def expr(s):
    """Create an Expr representing a logic expression by parsing the input
    string. Symbols and numbers are automatically converted to Exprs.
    In addition you can use alternative spellings of these operators:
      'x ==> y'   parses as   (x >> y)    # Implication
      'x <== y'   parses as   (x << y)    # Reverse implication
      'x <=> y'   parses as   (x % y)     # Logical equivalence
      'x =/= y'   parses as   (x ^ y)     # Logical disequality (xor)
    But BE CAREFUL; precedence of implication is wrong. expr('P & Q ==> R & S')
    is ((P & (Q >> R)) & S); so you must use expr('(P & Q) ==> (R & S)').
    >>> expr('P <=> Q(1)')
    (P <=> Q(1))
    >>> expr('P & Q | ~R(x, F(x))')
    ((P & Q) | ~R(x, F(x)))
    """
    if isinstance(s, Expr): return s
    if isnumber(s): return Expr(s)
    ## Replace the alternative spellings of operators with canonical spellings
    s = s.replace('==>', '>>').replace('<==', '<<')
    s = s.replace('<=>', '%').replace('=/=', '^')
    ## Replace a symbol or number, such as 'P' with 'Expr("P")'
    s = re.sub(r'([a-zA-Z0-9_.]+)', r'Expr("\1")', s)
    ## Now eval the string.  (A security hole; do not use with an adversary.)
    return eval(s, {'Expr':Expr})
def expr(s):
    """Create an Expr representing a logic expression by parsing the input
    string. Symbols and numbers are automatically converted to Exprs.
    In addition you can use alternative spellings of these operators:
      'x ==> y'   parses as   (x >> y)    # Implication
      'x <== y'   parses as   (x << y)    # Reverse implication
      'x <=> y'   parses as   (x % y)     # Logical equivalence
      'x =/= y'   parses as   (x ^ y)     # Logical disequality (xor)
    But BE CAREFUL; precedence of implication is wrong. expr('P & Q ==> R & S')
    is ((P & (Q >> R)) & S); so you must use expr('(P & Q) ==> (R & S)').
    >>> expr('P <=> Q(1)')
    (P <=> Q(1))
    >>> expr('P & Q | ~R(x, F(x))')
    ((P & Q) | ~R(x, F(x)))
    """
    if isinstance(s, Expr): return s
    if isnumber(s): return Expr(s)
    ## Replace the alternative spellings of operators with canonical spellings
    s = s.replace('==>', '>>').replace('<==', '<<')
    s = s.replace('<=>', '%').replace('=/=', '^')
    ## Replace a symbol or number, such as 'P' with 'Expr("P")'
    s = re.sub(r'([a-zA-Z0-9_.]+)', r'Expr("\1")', s)
    ## Now eval the string.  (A security hole; do not use with an adversary.)
    return eval(s, {'Expr':Expr})
def expr ( s ) : if isinstance ( s , Expr ) : return s if isnumber ( s ) : return Expr ( s ) ## Replace the alternative spellings of operators with canonical spellings s = s . replace ( '==>' , '>>' ) . replace ( '<==' , '<<' ) s = s . replace ( '<=>' , '%' ) . replace ( '=/=' , '^' ) ## Replace a symbol or number, such as 'P' with 'Expr("P")' s = re . sub ( r'([a-zA-Z0-9_.]+)' , r'Expr("\1")' , s ) ## Now eval the string.  (A security hole; do not use with an adversary.) return eval ( s , { 'Expr' : Expr } )
Create an Expr representing a logic expression by parsing the input
    string. Symbols and numbers are automatically converted to Exprs.
    In addition you can use alternative spellings of these operators:
      'x ==> y'   parses as   (x >> y)    # Implication
      'x <== y'   parses as   (x << y)    # Reverse implication
      'x <=> y'   parses as   (x % y)     # Logical equivalence
      'x =/= y'   parses as   (x ^ y)     # Logical disequality (xor)
    But BE CAREFUL; precedence of implication is wrong. expr('P & Q ==> R & S')
    is ((P & (Q >> R)) & S); so you must use expr('(P & Q) ==> (R & S)').
    >>> expr('P <=> Q(1)')
    (P <=> Q(1))
    >>> expr('P & Q | ~R(x, F(x))')
    ((P & Q) | ~R(x, F(x)))
Create an Expr representing a logic expression by parsing the input string . Symbols and numbers are automatically converted to Exprs . In addition you can use alternative spellings of these operators : x == > y parses as ( x >> y ) # Implication x < == y parses as ( x << y ) # Reverse implication x < = > y parses as ( x % y ) # Logical equivalence x = / = y parses as ( x ^ y ) # Logical disequality ( xor ) But BE CAREFUL ; precedence of implication is wrong . expr ( P & Q == > R & S ) is (( P & ( Q >> R )) & S ) ; so you must use expr ( ( P & Q ) == > ( R & S ) ) . >>> expr ( P < = > Q ( 1 ) ) ( P < = > Q ( 1 )) >>> expr ( P & Q | ~R ( x F ( x )) ) (( P & Q ) | ~R ( x F ( x )))