The result of expression '(k ^ ~v)  & k' is exactly
the same with  'k & v', so simplify it.
(k ^ ~v) & k == k & v
The truth table (in non table form):
k == 0, v == 0:
  (k ^ ~v) & k == (0 ^ ~0) & 0 == (0 ^ 1) & 0 == 1 & 0 == 0
  k & v == 0 & 0 == 0

k == 0, v == 1:
  (k ^ ~v) & k == (0 ^ ~1) & 0 == (0 ^ 0) & 0 == 1 & 0 == 0
  k & v == 0 & 1 == 0

k == 1, v == 0:
  (k ^ ~v) & k == (1 ^ ~0) & 1 == (1 ^ 1) & 1 == 0 & 1 == 0
  k & v == 1 & 0 == 0

k == 1, v == 1:
  (k ^ ~v) & k == (1 ^ ~1) & 1 == (1 ^ 0) & 1 == 1 & 1 == 1
  k & v == 1 & 1 == 1
Signed-off-by: Jian Shen <shenjian15@huawei.com>
Signed-off-by: Hao Lan <lanhao@huawei.com>
Signed-off-by: Paolo Abeni <pabeni@redhat.com>