YES

After renaming modulo { a->0, b->1 }, 
it remains to prove termination of the 1-rule system
{ 0 1 0 0 0 0 -> 0 0 0 0 0 1 0 0 1 0 0 1 }


The system was reversed.

After renaming modulo { 0->0, 1->1 }, 
it remains to prove termination of the 1-rule system
{ 0 0 0 0 1 0 -> 1 0 0 1 0 0 1 0 0 0 0 0 }


Applying the dependency pairs transformation.

After renaming modulo { (0,true)->0, (0,false)->1, (1,false)->2 },
it remains to prove termination of the 10-rule system
{ 0 1 1 1 2 1 -> 0 1 2 1 1 2 1 1 1 1 1 ,
  0 1 1 1 2 1 -> 0 2 1 1 2 1 1 1 1 1 ,
  0 1 1 1 2 1 -> 0 1 2 1 1 1 1 1 ,
  0 1 1 1 2 1 -> 0 2 1 1 1 1 1 ,
  0 1 1 1 2 1 -> 0 1 1 1 1 ,
  0 1 1 1 2 1 -> 0 1 1 1 ,
  0 1 1 1 2 1 -> 0 1 1 ,
  0 1 1 1 2 1 -> 0 1 ,
  0 1 1 1 2 1 -> 0 ,
  1 1 1 1 2 1 ->= 2 1 1 2 1 1 2 1 1 1 1 1 }


The system was filtered by the following matrix interpretation
of type E_J with J = {1,...,2} and dimension 7:

0 is interpreted by
 /             \
| 1 0 1 0 0 0 0 |
| 0 1 0 0 0 0 0 |
| 0 0 0 0 0 0 0 |
| 0 0 0 0 0 0 0 |
| 0 0 0 0 0 0 0 |
| 0 0 0 0 0 0 0 |
| 0 0 0 0 0 0 0 |
 \             /
1 is interpreted by
 /             \
| 1 0 0 0 0 0 0 |
| 0 1 0 0 0 0 0 |
| 0 0 0 3 0 0 0 |
| 0 0 0 0 1 0 0 |
| 0 0 0 0 0 1 0 |
| 0 0 1 0 0 0 0 |
| 0 1 1 0 0 0 0 |
 \             /
2 is interpreted by
 /             \
| 1 0 0 0 0 0 0 |
| 0 1 0 0 0 0 0 |
| 0 0 0 0 0 0 0 |
| 0 0 0 0 0 0 0 |
| 0 0 0 0 0 0 0 |
| 0 0 1 1 1 0 1 |
| 0 0 0 0 0 0 0 |
 \             /

After renaming modulo { 1->0, 2->1 }, 
it remains to prove termination of the 1-rule system
{ 0 0 0 0 1 0 ->= 1 0 0 1 0 0 1 0 0 0 0 0 }

The system is trivially terminating.