YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
b(b(x0)) |
→ |
a(a(a(x0))) |
b(a(b(x0))) |
→ |
a(x0) |
b(a(a(x0))) |
→ |
b(a(b(x0))) |
Proof
1 Rule Removal
Using the
linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1
over the arctic semiring over the integers
[a(x1)] |
= |
·
x1 +
|
[b(x1)] |
= |
·
x1 +
|
the
rules
b(b(x0)) |
→ |
a(a(a(x0))) |
b(a(a(x0))) |
→ |
b(a(b(x0))) |
remain.
1.1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
b(b(x0)) |
→ |
a(a(a(x0))) |
a(a(b(x0))) |
→ |
b(a(b(x0))) |
1.1.1 Bounds
The given TRS is
match-bounded by 3.
This is shown by the following automaton.
-
final states:
{5, 1}
-
transitions:
15 |
→ |
7 |
15 |
→ |
6 |
15 |
→ |
1 |
5 |
→ |
4 |
5 |
→ |
3 |
34 |
→ |
44 |
31 |
→ |
13 |
31 |
→ |
33 |
1 |
→ |
6 |
14 |
→ |
16 |
14 |
→ |
28 |
12 |
→ |
32 |
7 |
→ |
12 |
35 |
→ |
17 |
35 |
→ |
29 |
47 |
→ |
13 |
47 |
→ |
4 |
47 |
→ |
5 |
47 |
→ |
19 |
47 |
→ |
31 |
19 |
→ |
5 |
a1(18) |
→ |
19 |
a1(16) |
→ |
17 |
a1(13) |
→ |
14 |
a1(17) |
→ |
18 |
b0(7) |
→ |
5 |
b0(2) |
→ |
6 |
b2(34) |
→ |
35 |
b2(32) |
→ |
33 |
b3(46) |
→ |
47 |
b3(44) |
→ |
45 |
a3(45) |
→ |
46 |
a2(28) |
→ |
29 |
a2(30) |
→ |
31 |
a2(29) |
→ |
30 |
a2(33) |
→ |
34 |
a0(3) |
→ |
4 |
a0(4) |
→ |
1 |
a0(2) |
→ |
3 |
a0(6) |
→ |
7 |
f20
|
→ |
2 |
b1(14) |
→ |
15 |
b1(12) |
→ |
13 |