Certification Problem
Input (TPDB TRS_Relative/INVY_15/#3.35_rand)
The relative rewrite relation R/S is considered where R is the following TRS
|
g(s(x)) |
→ |
f(x) |
(1) |
|
f(0) |
→ |
s(0) |
(2) |
|
f(s(x)) |
→ |
s(s(g(x))) |
(3) |
|
g(0) |
→ |
0 |
(4) |
and S is the following TRS.
|
rand(x) |
→ |
rand(s(x)) |
(5) |
|
rand(x) |
→ |
x |
(6) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [g(x1)] |
= |
1 · x1
|
| [s(x1)] |
= |
1 · x1
|
| [f(x1)] |
= |
1 · x1
|
| [0] |
= |
0 |
| [rand(x1)] |
= |
1 + 1 · x1
|
all of the following rules can be deleted.
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [g(x1)] |
= |
1 + 1 · x1
|
| [s(x1)] |
= |
1 · x1
|
| [f(x1)] |
= |
1 + 1 · x1
|
| [0] |
= |
0 |
| [rand(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
f(0) |
→ |
s(0) |
(2) |
|
g(0) |
→ |
0 |
(4) |
1.1.1 Rule Removal
Using the
matrix interpretations of dimension 2 with strict dimension 1 over the integers
| [g(x1)] |
= |
+ · x1
|
| [s(x1)] |
= |
+ · x1
|
| [f(x1)] |
= |
+ · x1
|
| [rand(x1)] |
= |
+ · x1
|
all of the following rules can be deleted.
|
g(s(x)) |
→ |
f(x) |
(1) |
|
f(s(x)) |
→ |
s(s(g(x))) |
(3) |
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.