Certification Problem
Input (TPDB TRS_Innermost/AG01_innermost/#4.20a)
The rewrite relation of the following TRS is considered.
|
f(f(x)) |
→ |
f(x) |
(1) |
|
f(s(x)) |
→ |
f(x) |
(2) |
|
g(s(0)) |
→ |
g(f(s(0))) |
(3) |
The evaluation strategy is innermost.Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Semantic Labeling
Root-labeling is applied.
We obtain the labeled TRS
|
ff(ff(x)) |
→ |
ff(x) |
(4) |
|
ff(fs(x)) |
→ |
fs(x) |
(5) |
|
ff(fg(x)) |
→ |
fg(x) |
(6) |
|
ff(f0(x)) |
→ |
f0(x) |
(7) |
|
fs(sf(x)) |
→ |
ff(x) |
(8) |
|
fs(ss(x)) |
→ |
fs(x) |
(9) |
|
fs(sg(x)) |
→ |
fg(x) |
(10) |
|
fs(s0(x)) |
→ |
f0(x) |
(11) |
|
gs(s0(0)) |
→ |
gf(fs(s0(0))) |
(12) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [ff(x1)] |
= |
1 · x1
|
| [fs(x1)] |
= |
1 · x1
|
| [fg(x1)] |
= |
1 · x1
|
| [f0(x1)] |
= |
1 · x1
|
| [sf(x1)] |
= |
1 · x1 + 1 |
| [ss(x1)] |
= |
1 · x1 + 1 |
| [sg(x1)] |
= |
1 · x1 + 1 |
| [s0(x1)] |
= |
1 · x1 + 1 |
| [gs(x1)] |
= |
1 · x1 + 1 |
| [0] |
= |
0 |
| [gf(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
fs(sf(x)) |
→ |
ff(x) |
(8) |
|
fs(ss(x)) |
→ |
fs(x) |
(9) |
|
fs(sg(x)) |
→ |
fg(x) |
(10) |
|
fs(s0(x)) |
→ |
f0(x) |
(11) |
|
gs(s0(0)) |
→ |
gf(fs(s0(0))) |
(12) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [ff(x1)] |
= |
1 · x1 + 1 |
| [fs(x1)] |
= |
1 · x1
|
| [fg(x1)] |
= |
1 · x1
|
| [f0(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
ff(ff(x)) |
→ |
ff(x) |
(4) |
|
ff(fs(x)) |
→ |
fs(x) |
(5) |
|
ff(fg(x)) |
→ |
fg(x) |
(6) |
|
ff(f0(x)) |
→ |
f0(x) |
(7) |
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.