Predicate logic; 3 variables; implications (Python)

# This scripts was used to find the edges in the following Hasse diagram:
# https://commons.wikimedia.org/wiki/File:Predicate_logic;_3_variables;_implications.png

import numpy as np

# The "octal" representation is the base of the calculation.
# The octal notation is motivated by the similarity of the following Hasse diagram with the cube:
# https://commons.wikimedia.org/wiki/File:Predicate_logic;_2_variables;_implications;_pairs.svg
octal = [
    '777',
    '733', '376', '667',
    '731', '713', '374', '176', '647', '467',
    '711', '174', '447',
    '330', '603', '066',
    '310', '130', '601', '403', '064', '046',
    '110', '401', '044',
    '000',
    '577', '533', '266', '511', '244', '200',
    '757', '356', '623', '154', '421', '020',
    '775', '665', '332', '445', '112', '002',
    '555', '222'
]

# The abbreviations are just used in the output.
abbr = [
    'e123',
    'a3e12', 'a2e13', 'a1e23',
    'e2a3e1', 'e1a3e2', 'e3a2e1', 'e1a2e3', 'e3a1e2', 'e2a1e3',
    'e12a3', 'e13a2', 'e23a1',
    'a23e1', 'a13e2', 'a12e3',
    'a2e1a3', 'a3e1a2', 'a1e2a3', 'a3e2a1', 'a1e3a2', 'a2e3a1',
    'e1a23', 'e2a13', 'e3a12',
    'a123',
    'e(12)3', 'a3e(12)', 'a(12)e3', 'e(12)a3', 'e3a(12)', 'a(12)3',
    'e(13)2', 'a2e(13)', 'a(13)e2', 'e(13)a2', 'e2a(13)', 'a(13)2',
    'e1(23)', 'a1e(23)', 'a(23)e1', 'e(23)a1', 'e1a(23)', 'a1(23)',
    'e(123)', 'a(123)'
]

length = len(octal)

# This function checks if v1 implies v2.
def follows(v1, v2):
    def digit(d1, d2):
        d1 = int(d1)
        d2 = int(d2)
        if (d1 & d2 == d1) or (d1 == 2 and d2 == 5):  # bitwise smaller or central arrow
            return True
        else:
            return False

    result = True
    for i in range(3):
        if not digit(v1[i], v2[i]):
            result = False

    return result

hasse = np.zeros((length, length), np.int8)

for (i, v1) in enumerate(octal):
    for (j, v2) in enumerate(octal):
        hasse[i, j] = follows(v1, v2)

for i in range(length):
    hasse[i, i] = 0

# A Hasse diagram is supposed to contain no redundant edges,
# so if i implies k and k implies j, there shall be no edge from i to j.
def clean(mat):
    for i in range(length):
        for j in range(length):
            if mat[i, j] == 1:
                for k in range(length):
                    if hasse[i, k] == 1 and hasse[k, j] == 1:
                        mat[i, j] = 0
    return mat

hasse = clean(hasse)

if np.array_equal(hasse, clean(hasse)):
    print 'cleaning successful'

# Print the number of edges in the Hasse diagram (139).
print hasse.sum()

# Print all edges of the Hasse diagram.
for i in range(length):
    for j in range(length):
        if hasse[i, j] == 1:
            print abbr[i] + ' --> ' + abbr[j]

# The output can the found here: http://pastebin.com/ADwKEkgx