Certification Problem
Input (TPDB TRS_Standard/Transformed_CSR_04/Ex15_Luc06_FR)
The rewrite relation of the following TRS is considered.
f(n__f(n__a)) |
→ |
f(n__g(n__f(n__a))) |
(1) |
f(X) |
→ |
n__f(X) |
(2) |
a |
→ |
n__a |
(3) |
g(X) |
→ |
n__g(X) |
(4) |
activate(n__f(X)) |
→ |
f(X) |
(5) |
activate(n__a) |
→ |
a |
(6) |
activate(n__g(X)) |
→ |
g(activate(X)) |
(7) |
activate(X) |
→ |
X |
(8) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
[a] |
= |
|
[n__f(x1)] |
= |
· x1 +
|
[n__g(x1)] |
= |
· x1 +
|
[g(x1)] |
= |
· x1 +
|
[n__a] |
= |
|
[activate(x1)] |
= |
· x1 +
|
[f(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
f(X) |
→ |
n__f(X) |
(2) |
a |
→ |
n__a |
(3) |
g(X) |
→ |
n__g(X) |
(4) |
activate(n__f(X)) |
→ |
f(X) |
(5) |
activate(n__a) |
→ |
a |
(6) |
activate(n__g(X)) |
→ |
g(activate(X)) |
(7) |
activate(X) |
→ |
X |
(8) |
1.1 Rule Removal
Using the
linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1
over the naturals
[n__f(x1)] |
= |
|
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
|
|
· x1 +
|
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
|
|
[n__g(x1)] |
= |
|
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
|
· x1 +
|
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
|
|
[n__a] |
= |
|
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
|
|
|
[f(x1)] |
= |
|
1 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
|
· x1 +
|
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
|
|
all of the following rules can be deleted.
f(n__f(n__a)) |
→ |
f(n__g(n__f(n__a))) |
(1) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.