YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
| 
a(p(x0)) | 
→ | 
p(a(A(x0))) | 
| 
a(A(x0)) | 
→ | 
A(a(x0)) | 
| 
p(A(A(x0))) | 
→ | 
a(p(x0)) | 
Proof
1 String Reversal
        Since only unary symbols occur, one can reverse all terms and obtains the TRS        
        
| 
p(a(x0)) | 
→ | 
A(a(p(x0))) | 
| 
A(a(x0)) | 
→ | 
a(A(x0)) | 
| 
A(A(p(x0))) | 
→ | 
p(a(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 + 
 | 
| [p(x1)] | 
 =  | 
 · 
                    x1 + 
 | 
| [A(x1)] | 
 =  | 
 · 
                    x1 + 
 | 
          the
          rules
| 
p(a(x0)) | 
→ | 
A(a(p(x0))) | 
| 
A(a(x0)) | 
→ | 
a(A(x0)) | 
          remain.
        1.1.1 String Reversal
        Since only unary symbols occur, one can reverse all terms and obtains the TRS        
        
| 
a(p(x0)) | 
→ | 
p(a(A(x0))) | 
| 
a(A(x0)) | 
→ | 
A(a(x0)) | 
1.1.1.1 Bounds
        The given TRS is 
        match-bounded by 1.
        This is shown by the following automaton.
        
- 
final states:
{5, 1}
 
- 
transitions:
| 9 | 
 →  | 
4 | 
| 5 | 
 →  | 
6 | 
| 5 | 
 →  | 
8 | 
| 1 | 
 →  | 
6 | 
| 1 | 
 →  | 
8 | 
| 2 | 
 →  | 
7 | 
| 
a1(7) | 
 →  | 
8 | 
| 
f30
 | 
 →  | 
2 | 
| 
p0(4) | 
 →  | 
1 | 
| 
A0(2) | 
 →  | 
3 | 
| 
A0(6) | 
 →  | 
5 | 
| 
A1(8) | 
 →  | 
9 | 
| 
a0(2) | 
 →  | 
6 | 
| 
a0(3) | 
 →  | 
4 |