Certification Problem
Input (TPDB TRS_Standard/SK90/4.48)
The rewrite relation of the following TRS is considered.
|
f(f(x,y,z),u,f(x,y,v)) |
→ |
f(x,y,f(z,u,v)) |
(1) |
|
f(x,y,y) |
→ |
y |
(2) |
|
f(x,y,g(y)) |
→ |
x |
(3) |
|
f(x,x,y) |
→ |
x |
(4) |
|
f(g(x),x,y) |
→ |
y |
(5) |
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) |
= |
1 |
|
weight(g) |
= |
2 |
|
|
|
| prec(f) |
= |
0 |
|
weight(f) |
= |
1 |
|
|
|
all of the following rules can be deleted.
|
f(f(x,y,z),u,f(x,y,v)) |
→ |
f(x,y,f(z,u,v)) |
(1) |
|
f(x,y,y) |
→ |
y |
(2) |
|
f(x,y,g(y)) |
→ |
x |
(3) |
|
f(x,x,y) |
→ |
x |
(4) |
|
f(g(x),x,y) |
→ |
y |
(5) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.