YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
a(a(d(d(x0)))) |
→ |
d(d(b(b(x0)))) |
a(a(x0)) |
→ |
b(b(b(b(b(b(x0)))))) |
b(b(d(d(b(b(x0)))))) |
→ |
a(a(c(c(x0)))) |
c(c(x0)) |
→ |
d(d(x0)) |
Proof
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
[a(x1)] |
= |
1 ·
x1 + 4 |
[c(x1)] |
= |
2 ·
x1 + 0 |
[d(x1)] |
= |
2 ·
x1 + 0 |
[b(x1)] |
= |
1 ·
x1 + 1 |
the
rules
a(a(d(d(x0)))) |
→ |
d(d(b(b(x0)))) |
c(c(x0)) |
→ |
d(d(x0)) |
remain.
1.1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
[a(x1)] |
= |
12 ·
x1 +
-∞
|
[c(x1)] |
= |
0 ·
x1 +
-∞
|
[d(x1)] |
= |
0 ·
x1 +
-∞
|
[b(x1)] |
= |
7 ·
x1 +
-∞
|
the
rule
remains.
1.1.1 Rule Removal
Using the
linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1
over the arctic semiring over the integers
[c(x1)] |
= |
·
x1 +
|
[d(x1)] |
= |
·
x1 +
|
all rules could be removed.
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.