Certification Problem
Input (TPDB TRS_Standard/SK90/2.58)
The rewrite relation of the following TRS is considered.
|
f(x,y) |
→ |
x |
(1) |
|
g(a) |
→ |
h(a,b,a) |
(2) |
|
i(x) |
→ |
f(x,x) |
(3) |
|
h(x,x,y) |
→ |
g(x) |
(4) |
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)] |
= |
2 + 1 · x1 + 1 · x2
|
| [g(x1)] |
= |
1 · x1
|
| [a] |
= |
0 |
| [h(x1, x2, x3)] |
= |
2 · x1 + 2 · x2 + 1 · x3
|
| [b] |
= |
0 |
| [i(x1)] |
= |
2 + 2 · x1
|
all of the following rules can be deleted.
1.1 Rule Removal
Using the
| prec(g) |
= |
2 |
|
stat(g) |
= |
lex
|
| prec(a) |
= |
2 |
|
stat(a) |
= |
mul
|
| prec(h) |
= |
2 |
|
stat(h) |
= |
lex
|
| prec(b) |
= |
1 |
|
stat(b) |
= |
mul
|
| prec(i) |
= |
3 |
|
stat(i) |
= |
mul
|
| prec(f) |
= |
0 |
|
stat(f) |
= |
mul
|
| π(g) |
= |
[1] |
| π(a) |
= |
[] |
| π(h) |
= |
[2,1,3] |
| π(b) |
= |
[] |
| π(i) |
= |
[1] |
| π(f) |
= |
[1,2] |
all of the following rules can be deleted.
|
g(a) |
→ |
h(a,b,a) |
(2) |
|
i(x) |
→ |
f(x,x) |
(3) |
|
h(x,x,y) |
→ |
g(x) |
(4) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.