Certification Problem
Input (TPDB TRS_Relative/Relative_05/rt1-3)
The relative rewrite relation R/S is considered where R is the following TRS
|
t(f(x),g(y),f(z)) |
→ |
t(z,g(x),g(y)) |
(1) |
|
t(g(x),g(y),f(z)) |
→ |
t(f(y),f(z),x) |
(2) |
and S is the following TRS.
|
g(g(x)) |
→ |
f(f(x)) |
(3) |
|
f(f(x)) |
→ |
g(g(x)) |
(4) |
|
g(f(x)) |
→ |
f(g(x)) |
(5) |
|
f(g(x)) |
→ |
g(f(x)) |
(6) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
matrix interpretations of dimension 2 with strict dimension 1 over the integers
| [t(x1, x2, x3)] |
= |
+ · x1 + · x2 + · x3
|
| [f(x1)] |
= |
+ · x1
|
| [g(x1)] |
= |
+ · x1
|
all of the following rules can be deleted.
|
t(g(x),g(y),f(z)) |
→ |
t(f(y),f(z),x) |
(2) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [t(x1, x2, x3)] |
= |
1 · x1 + 1 · x2 + 1 · x3
|
| [f(x1)] |
= |
1 + 1 · x1
|
| [g(x1)] |
= |
1 + 1 · x1
|
all of the following rules can be deleted.
|
t(f(x),g(y),f(z)) |
→ |
t(z,g(x),g(y)) |
(1) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.