Certification Problem
Input (TPDB TRS_Equational/AProVE_AC_04/AC11)
The rewrite relation of the following equational TRS is considered.
|
f(g(f(h(a),a)),a) |
→ |
f(h(a),f(a,a)) |
(1) |
|
f(h(a),g(a)) |
→ |
f(g(h(a)),a) |
(2) |
|
f(g(h(a)),f(b,f(b,b))) |
→ |
f(g(f(h(a),a)),a) |
(3) |
|
f(h(a),a) |
→ |
f(h(a),b) |
(4) |
Associative symbols: f
Commutative symbols: f
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 AC Rule Removal
Using the
linear polynomial interpretation over the naturals
| [f(x1, x2)] |
= |
1 · x2 + 1 · x1
|
| [g(x1)] |
= |
1 + 1 · x1
|
| [h(x1)] |
= |
1 · x1
|
| [a] |
= |
0 |
| [b] |
= |
0 |
the
rule
|
f(g(f(h(a),a)),a) |
→ |
f(h(a),f(a,a)) |
(1) |
can be deleted.
1.1 AC Rule Removal
Using the
linear polynomial interpretation over the naturals
| [f(x1, x2)] |
= |
2 + 1 · x2 + 1 · x1
|
| [h(x1)] |
= |
2 + 1 · x1
|
| [a] |
= |
0 |
| [g(x1)] |
= |
1 · x1
|
| [b] |
= |
0 |
the
rule
|
f(g(h(a)),f(b,f(b,b))) |
→ |
f(g(f(h(a),a)),a) |
(3) |
can be deleted.
1.1.1 AC Rule Removal
Using the
linear polynomial interpretation over the naturals
| [f(x1, x2)] |
= |
3 + 1 · x2 + 1 · x1
|
| [h(x1)] |
= |
2 + 1 · x1
|
| [a] |
= |
2 |
| [g(x1)] |
= |
1 · x1
|
| [b] |
= |
1 |
the
rule
|
f(h(a),a) |
→ |
f(h(a),b) |
(4) |
can be deleted.
1.1.1.1 AC Rule Removal
Using the
non-linear polynomial interpretation over the naturals
| [f(x1, x2)] |
= |
1 · x2 + 1 · x1 + 2 · x1 · x2
|
| [h(x1)] |
= |
2 + 1 · x1
|
| [a] |
= |
0 |
| [g(x1)] |
= |
2 + 1 · x1
|
the
rule
|
f(h(a),g(a)) |
→ |
f(g(h(a)),a) |
(2) |
can be deleted.
1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is AC-terminating.