Certification Problem
Input (TPDB TRS_Standard/Transformed_CSR_04/ExSec11_1_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) |
half(0) |
→ |
0 |
(10) |
half(s(0)) |
→ |
0 |
(11) |
half(s(s(X))) |
→ |
s(half(X)) |
(12) |
half(dbl(X)) |
→ |
X |
(13) |
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(recip) |
= |
2 |
|
stat(recip) |
= |
mul
|
prec(sqr) |
= |
5 |
|
stat(sqr) |
= |
mul
|
prec(0) |
= |
3 |
|
stat(0) |
= |
mul
|
prec(s) |
= |
0 |
|
stat(s) |
= |
mul
|
prec(add) |
= |
4 |
|
stat(add) |
= |
lex
|
prec(dbl) |
= |
5 |
|
stat(dbl) |
= |
mul
|
prec(first) |
= |
7 |
|
stat(first) |
= |
mul
|
prec(nil) |
= |
3 |
|
stat(nil) |
= |
mul
|
prec(half) |
= |
3 |
|
stat(half) |
= |
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) |
= |
[] |
π(half) |
= |
[1] |
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) |
half(0) |
→ |
0 |
(10) |
half(s(0)) |
→ |
0 |
(11) |
half(s(s(X))) |
→ |
s(half(X)) |
(12) |
half(dbl(X)) |
→ |
X |
(13) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.