Certification Problem
Input (TPDB TRS_Standard/Various_04/06)
The rewrite relation of the following TRS is considered.
|
f(x,y) |
→ |
g1(x,x,y) |
(1) |
|
f(x,y) |
→ |
g1(y,x,x) |
(2) |
|
f(x,y) |
→ |
g2(x,y,y) |
(3) |
|
f(x,y) |
→ |
g2(y,y,x) |
(4) |
|
g1(x,x,y) |
→ |
h(x,y) |
(5) |
|
g1(y,x,x) |
→ |
h(x,y) |
(6) |
|
g2(x,y,y) |
→ |
h(x,y) |
(7) |
|
g2(y,y,x) |
→ |
h(x,y) |
(8) |
|
h(x,x) |
→ |
x |
(9) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [f(x1, x2)] |
= |
1 + 2 · x1 + 2 · x2
|
| [g1(x1, x2, x3)] |
= |
1 + 1 · x1 + 1 · x2 + 1 · x3
|
| [g2(x1, x2, x3)] |
= |
1 + 1 · x1 + 1 · x2 + 1 · x3
|
| [h(x1, x2)] |
= |
1 + 1 · x1 + 1 · x2
|
all of the following rules can be deleted.
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [f(x1, x2)] |
= |
2 + 2 · x1 + 2 · x2
|
| [g1(x1, x2, x3)] |
= |
1 + 1 · x1 + 1 · x2 + 1 · x3
|
| [g2(x1, x2, x3)] |
= |
1 + 1 · x1 + 1 · x2 + 1 · x3
|
| [h(x1, x2)] |
= |
1 + 1 · x1 + 1 · x2
|
all of the following rules can be deleted.
|
f(x,y) |
→ |
g1(x,x,y) |
(1) |
|
f(x,y) |
→ |
g1(y,x,x) |
(2) |
|
f(x,y) |
→ |
g2(x,y,y) |
(3) |
|
f(x,y) |
→ |
g2(y,y,x) |
(4) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [g1(x1, x2, x3)] |
= |
1 + 1 · x1 + 1 · x2 + 1 · x3
|
| [h(x1, x2)] |
= |
1 + 2 · x1 + 1 · x2
|
| [g2(x1, x2, x3)] |
= |
2 + 2 · x1 + 2 · x2 + 2 · x3
|
all of the following rules can be deleted.
|
g2(x,y,y) |
→ |
h(x,y) |
(7) |
|
g2(y,y,x) |
→ |
h(x,y) |
(8) |
1.1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(g1) |
= |
1 |
|
weight(g1) |
= |
0 |
|
|
|
| prec(h) |
= |
0 |
|
weight(h) |
= |
1 |
|
|
|
all of the following rules can be deleted.
|
g1(x,x,y) |
→ |
h(x,y) |
(5) |
|
g1(y,x,x) |
→ |
h(x,y) |
(6) |
1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.