YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
| 
a(b(x0)) | 
→ | 
b(b(a(x0))) | 
| 
c(b(x0)) | 
→ | 
b(c(c(x0))) | 
Proof
1 String Reversal
        Since only unary symbols occur, one can reverse all terms and obtains the TRS        
        
| 
b(a(x0)) | 
→ | 
a(b(b(x0))) | 
| 
b(c(x0)) | 
→ | 
c(c(b(x0))) | 
1.1 Rule Removal
      Using the
      linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 
            over the naturals
| [a(x1)] | 
 =  | 
 · 
                    x1 + 
 | 
| [b(x1)] | 
 =  | 
 · 
                    x1 + 
 | 
| [c(x1)] | 
 =  | 
 · 
                    x1 + 
 | 
          the
          rule
          remains.
        1.1.1 Bounds
        The given TRS is 
        match-bounded by 0.
        This is shown by the following automaton.
        
- 
final states:
{1}
 
- 
transitions:
| 1 | 
 →  | 
3 | 
| 
f30
 | 
 →  | 
2 | 
| 
c0(4) | 
 →  | 
1 | 
| 
c0(3) | 
 →  | 
4 | 
| 
b0(2) | 
 →  | 
3 |