Certification Problem
Input (TPDB TRS_Standard/CiME_04/list-sum-prod)
The rewrite relation of the following TRS is considered.
|
+(x,0) |
→ |
x |
(1) |
|
+(0,x) |
→ |
x |
(2) |
|
+(s(x),s(y)) |
→ |
s(s(+(x,y))) |
(3) |
|
*(x,0) |
→ |
0 |
(4) |
|
*(0,x) |
→ |
0 |
(5) |
|
*(s(x),s(y)) |
→ |
s(+(*(x,y),+(x,y))) |
(6) |
|
sum(nil) |
→ |
0 |
(7) |
|
sum(cons(x,l)) |
→ |
+(x,sum(l)) |
(8) |
|
prod(nil) |
→ |
s(0) |
(9) |
|
prod(cons(x,l)) |
→ |
*(x,prod(l)) |
(10) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(+) |
= |
1 |
|
stat(+) |
= |
mul
|
| prec(0) |
= |
2 |
|
stat(0) |
= |
mul
|
| prec(s) |
= |
0 |
|
stat(s) |
= |
mul
|
| prec(*) |
= |
2 |
|
stat(*) |
= |
lex
|
| prec(sum) |
= |
3 |
|
stat(sum) |
= |
mul
|
| prec(nil) |
= |
4 |
|
stat(nil) |
= |
mul
|
| prec(cons) |
= |
3 |
|
stat(cons) |
= |
mul
|
| π(+) |
= |
[1,2] |
| π(0) |
= |
[] |
| π(s) |
= |
[1] |
| π(*) |
= |
[2,1] |
| π(sum) |
= |
[1] |
| π(nil) |
= |
[] |
| π(cons) |
= |
[1,2] |
| π(prod) |
= |
1 |
all of the following rules can be deleted.
|
+(x,0) |
→ |
x |
(1) |
|
+(0,x) |
→ |
x |
(2) |
|
+(s(x),s(y)) |
→ |
s(s(+(x,y))) |
(3) |
|
*(x,0) |
→ |
0 |
(4) |
|
*(0,x) |
→ |
0 |
(5) |
|
*(s(x),s(y)) |
→ |
s(+(*(x,y),+(x,y))) |
(6) |
|
sum(nil) |
→ |
0 |
(7) |
|
sum(cons(x,l)) |
→ |
+(x,sum(l)) |
(8) |
|
prod(nil) |
→ |
s(0) |
(9) |
|
prod(cons(x,l)) |
→ |
*(x,prod(l)) |
(10) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.