Certification Problem
Input (TPDB TRS_Standard/Der95/08)
The rewrite relation of the following TRS is considered.
|
D(t) |
→ |
1 |
(1) |
|
D(constant) |
→ |
0 |
(2) |
|
D(+(x,y)) |
→ |
+(D(x),D(y)) |
(3) |
|
D(*(x,y)) |
→ |
+(*(y,D(x)),*(x,D(y))) |
(4) |
|
D(-(x,y)) |
→ |
-(D(x),D(y)) |
(5) |
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(-) |
= |
0 |
|
status(-) |
= |
[2, 1] |
|
list-extension(-) |
= |
Lex |
| prec(*) |
= |
0 |
|
status(*) |
= |
[1, 2] |
|
list-extension(*) |
= |
Lex |
| prec(+) |
= |
0 |
|
status(+) |
= |
[1, 2] |
|
list-extension(+) |
= |
Lex |
| prec(0) |
= |
0 |
|
status(0) |
= |
[] |
|
list-extension(0) |
= |
Lex |
| prec(constant) |
= |
0 |
|
status(constant) |
= |
[] |
|
list-extension(constant) |
= |
Lex |
| prec(1) |
= |
0 |
|
status(1) |
= |
[] |
|
list-extension(1) |
= |
Lex |
| prec(D) |
= |
5 |
|
status(D) |
= |
[1] |
|
list-extension(D) |
= |
Lex |
| prec(t) |
= |
0 |
|
status(t) |
= |
[] |
|
list-extension(t) |
= |
Lex |
and the following
Max-polynomial interpretation
| [-(x1, x2)] |
=
|
max(2, 0 + 1 · x1, 2 + 1 · x2) |
| [*(x1, x2)] |
=
|
0 + 1 · x1 + 1 · x2
|
| [+(x1, x2)] |
=
|
max(0, 0 + 1 · x1, 0 + 1 · x2) |
| [0] |
=
|
max(0) |
| [constant] |
=
|
max(4) |
| [1] |
=
|
max(0) |
| [D(x1)] |
=
|
max(0, 0 + 1 · x1) |
| [t] |
=
|
max(5) |
all of the following rules can be deleted.
|
D(t) |
→ |
1 |
(1) |
|
D(constant) |
→ |
0 |
(2) |
|
D(+(x,y)) |
→ |
+(D(x),D(y)) |
(3) |
|
D(*(x,y)) |
→ |
+(*(y,D(x)),*(x,D(y))) |
(4) |
|
D(-(x,y)) |
→ |
-(D(x),D(y)) |
(5) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.