Certification Problem
Input (TPDB TRS_Standard/SK90/4.03)
The rewrite relation of the following TRS is considered.
|
+(x,0) |
→ |
x |
(1) |
|
+(minus(x),x) |
→ |
0 |
(2) |
|
minus(0) |
→ |
0 |
(3) |
|
minus(minus(x)) |
→ |
x |
(4) |
|
minus(+(x,y)) |
→ |
+(minus(y),minus(x)) |
(5) |
|
*(x,1) |
→ |
x |
(6) |
|
*(x,0) |
→ |
0 |
(7) |
|
*(x,+(y,z)) |
→ |
+(*(x,y),*(x,z)) |
(8) |
|
*(x,minus(y)) |
→ |
minus(*(x,y)) |
(9) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
Weighted Path Order with the following precedence and status
| prec(*) |
= |
3 |
|
status(*) |
= |
[1, 2] |
|
list-extension(*) |
= |
Lex |
| prec(1) |
= |
0 |
|
status(1) |
= |
[] |
|
list-extension(1) |
= |
Lex |
| prec(minus) |
= |
1 |
|
status(minus) |
= |
[1] |
|
list-extension(minus) |
= |
Lex |
| prec(+) |
= |
0 |
|
status(+) |
= |
[1, 2] |
|
list-extension(+) |
= |
Lex |
| prec(0) |
= |
0 |
|
status(0) |
= |
[] |
|
list-extension(0) |
= |
Lex |
and the following
Max-polynomial interpretation
| [*(x1, x2)] |
=
|
0 + 1 · x1 + 1 · x2
|
| [1] |
=
|
1 |
| [minus(x1)] |
=
|
max(1, 0 + 1 · x1) |
| [+(x1, x2)] |
=
|
max(3, 2 + 1 · x1, 2 + 1 · x2) |
| [0] |
=
|
max(0) |
all of the following rules can be deleted.
|
+(x,0) |
→ |
x |
(1) |
|
+(minus(x),x) |
→ |
0 |
(2) |
|
minus(0) |
→ |
0 |
(3) |
|
minus(minus(x)) |
→ |
x |
(4) |
|
minus(+(x,y)) |
→ |
+(minus(y),minus(x)) |
(5) |
|
*(x,1) |
→ |
x |
(6) |
|
*(x,0) |
→ |
0 |
(7) |
|
*(x,+(y,z)) |
→ |
+(*(x,y),*(x,z)) |
(8) |
|
*(x,minus(y)) |
→ |
minus(*(x,y)) |
(9) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.