Certification Problem
Input (TPDB TRS_Standard/Transformed_CSR_04/ExSec11_1_Luc02a_Z)
The rewrite relation of the following TRS is considered.
|
terms(N) |
→ |
cons(recip(sqr(N)),n__terms(s(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,Z)) |
→ |
cons(Y,n__first(X,activate(Z))) |
(9) |
|
half(0) |
→ |
0 |
(10) |
|
half(s(0)) |
→ |
0 |
(11) |
|
half(s(s(X))) |
→ |
s(half(X)) |
(12) |
|
half(dbl(X)) |
→ |
X |
(13) |
|
terms(X) |
→ |
n__terms(X) |
(14) |
|
first(X1,X2) |
→ |
n__first(X1,X2) |
(15) |
|
activate(n__terms(X)) |
→ |
terms(X) |
(16) |
|
activate(n__first(X1,X2)) |
→ |
first(X1,X2) |
(17) |
|
activate(X) |
→ |
X |
(18) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(terms) |
= |
6 |
|
stat(terms) |
= |
mul
|
| prec(cons) |
= |
1 |
|
stat(cons) |
= |
mul
|
| prec(sqr) |
= |
5 |
|
stat(sqr) |
= |
mul
|
| prec(s) |
= |
3 |
|
stat(s) |
= |
mul
|
| prec(0) |
= |
2 |
|
stat(0) |
= |
mul
|
| prec(add) |
= |
4 |
|
stat(add) |
= |
lex
|
| prec(dbl) |
= |
5 |
|
stat(dbl) |
= |
mul
|
| prec(first) |
= |
6 |
|
stat(first) |
= |
mul
|
| prec(nil) |
= |
0 |
|
stat(nil) |
= |
mul
|
| prec(n__first) |
= |
1 |
|
stat(n__first) |
= |
mul
|
| prec(activate) |
= |
6 |
|
stat(activate) |
= |
mul
|
| π(terms) |
= |
[1] |
| π(cons) |
= |
[1,2] |
| π(recip) |
= |
1 |
| π(sqr) |
= |
[1] |
| π(n__terms) |
= |
1 |
| π(s) |
= |
[1] |
| π(0) |
= |
[] |
| π(add) |
= |
[2,1] |
| π(dbl) |
= |
[1] |
| π(first) |
= |
[1,2] |
| π(nil) |
= |
[] |
| π(n__first) |
= |
[1,2] |
| π(activate) |
= |
[1] |
| π(half) |
= |
1 |
all of the following rules can be deleted.
|
terms(N) |
→ |
cons(recip(sqr(N)),n__terms(s(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,Z)) |
→ |
cons(Y,n__first(X,activate(Z))) |
(9) |
|
half(s(0)) |
→ |
0 |
(11) |
|
half(s(s(X))) |
→ |
s(half(X)) |
(12) |
|
half(dbl(X)) |
→ |
X |
(13) |
|
terms(X) |
→ |
n__terms(X) |
(14) |
|
first(X1,X2) |
→ |
n__first(X1,X2) |
(15) |
|
activate(n__first(X1,X2)) |
→ |
first(X1,X2) |
(17) |
|
activate(X) |
→ |
X |
(18) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [half(x1)] |
= |
2 + 1 · x1
|
| [0] |
= |
2 |
| [activate(x1)] |
= |
2 + 1 · x1
|
| [n__terms(x1)] |
= |
2 + 2 · x1
|
| [terms(x1)] |
= |
2 · x1
|
all of the following rules can be deleted.
|
half(0) |
→ |
0 |
(10) |
|
activate(n__terms(X)) |
→ |
terms(X) |
(16) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.