logic.py

    ((F(A) & G(B)) & H(z))
    >>> subst(s2, p)
    ((F(C) & G(y)) & H(x))
    
    >>> subst(s2, subst(s1, p))
    ((F(A) & G(B)) & H(x))
    >>> subst(subst_compose(s1, s2), p)
    ((F(A) & G(B)) & H(x))

    >>> subst(s1, subst(s2, p))
    ((F(C) & G(B)) & H(A))
    >>> subst(subst_compose(s2, s1), p)
    ((F(C) & G(B)) & H(A))
    >>> ppsubst(subst_compose(s1, s2))
    {x: A, y: B, z: x}
    >>> ppsubst(subst_compose(s2, s1))
    {x: C, y: B, z: A}
    >>> subst(subst_compose(s1, s2), p) == subst(s2, subst(s1, p))
    True
    >>> subst(subst_compose(s2, s1), p) == subst(s1, subst(s2, p))
    True
    """
    sc = {}
    for x, v in s1.items():
        if s2.has_key(v):
            w = s2[v]
            sc[x] = w # x -> v -> w
        else:
            sc[x] = v
    for x, v in s2.items():
        if not (s1.has_key(x)):
            sc[x] = v
        # otherwise s1[x] preemptys s2[x]
    return sc

#______________________________________________________________________________

# Example application (not in the book).
# You can use the Expr class to do symbolic differentiation.  This used to be
# a part of AI; now it is considered a separate field, Symbolic Algebra.

def diff(y, x):
    """Return the symbolic derivative, dy/dx, as an Expr.
    However, you probably want to simplify the results with simp.
    >>> diff(x * x, x)
    ((x * 1) + (x * 1))
    >>> simp(diff(x * x, x))
    (2 * x)
    """
    if y == x: return ONE
    elif not y.args: return ZERO
    else:
        u, op, v = y.args[0], y.op, y.args[-1]
        if op == '+': return diff(u, x) + diff(v, x)
        elif op == '-' and len(args) == 1: return -diff(u, x)
        elif op == '-': return diff(u, x) - diff(v, x)
        elif op == '*': return u * diff(v, x) + v * diff(u, x)
        elif op == '/': return (v*diff(u, x) - u*diff(v, x)) / (v * v)
        elif op == '**' and isnumber(x.op):
            return (v * u ** (v - 1) * diff(u, x))
        elif op == '**': return (v * u ** (v - 1) * diff(u, x)
                                 + u ** v * Expr('log')(u) * diff(v, x))
        elif op == 'log': return diff(u, x) / u
        else: raise ValueError("Unknown op: %s in diff(%s, %s)" % (op, y, x))

def simp(x):
    if not x.args: return x
    args = map(simp, x.args)
    u, op, v = args[0], x.op, args[-1]
    if op == '+': 
        if v == ZERO: return u
        if u == ZERO: return v
        if u == v: return TWO * u
        if u == -v or v == -u: return ZERO
    elif op == '-' and len(args) == 1: 
        if u.op == '-' and len(u.args) == 1: return u.args[0] ## --y ==> y
    elif op == '-': 
        if v == ZERO: return u
        if u == ZERO: return -v
        if u == v: return ZERO
        if u == -v or v == -u: return ZERO
    elif op == '*': 
        if u == ZERO or v == ZERO: return ZERO
        if u == ONE: return v
        if v == ONE: return u
        if u == v: return u ** 2
    elif op == '/': 
        if u == ZERO: return ZERO
        if v == ZERO: return Expr('Undefined')
        if u == v: return ONE
        if u == -v or v == -u: return ZERO
    elif op == '**': 
        if u == ZERO: return ZERO
        if v == ZERO: return ONE
        if u == ONE: return ONE
        if v == ONE: return u
    elif op == 'log': 
        if u == ONE: return ZERO
    else: raise ValueError("Unknown op: " + op)
    ## If we fall through to here, we can not simplify further
    return Expr(op, *args)

def d(y, x):
    "Differentiate and then simplify."
    return simp(diff(y, x))    

#_______________________________________________________________________________

# Utilities for doctest cases
# These functions print their arguments in a standard order
# to compensate for the random order in the standard representation

def ppsubst(s):
    """Print substitution s"""
    ppdict(s)

def ppdict(d):
    """Print the dictionary d.
    
    Prints a string representation of the dictionary
    with keys in sorted order according to their string
    representation: {a: A, d: D, ...}.
    >>> ppdict({'m': 'M', 'a': 'A', 'r': 'R', 'k': 'K'})
    {'a': 'A', 'k': 'K', 'm': 'M', 'r': 'R'}
    >>> ppdict({z: C, y: B, x: A})
    {x: A, y: B, z: C}
    """

    def format(k, v):
        return "%s: %s" % (repr(k), repr(v))

    ditems = d.items()
    ditems.sort(key=str)
    k, v = ditems[0]
    dpairs = format(k, v)
    for (k, v) in ditems[1:]:
        dpairs += (', ' + format(k, v))
    print '{%s}' % dpairs

def ppset(s):
    """Print the set s.

    >>> ppset(set(['A', 'Q', 'F', 'K', 'Y', 'B']))
    set(['A', 'B', 'F', 'K', 'Q', 'Y'])
    >>> ppset(set([z, y, x]))
    set([x, y, z])
    """

    slist = list(s)
    slist.sort(key=str)
    print 'set(%s)' % slist

# Debug this expression:

e = expr('Human(x)')
print e

debug = False

if debug:
    fol_bc_ask(test_kb, [e])