Certification Problem
Input (TPDB TRS_Standard/Der95/27)
The rewrite relation of the following TRS is considered.
|
h(f(x),y) |
→ |
f(g(x,y)) |
(1) |
|
g(x,y) |
→ |
h(x,y) |
(2) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(g) |
= |
2 |
|
weight(g) |
= |
0 |
|
|
|
| prec(h) |
= |
1 |
|
weight(h) |
= |
0 |
|
|
|
| prec(f) |
= |
0 |
|
weight(f) |
= |
2 |
|
|
|
all of the following rules can be deleted.
|
h(f(x),y) |
→ |
f(g(x,y)) |
(1) |
|
g(x,y) |
→ |
h(x,y) |
(2) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.