Certification Problem
Input (TPDB TRS_Standard/SK90/2.19)
The rewrite relation of the following TRS is considered.
|
sqr(0) |
→ |
0 |
(1) |
|
sqr(s(x)) |
→ |
+(sqr(x),s(double(x))) |
(2) |
|
double(0) |
→ |
0 |
(3) |
|
double(s(x)) |
→ |
s(s(double(x))) |
(4) |
|
+(x,0) |
→ |
x |
(5) |
|
+(x,s(y)) |
→ |
s(+(x,y)) |
(6) |
|
sqr(s(x)) |
→ |
s(+(sqr(x),double(x))) |
(7) |
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(+) |
= |
2 |
|
status(+) |
= |
[2, 1] |
|
list-extension(+) |
= |
Lex |
| prec(double) |
= |
2 |
|
status(double) |
= |
[1] |
|
list-extension(double) |
= |
Lex |
| prec(s) |
= |
0 |
|
status(s) |
= |
[1] |
|
list-extension(s) |
= |
Lex |
| prec(sqr) |
= |
3 |
|
status(sqr) |
= |
[1] |
|
list-extension(sqr) |
= |
Lex |
| prec(0) |
= |
0 |
|
status(0) |
= |
[] |
|
list-extension(0) |
= |
Lex |
and the following
Max-polynomial interpretation
| [+(x1, x2)] |
=
|
max(4, 0 + 1 · x1, 0 + 1 · x2) |
| [double(x1)] |
=
|
max(2, 4 + 1 · x1) |
| [s(x1)] |
=
|
0 + 1 · x1
|
| [sqr(x1)] |
=
|
max(4, 4 + 1 · x1) |
| [0] |
=
|
max(2) |
all of the following rules can be deleted.
|
sqr(0) |
→ |
0 |
(1) |
|
sqr(s(x)) |
→ |
+(sqr(x),s(double(x))) |
(2) |
|
double(0) |
→ |
0 |
(3) |
|
double(s(x)) |
→ |
s(s(double(x))) |
(4) |
|
+(x,0) |
→ |
x |
(5) |
|
+(x,s(y)) |
→ |
s(+(x,y)) |
(6) |
|
sqr(s(x)) |
→ |
s(+(sqr(x),double(x))) |
(7) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.