Certification Problem
Input (TPDB TRS_Equational/Mixed_AC/YWHM14_1)
The rewrite relation of the following equational TRS is considered.
|
f(plus(a,a)) |
→ |
plus(f(a),f(a)) |
(1) |
|
plus(a,f(f(a))) |
→ |
plus(f(a),f(a)) |
(2) |
Associative symbols: plus
Commutative symbols: plus
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
| [plus(x1, x2)] |
= |
1 + 2 · x2 + 2 · x1 + 2 · x1 · x2
|
| [f(x1)] |
= |
2 · x1
|
| [a] |
= |
0 |
the
rule
|
f(plus(a,a)) |
→ |
plus(f(a),f(a)) |
(1) |
can be deleted.
1.1 AC Rule Removal
Using the
non-linear polynomial interpretation over the naturals
| [plus(x1, x2)] |
= |
1 + 2 · x2 + 2 · x1 + 2 · x1 · x2
|
| [a] |
= |
2 |
| [f(x1)] |
= |
2 · x1
|
the
rule
|
plus(a,f(f(a))) |
→ |
plus(f(a),f(a)) |
(2) |
can be deleted.
1.1.1 R is empty
There are no rules in the TRS. Hence, it is AC-terminating.