Certification Problem
Input (TPDB TRS_Standard/SK90/2.23)
The rewrite relation of the following TRS is considered.
|
fac(0) |
→ |
1 |
(1) |
|
fac(s(x)) |
→ |
*(s(x),fac(x)) |
(2) |
|
floop(0,y) |
→ |
y |
(3) |
|
floop(s(x),y) |
→ |
floop(x,*(s(x),y)) |
(4) |
|
*(x,0) |
→ |
0 |
(5) |
|
*(x,s(y)) |
→ |
+(*(x,y),x) |
(6) |
|
+(x,0) |
→ |
x |
(7) |
|
+(x,s(y)) |
→ |
s(+(x,y)) |
(8) |
| 1 |
→ |
s(0) |
(9) |
|
fac(0) |
→ |
s(0) |
(10) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(fac) |
= |
5 |
|
stat(fac) |
= |
mul
|
| prec(0) |
= |
0 |
|
stat(0) |
= |
mul
|
| prec(1) |
= |
2 |
|
stat(1) |
= |
mul
|
| prec(s) |
= |
1 |
|
stat(s) |
= |
mul
|
| prec(*) |
= |
4 |
|
stat(*) |
= |
mul
|
| prec(floop) |
= |
6 |
|
stat(floop) |
= |
lex
|
| prec(+) |
= |
3 |
|
stat(+) |
= |
mul
|
| π(fac) |
= |
[1] |
| π(0) |
= |
[] |
| π(1) |
= |
[] |
| π(s) |
= |
[1] |
| π(*) |
= |
[1,2] |
| π(floop) |
= |
[1,2] |
| π(+) |
= |
[1,2] |
all of the following rules can be deleted.
|
fac(0) |
→ |
1 |
(1) |
|
fac(s(x)) |
→ |
*(s(x),fac(x)) |
(2) |
|
floop(0,y) |
→ |
y |
(3) |
|
floop(s(x),y) |
→ |
floop(x,*(s(x),y)) |
(4) |
|
*(x,0) |
→ |
0 |
(5) |
|
*(x,s(y)) |
→ |
+(*(x,y),x) |
(6) |
|
+(x,0) |
→ |
x |
(7) |
|
+(x,s(y)) |
→ |
s(+(x,y)) |
(8) |
| 1 |
→ |
s(0) |
(9) |
|
fac(0) |
→ |
s(0) |
(10) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.