Certification Problem
Input (TPDB TRS_Equational/Mixed_AC/YWHM14_4)
The rewrite relation of the following equational TRS is considered.
|
ac1(a,ac2(b,c)) |
→ |
ac1(b,f(ac2(a,c))) |
(1) |
|
ac2(a,ac1(b,c)) |
→ |
ac2(b,f(ac1(a,c))) |
(2) |
Associative symbols: ac1, ac2
Commutative symbols: ac1, ac2
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 AC Rule Removal
Using the
non-linear polynomial interpretation over the naturals
| [ac1(x1, x2)] |
= |
2 + 2 · x2 + 2 · x1 + 1 · x1 · x2
|
| [ac2(x1, x2)] |
= |
1 · x2 + 1 · x1
|
| [a] |
= |
0 |
| [b] |
= |
2 |
| [c] |
= |
0 |
| [f(x1)] |
= |
1 · x1
|
the
rule
|
ac2(a,ac1(b,c)) |
→ |
ac2(b,f(ac1(a,c))) |
(2) |
can be deleted.
1.1 AC Rule Removal
Using the
non-linear polynomial interpretation over the naturals
| [ac1(x1, x2)] |
= |
1 + 2 · x2 + 2 · x1 + 2 · x1 · x2
|
| [ac2(x1, x2)] |
= |
2 + 2 · x2 + 2 · x1 + 1 · x1 · x2
|
| [a] |
= |
2 |
| [b] |
= |
0 |
| [c] |
= |
0 |
| [f(x1)] |
= |
1 · x1
|
the
rule
|
ac1(a,ac2(b,c)) |
→ |
ac1(b,f(ac2(a,c))) |
(1) |
can be deleted.
1.1.1 R is empty
There are no rules in the TRS. Hence, it is AC-terminating.