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 |