def implication(a, b):
    return not a or b

# полная таблица истинности (как в вашем коде)
table = []
for x in (0, 1):
    for y in (0, 1):
        for z in (0, 1):
            for w in (0, 1):
                f1 = int(implication(w, y) == implication(z, x))
                f2 = int(implication(w, y) and (not x) == z)
                table.append({'x': x, 'y': y, 'z': z, 'w': w, 'f1': f1, 'f2': f2})

# фрагмент из задания ('?' - пусто)
rows = [
    [0, '?', 0, 0, 0, 1],
    [0, 0, 0, '?', 0, '?'],
    [0, 1, 1, '?', '?', 0],
]

# перебираем, какая переменная в каком столбце
for a in 'xyzw':
    for b in 'xyzw':
        for c in 'xyzw':
            for d in 'xyzw':
                if len({a, b, c, d}) < 4:
                    continue
                cols = [a, b, c, d, 'f1', 'f2']
                ok = True
                for row in rows:
                    found = False
                    for t in table:
                        match = True
                        for i in range(6):
                            if row[i] != '?' and row[i] != t[cols[i]]:
                                match = False
                        if match:
                            found = True
                    if not found:
                        ok = False
                if ok:
                    print(a + b + c + d)