Certification Problem
Input (TPDB TRS_Standard/SK90/2.27)
The rewrite relation of the following TRS is considered.
|
fib(0) |
→ |
0 |
(1) |
|
fib(s(0)) |
→ |
s(0) |
(2) |
|
fib(s(s(0))) |
→ |
s(0) |
(3) |
|
fib(s(s(x))) |
→ |
sp(g(x)) |
(4) |
|
g(0) |
→ |
pair(s(0),0) |
(5) |
|
g(s(0)) |
→ |
pair(s(0),s(0)) |
(6) |
|
g(s(x)) |
→ |
np(g(x)) |
(7) |
|
sp(pair(x,y)) |
→ |
+(x,y) |
(8) |
|
np(pair(x,y)) |
→ |
pair(+(x,y),x) |
(9) |
|
+(x,0) |
→ |
x |
(10) |
|
+(x,s(y)) |
→ |
s(+(x,y)) |
(11) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(fib) |
= |
3 |
|
stat(fib) |
= |
mul
|
| prec(0) |
= |
0 |
|
stat(0) |
= |
mul
|
| prec(s) |
= |
1 |
|
stat(s) |
= |
mul
|
| prec(sp) |
= |
2 |
|
stat(sp) |
= |
mul
|
| prec(g) |
= |
3 |
|
stat(g) |
= |
mul
|
| prec(pair) |
= |
0 |
|
stat(pair) |
= |
mul
|
| prec(np) |
= |
2 |
|
stat(np) |
= |
mul
|
| prec(+) |
= |
2 |
|
stat(+) |
= |
mul
|
| π(fib) |
= |
[1] |
| π(0) |
= |
[] |
| π(s) |
= |
[1] |
| π(sp) |
= |
[1] |
| π(g) |
= |
[1] |
| π(pair) |
= |
[1,2] |
| π(np) |
= |
[1] |
| π(+) |
= |
[1,2] |
all of the following rules can be deleted.
|
fib(0) |
→ |
0 |
(1) |
|
fib(s(0)) |
→ |
s(0) |
(2) |
|
fib(s(s(0))) |
→ |
s(0) |
(3) |
|
fib(s(s(x))) |
→ |
sp(g(x)) |
(4) |
|
g(0) |
→ |
pair(s(0),0) |
(5) |
|
g(s(0)) |
→ |
pair(s(0),s(0)) |
(6) |
|
g(s(x)) |
→ |
np(g(x)) |
(7) |
|
sp(pair(x,y)) |
→ |
+(x,y) |
(8) |
|
np(pair(x,y)) |
→ |
pair(+(x,y),x) |
(9) |
|
+(x,0) |
→ |
x |
(10) |
|
+(x,s(y)) |
→ |
s(+(x,y)) |
(11) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.