUTPUT:
The full syntax parse tree as a nested list
EXAMPLES:
This example shows how to find the full syntax parse tree of a formula.
sage: import sage.logic.propcalc as propcalc
sage: s = propcalc.formula("a->(b&c)")
sage: s.full_tree()
['->', 'a', ['&', 'b', 'c']]
sage: t = propcalc.formula("a & ((~b | c) ^ a -> c) <-> ~b")
sage: t.full_tree()
['<->', ['&', 'a', ['->', ['^', ['|', ['~', 'b'], 'c'], 'a'], 'c']], ['~', 'b']]
sage: f = propcalc.formula("~~(a&~b)")
sage: f.full_tree()
['~', ['~', ['&', 'a', ['~', 'b']]]]
Note
This function is used by other functions in the logic module that perform syntactic operations on a boolean formula.
AUTHORS:
Determine if bit c of the number x is 1.
INPUT:
OUTPUT:
A boolean to be determined as follows:
True - if bit c of x is 1
False - if bit c of x is not 1
EXAMPLES:
This example illustrates the use of get_bit().
sage: import sage.logic.propcalc as propcalc
sage: s = propcalc.formula("a&b")
sage: s.get_bit(2, 1)
True
sage: s.get_bit(8, 0)
False
It is not an error to have a bit out of range.
sage: s.get_bit(64, 7)
False
Nor is it an error to use a negative number.
sage: s.get_bit(-1, 3)
False
sage: s.get_bit(64, -1)
True
sage: s.get_bit(64, -2)
False
Note
The 0 bit is the low order bit. Errors should be handled gracefully by a return of false, and negative numbers x always return false while a negative c will index from the high order bit.
Return the next operator in a string.
INPUT:
OUTPUT:
The next operator as a string
EXAMPLES:
This example illustrates how to find the next operator in a formula.
sage: import sage.logic.propcalc as propcalc
sage: s = propcalc.formula("f&p")
sage: s.get_next_op("abra|cadabra")
'|'
Note
The parameter str is not necessarily the string representation of the calling object.
Combine two formulas with the <-> operator.
INPUT:
OUTPUT:
A boolean formula of the following form:
self <-> other
EXAMPLES:
This example illustrates how to combine two formulas with ‘<->’.
sage: import sage.logic.propcalc as propcalc
sage: s = propcalc.formula("a&b")
sage: f = propcalc.formula("c^d")
sage: s.iff(f)
(a&b)<->(c^d)
Combine two formulas with the -> operator.
INPUT:
- self – calling object. This is the formula on the left side of the operator.
- other – a boolean formula. This is the formula on the right side of the operator.
OUTPUT:
A boolean formula of the following form:
self -> other
EXAMPLES:
This example illustrates how to combine two formulas with ‘->’.
sage: import sage.logic.propcalc as propcalc
sage: s = propcalc.formula("a&b")
sage: f = propcalc.formula("c^d")
sage: s.ifthen(f)
(a&b)->(c^d)
Determine if the formula is always False.
INPUT:
OUTPUT:
A boolean value to be determined as follows:
True - if the formula is a contradiction
False - if the formula is not a contradiction
EXAMPLES:
This example illustrates how to check if a formula is a contradiction.
sage: import sage.logic.propcalc as propcalc
sage: f = propcalc.formula("a&~a")
sage: f.is_contradiction()
True
sage: f = propcalc.formula("a|~a")
sage: f.is_contradiction()
False
sage: f = propcalc.formula("a|b")
sage: f.is_contradiction()
False
Determine if the formula is True for some assignment of values.
INPUT:
OUTPUT:
A boolean value to be determined as follows:
True - if there is an assignment of values that makes the formula True
False - if the formula cannot be made True by any assignment of values
EXAMPLES:
This example illustrates how to check a formula for satisfiability.
sage: import sage.logic.propcalc as propcalc
sage: f = propcalc.formula("a|b")
sage: f.is_satisfiable()
True
sage: g = f & (~f)
sage: g.is_satisfiable()
False
Determine if the formula is always True.
INPUT:
OUTPUT:
A boolean value to be determined as follows:
True - if the formula is a tautology
False - if the formula is not a tautology
EXAMPLES:
This example illustrates how to check if a formula is a tautology.
sage: import sage.logic.propcalc as propcalc
sage: f = propcalc.formula("a|~a")
sage: f.is_tautology()
True
sage: f = propcalc.formula("a&~a")
sage: f.is_tautology()
False
sage: f = propcalc.formula("a&b")
sage: f.is_tautology()
False
Convert the calling boolean formula into polish notation
INPUT:
OUTPUT:
A string representation of the formula in polish notation.
EXAMPLES:
This example illustrates converting a formula to polish notation.
sage: import sage.logic.propcalc as propcalc
sage: f = propcalc.formula("~~a|(c->b)")
sage: f.polish_notation()
'|~~a->cb'
sage: g = propcalc.formula("(a|~b)->c")
sage: g.polish_notation()
'->|a~bc'
AUTHORS:
Convert if-and-only-if, if-then, and xor operations to operations only involving and/or operations.
INPUT:
OUTPUT:
A new list with no ^, ->, or <-> as first element of list.
EXAMPLES:
This example illustrates the use of reduce_op() with apply_func().
sage: import sage.logic.propcalc as propcalc, sage.logic.logicparser as logicparser
sage: s = propcalc.formula("a->b^c")
sage: tree = ['->', 'a', ['^', 'b', 'c']]
sage: logicparser.apply_func(tree, s.reduce_op)
['|', ['~', 'a', None], ['&', ['|', 'b', 'c'], ['~', ['&', 'b', 'c'], None]]]
Note
This function only operates on a single branch of a parse tree. To apply the function to an entire parse tree, pass the function as an argument to apply_func() in logicparser.py.
Return the satformat representation of a boolean formula.
INPUT:
OUTPUT:
The satformat of the formula as a string
EXAMPLES:
This example illustrates how to find the satformat of a formula.
sage: import sage.logic.propcalc as propcalc
sage: f = propcalc.formula("a&((b|c)^a->c)<->b")
sage: f.convert_cnf()
sage: f
(a|~b|c)&(a|~b|~c)&(~a|b|~c)
sage: f.satformat()
'p cnf 3 0\n1 -2 3 0 1 -2 -3 \n0 -1 2 -3'
Note
See www.cs.ubc.ca/~hoos/SATLIB/Benchmarks/SAT/satformat.ps for a description of satformat.
If the instance of boolean formula has not been converted to CNF form by a call to convert_cnf() or convert_cnf_recur() satformat() will call convert_cnf(). Please see the notes for convert_cnf() and convert_cnf_recur() for performance issues.
Convert a parse tree from prefix to infix form.
INPUT:
OUTPUT:
A new list
EXAMPLES:
This example shows how to convert a parse tree from prefix to infix form.
sage: import sage.logic.propcalc as propcalc, sage.logic.logicparser as logicparser
sage: s = propcalc.formula("(a&b)|(a&c)")
sage: tree = ['|', ['&', 'a', 'b'], ['&', 'a', 'c']]
sage: logicparser.apply_func(tree, s.to_infix)
[['a', '&', 'b'], '|', ['a', '&', 'c']]
Note
This function only operates on a single branch of a parse tree. To apply the function to an entire parse tree, pass the function as an argument to apply_func() in logicparser.py.
Return the parse tree of this boolean expression.
INPUT:
OUTPUT:
The parse tree as a nested list
EXAMPLES:
This example illustrates how to find the parse tree of a boolean formula.
sage: import sage.logic.propcalc as propcalc
sage: s = propcalc.formula("man -> monkey & human")
sage: s.tree()
['->', 'man', ['&', 'monkey', 'human']]
sage: f = propcalc.formula("a & ((~b | c) ^ a -> c) <-> ~b")
sage: f.tree()
['<->',
['&', 'a', ['->', ['^', ['|', ['~', 'b', None], 'c'], 'a'], 'c']],
['~', 'b', None]]
Note
This function is used by other functions in the logic module that perform semantic operations on a boolean formula.
Return a truth table for the calling formula.
INPUT:
OUTPUT:
The truth table as a 2-D array
EXAMPLES:
This example illustrates the creation of a truth table.
sage: import sage.logic.propcalc as propcalc
sage: s = propcalc.formula("a&b|~(c|a)")
sage: s.truthtable()
a b c value
False False False True
False False True False
False True False True
False True True False
True False False False
True False True False
True True False True
True True True True
We can now create a truthtable of rows 1 to 4, inclusive.
sage: s.truthtable(1, 5)
a b c value
False False True False
False True False True
False True True False
True False False False
Note
Each row of the table corresponds to a binary number, with each variable associated to a column of the number, and taking on a true value if that column has a value of 1. Please see the logictable module for details. The function returns a table that start inclusive and end exclusive so truthtable(0, 2) will include row 0, but not row 2.
When sent with no start or end parameters, this is an exponential time function requiring O(2**n) time, where n is the number of variables in the expression.