Certification Problem
Input (TPDB TRS_Relative/INVY_15/#3.56_rand)
The relative rewrite relation R/S is considered where R is the following TRS
g(c(x,s(y))) |
→ |
g(c(s(x),y)) |
(1) |
f(c(s(x),y)) |
→ |
f(c(x,s(y))) |
(2) |
f(f(x)) |
→ |
f(d(f(x))) |
(3) |
f(x) |
→ |
x |
(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
|
[c(x1, x2)] |
= |
1 · x1 + 1 · x2
|
[s(x1)] |
= |
1 · x1
|
[f(x1)] |
= |
1 + 1 · x1
|
[d(x1)] |
= |
1 · x1
|
[rand(x1)] |
= |
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 · x1
|
[c(x1, x2)] |
= |
1 · x1 + 1 · x2
|
[s(x1)] |
= |
1 · x1
|
[f(x1)] |
= |
1 · x1
|
[d(x1)] |
= |
1 · x1
|
[rand(x1)] |
= |
1 + 1 · x1
|
all of the following rules can be deleted.
1.1.1 Rule Removal
Using the
matrix interpretations of dimension 2 with strict dimension 1 over the integers
[g(x1)] |
= |
+ · x1
|
[c(x1, x2)] |
= |
+ · x1 + · x2
|
[s(x1)] |
= |
+ · x1
|
[f(x1)] |
= |
+ · x1
|
[d(x1)] |
= |
+ · x1
|
[rand(x1)] |
= |
+ · x1
|
all of the following rules can be deleted.
1.1.1.1 Rule Removal
Using the
matrix interpretations of dimension 2 with strict dimension 1 over the integers
[g(x1)] |
= |
+ · x1
|
[c(x1, x2)] |
= |
+ · x1 + · x2
|
[s(x1)] |
= |
+ · x1
|
[f(x1)] |
= |
+ · x1
|
[rand(x1)] |
= |
+ · x1
|
all of the following rules can be deleted.
g(c(x,s(y))) |
→ |
g(c(s(x),y)) |
(1) |
1.1.1.1.1 Rule Removal
Using the
matrix interpretations of dimension 2 with strict dimension 1 over the integers
[f(x1)] |
= |
+ · x1
|
[c(x1, x2)] |
= |
+ · x1 + · x2
|
[s(x1)] |
= |
+ · x1
|
[rand(x1)] |
= |
+ · x1
|
all of the following rules can be deleted.
f(c(s(x),y)) |
→ |
f(c(x,s(y))) |
(2) |
1.1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.