YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
| 
a(a(a(x0))) | 
→ | 
b(b(b(x0))) | 
| 
b(a(a(b(x0)))) | 
→ | 
x0 | 
| 
b(a(a(b(x0)))) | 
→ | 
b(a(a(a(b(x0))))) | 
Proof
1 Rule Removal
      Using the
      linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 
            over the arctic semiring over the integers
| [b(x1)] | 
 =  | 
 · 
                    x1 + 
| 
                -∞
             | 
                -∞
             | 
                -∞
             | 
 
| 
                -∞
             | 
                -∞
             | 
                -∞
             | 
 
| 
                -∞
             | 
                -∞
             | 
                -∞
             | 
 
 
 | 
| [a(x1)] | 
 =  | 
 · 
                    x1 + 
| 
                -∞
             | 
                -∞
             | 
                -∞
             | 
 
| 
                -∞
             | 
                -∞
             | 
                -∞
             | 
 
| 
                -∞
             | 
                -∞
             | 
                -∞
             | 
 
 
 | 
          the
          rules
| 
a(a(a(x0))) | 
→ | 
b(b(b(x0))) | 
| 
b(a(a(b(x0)))) | 
→ | 
b(a(a(a(b(x0))))) | 
          remain.
        1.1 Bounds
        The given TRS is 
        match-bounded by 1.
        This is shown by the following automaton.
        
- 
final states:
{5, 1}
 
- 
transitions:
| 3 | 
 →  | 
9 | 
| 5 | 
 →  | 
9 | 
| 5 | 
 →  | 
3 | 
| 12 | 
 →  | 
8 | 
| 
b0(3) | 
 →  | 
4 | 
| 
b0(8) | 
 →  | 
5 | 
| 
b0(4) | 
 →  | 
1 | 
| 
b0(2) | 
 →  | 
3 | 
| 
a0(7) | 
 →  | 
8 | 
| 
a0(6) | 
 →  | 
7 | 
| 
a0(3) | 
 →  | 
6 | 
| 
f20
 | 
 →  | 
2 | 
| 
b1(10) | 
 →  | 
11 | 
| 
b1(9) | 
 →  | 
10 | 
| 
b1(11) | 
 →  | 
12 |