Certification Problem
Input (TPDB TRS_Standard/SK90/4.52)
The rewrite relation of the following TRS is considered.
|
s(a) |
→ |
a |
(1) |
|
s(s(x)) |
→ |
x |
(2) |
|
s(f(x,y)) |
→ |
f(s(y),s(x)) |
(3) |
|
s(g(x,y)) |
→ |
g(s(x),s(y)) |
(4) |
|
f(x,a) |
→ |
x |
(5) |
|
f(a,y) |
→ |
y |
(6) |
|
f(g(x,y),g(u,v)) |
→ |
g(f(x,u),f(y,v)) |
(7) |
|
g(a,a) |
→ |
a |
(8) |
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) |
= |
0 |
|
weight(g) |
= |
0 |
|
|
|
| prec(f) |
= |
2 |
|
weight(f) |
= |
0 |
|
|
|
| prec(s) |
= |
3 |
|
weight(s) |
= |
0 |
|
|
|
| prec(a) |
= |
1 |
|
weight(a) |
= |
1 |
|
|
|
all of the following rules can be deleted.
|
s(a) |
→ |
a |
(1) |
|
s(s(x)) |
→ |
x |
(2) |
|
s(f(x,y)) |
→ |
f(s(y),s(x)) |
(3) |
|
s(g(x,y)) |
→ |
g(s(x),s(y)) |
(4) |
|
f(x,a) |
→ |
x |
(5) |
|
f(a,y) |
→ |
y |
(6) |
|
f(g(x,y),g(u,v)) |
→ |
g(f(x,u),f(y,v)) |
(7) |
|
g(a,a) |
→ |
a |
(8) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.