Certification Problem
Input (TPDB TRS_Innermost/Transformed_CSR_innermost_04/Ex2_Luc02a_L)
The rewrite relation of the following TRS is considered.
|
terms(N) |
→ |
cons(recip(sqr(N))) |
(1) |
|
sqr(0) |
→ |
0 |
(2) |
|
sqr(s(X)) |
→ |
s(add(sqr(X),dbl(X))) |
(3) |
|
dbl(0) |
→ |
0 |
(4) |
|
dbl(s(X)) |
→ |
s(s(dbl(X))) |
(5) |
|
add(0,X) |
→ |
X |
(6) |
|
add(s(X),Y) |
→ |
s(add(X,Y)) |
(7) |
|
first(0,X) |
→ |
nil |
(8) |
|
first(s(X),cons(Y)) |
→ |
cons(Y) |
(9) |
The evaluation strategy is innermost.Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(terms) |
= |
8 |
|
stat(terms) |
= |
mul
|
| prec(cons) |
= |
0 |
|
stat(cons) |
= |
mul
|
| prec(recip) |
= |
1 |
|
stat(recip) |
= |
mul
|
| prec(sqr) |
= |
7 |
|
stat(sqr) |
= |
mul
|
| prec(0) |
= |
3 |
|
stat(0) |
= |
mul
|
| prec(s) |
= |
4 |
|
stat(s) |
= |
mul
|
| prec(add) |
= |
5 |
|
stat(add) |
= |
mul
|
| prec(dbl) |
= |
6 |
|
stat(dbl) |
= |
mul
|
| prec(first) |
= |
0 |
|
stat(first) |
= |
mul
|
| prec(nil) |
= |
2 |
|
stat(nil) |
= |
mul
|
| π(terms) |
= |
[1] |
| π(cons) |
= |
[1] |
| π(recip) |
= |
[1] |
| π(sqr) |
= |
[1] |
| π(0) |
= |
[] |
| π(s) |
= |
[1] |
| π(add) |
= |
[1,2] |
| π(dbl) |
= |
[1] |
| π(first) |
= |
[1,2] |
| π(nil) |
= |
[] |
all of the following rules can be deleted.
|
terms(N) |
→ |
cons(recip(sqr(N))) |
(1) |
|
sqr(0) |
→ |
0 |
(2) |
|
sqr(s(X)) |
→ |
s(add(sqr(X),dbl(X))) |
(3) |
|
dbl(0) |
→ |
0 |
(4) |
|
dbl(s(X)) |
→ |
s(s(dbl(X))) |
(5) |
|
add(0,X) |
→ |
X |
(6) |
|
add(s(X),Y) |
→ |
s(add(X,Y)) |
(7) |
|
first(0,X) |
→ |
nil |
(8) |
|
first(s(X),cons(Y)) |
→ |
cons(Y) |
(9) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.