Certification Problem
Input (TPDB TRS_Standard/Various_04/19)
The rewrite relation of the following TRS is considered.
|
:(x,x) |
→ |
e |
(1) |
|
:(x,e) |
→ |
x |
(2) |
|
i(:(x,y)) |
→ |
:(y,x) |
(3) |
|
:(:(x,y),z) |
→ |
:(x,:(z,i(y))) |
(4) |
|
:(e,x) |
→ |
i(x) |
(5) |
|
i(i(x)) |
→ |
x |
(6) |
|
i(e) |
→ |
e |
(7) |
|
:(x,:(y,i(x))) |
→ |
i(y) |
(8) |
|
:(x,:(y,:(i(x),z))) |
→ |
:(i(z),y) |
(9) |
|
:(i(x),:(y,x)) |
→ |
i(y) |
(10) |
|
:(i(x),:(y,:(x,z))) |
→ |
:(i(z),y) |
(11) |
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(i) |
= |
3 |
|
weight(i) |
= |
0 |
|
|
|
| prec(e) |
= |
2 |
|
weight(e) |
= |
4 |
|
|
|
| prec(:) |
= |
0 |
|
weight(:) |
= |
5 |
|
|
|
all of the following rules can be deleted.
|
:(x,x) |
→ |
e |
(1) |
|
:(x,e) |
→ |
x |
(2) |
|
i(:(x,y)) |
→ |
:(y,x) |
(3) |
|
:(:(x,y),z) |
→ |
:(x,:(z,i(y))) |
(4) |
|
:(e,x) |
→ |
i(x) |
(5) |
|
i(i(x)) |
→ |
x |
(6) |
|
i(e) |
→ |
e |
(7) |
|
:(x,:(y,i(x))) |
→ |
i(y) |
(8) |
|
:(x,:(y,:(i(x),z))) |
→ |
:(i(z),y) |
(9) |
|
:(i(x),:(y,x)) |
→ |
i(y) |
(10) |
|
:(i(x),:(y,:(x,z))) |
→ |
:(i(z),y) |
(11) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.