Certification Problem
Input (TPDB TRS_Relative/Relative_05/rt2-5)
The relative rewrite relation R/S is considered where R is the following TRS
and S is the following TRS.
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Closure Under Flat Contexts
Using the flat contexts
{f(☐), g(☐)}
We obtain the transformed TRS
|
f(f(f(x))) |
→ |
f(x) |
(3) |
|
g(f(f(x))) |
→ |
g(x) |
(4) |
|
f(f(x)) |
→ |
f(g(f(g(f(x))))) |
(5) |
|
g(f(x)) |
→ |
g(g(f(g(f(x))))) |
(6) |
1.1 Semantic Labeling
Root-labeling is applied.
We obtain the labeled TRS
|
ff(ff(ff(x))) |
→ |
ff(x) |
(7) |
|
ff(ff(fg(x))) |
→ |
fg(x) |
(8) |
|
gf(ff(ff(x))) |
→ |
gf(x) |
(9) |
|
gf(ff(fg(x))) |
→ |
gg(x) |
(10) |
|
ff(ff(x)) |
→ |
fg(gf(fg(gf(ff(x))))) |
(11) |
|
ff(fg(x)) |
→ |
fg(gf(fg(gf(fg(x))))) |
(12) |
|
gf(ff(x)) |
→ |
gg(gf(fg(gf(ff(x))))) |
(13) |
|
gf(fg(x)) |
→ |
gg(gf(fg(gf(fg(x))))) |
(14) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [ff(x1)] |
= |
1 + 1 · x1
|
| [fg(x1)] |
= |
1 · x1
|
| [gf(x1)] |
= |
1 · x1
|
| [gg(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
ff(ff(ff(x))) |
→ |
ff(x) |
(7) |
|
ff(ff(fg(x))) |
→ |
fg(x) |
(8) |
|
gf(ff(ff(x))) |
→ |
gf(x) |
(9) |
|
gf(ff(fg(x))) |
→ |
gg(x) |
(10) |
|
ff(ff(x)) |
→ |
fg(gf(fg(gf(ff(x))))) |
(11) |
|
ff(fg(x)) |
→ |
fg(gf(fg(gf(fg(x))))) |
(12) |
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.