Certification Problem
Input (TPDB TRS_Standard/Der95/01)
The rewrite relation of the following TRS is considered.
|
\(x,x) |
→ |
e |
(1) |
|
/(x,x) |
→ |
e |
(2) |
|
.(e,x) |
→ |
x |
(3) |
|
.(x,e) |
→ |
x |
(4) |
|
\(e,x) |
→ |
x |
(5) |
|
/(x,e) |
→ |
x |
(6) |
|
.(x,\(x,y)) |
→ |
y |
(7) |
|
.(/(y,x),x) |
→ |
y |
(8) |
|
\(x,.(x,y)) |
→ |
y |
(9) |
|
/(.(y,x),x) |
→ |
y |
(10) |
|
/(x,\(y,x)) |
→ |
y |
(11) |
|
\(/(x,y),x) |
→ |
y |
(12) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(.) |
= |
2 |
|
weight(.) |
= |
0 |
|
|
|
| prec(/) |
= |
3 |
|
weight(/) |
= |
0 |
|
|
|
| prec(e) |
= |
0 |
|
weight(e) |
= |
2 |
|
|
|
| prec(\) |
= |
1 |
|
weight(\) |
= |
0 |
|
|
|
all of the following rules can be deleted.
|
\(x,x) |
→ |
e |
(1) |
|
/(x,x) |
→ |
e |
(2) |
|
.(e,x) |
→ |
x |
(3) |
|
.(x,e) |
→ |
x |
(4) |
|
\(e,x) |
→ |
x |
(5) |
|
/(x,e) |
→ |
x |
(6) |
|
.(x,\(x,y)) |
→ |
y |
(7) |
|
.(/(y,x),x) |
→ |
y |
(8) |
|
\(x,.(x,y)) |
→ |
y |
(9) |
|
/(.(y,x),x) |
→ |
y |
(10) |
|
/(x,\(y,x)) |
→ |
y |
(11) |
|
\(/(x,y),x) |
→ |
y |
(12) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.