Certification Problem
Input (TPDB TRS_Equational/Mixed_AC/YWHM14_3)
The rewrite relation of the following equational TRS is considered.
|
f(plus(x,y)) |
→ |
plus(f(x),y) |
(1) |
|
plus(g(x),y) |
→ |
g(plus(x,y)) |
(2) |
|
plus(f(a),g(b)) |
→ |
plus(f(b),g(a)) |
(3) |
|
h(a,b) |
→ |
h(b,a) |
(4) |
|
h(a,g(g(a))) |
→ |
h(g(a),f(a)) |
(5) |
|
h(g(a),a) |
→ |
h(a,g(b)) |
(6) |
|
h(g(a),b) |
→ |
h(a,g(a)) |
(7) |
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)] |
= |
2 + 2 · x2 + 2 · x1 + 1 · x1 · x2
|
| [f(x1)] |
= |
2 + 3 · x1
|
| [g(x1)] |
= |
3 + 1 · x1
|
| [a] |
= |
0 |
| [b] |
= |
0 |
| [h(x1, x2)] |
= |
1 · x2 + 1 · x1
|
the
rules
|
f(plus(x,y)) |
→ |
plus(f(x),y) |
(1) |
|
plus(g(x),y) |
→ |
g(plus(x,y)) |
(2) |
|
h(a,g(g(a))) |
→ |
h(g(a),f(a)) |
(5) |
can be deleted.
1.1 AC Rule Removal
Using the
non-linear polynomial interpretation over the naturals
| [plus(x1, x2)] |
= |
3 + 1 · x2 + 1 · x1
|
| [f(x1)] |
= |
2 · x1
|
| [a] |
= |
2 |
| [g(x1)] |
= |
1 · x1 · x1
|
| [b] |
= |
0 |
| [h(x1, x2)] |
= |
1 · x2 + 2 · x1
|
the
rules
|
h(a,b) |
→ |
h(b,a) |
(4) |
|
h(g(a),a) |
→ |
h(a,g(b)) |
(6) |
can be deleted.
1.1.1 AC Rule Removal
Using the
linear polynomial interpretation over the naturals
| [plus(x1, x2)] |
= |
3 + 1 · x2 + 1 · x1
|
| [f(x1)] |
= |
3 · x1
|
| [a] |
= |
2 |
| [g(x1)] |
= |
2 · x1
|
| [b] |
= |
0 |
| [h(x1, x2)] |
= |
1 · x2 + 2 · x1
|
the
rule
|
plus(f(a),g(b)) |
→ |
plus(f(b),g(a)) |
(3) |
can be deleted.
1.1.1.1 AC Rule Removal
Using the
non-linear polynomial interpretation over the naturals
| [plus(x1, x2)] |
= |
3 + 3 · x2 + 3 · x1 + 2 · x1 · x2
|
| [h(x1, x2)] |
= |
1 · x2 + 2 · x1
|
| [g(x1)] |
= |
1 · x1 · x1
|
| [a] |
= |
0 |
| [b] |
= |
1 |
the
rule
|
h(g(a),b) |
→ |
h(a,g(a)) |
(7) |
can be deleted.
1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is AC-terminating.