Certification Problem
Input (TPDB TRS_Standard/Transformed_CSR_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) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
Weighted Path Order with the following precedence and status
| prec(nil) |
= |
0 |
|
status(nil) |
= |
[] |
|
list-extension(nil) |
= |
Lex |
| prec(first) |
= |
1 |
|
status(first) |
= |
[2, 1] |
|
list-extension(first) |
= |
Lex |
| prec(add) |
= |
2 |
|
status(add) |
= |
[1, 2] |
|
list-extension(add) |
= |
Lex |
| prec(dbl) |
= |
2 |
|
status(dbl) |
= |
[1] |
|
list-extension(dbl) |
= |
Lex |
| prec(s) |
= |
0 |
|
status(s) |
= |
[1] |
|
list-extension(s) |
= |
Lex |
| prec(0) |
= |
0 |
|
status(0) |
= |
[] |
|
list-extension(0) |
= |
Lex |
| prec(cons) |
= |
0 |
|
status(cons) |
= |
[1] |
|
list-extension(cons) |
= |
Lex |
| prec(recip) |
= |
0 |
|
status(recip) |
= |
[1] |
|
list-extension(recip) |
= |
Lex |
| prec(sqr) |
= |
3 |
|
status(sqr) |
= |
[1] |
|
list-extension(sqr) |
= |
Lex |
| prec(terms) |
= |
0 |
|
status(terms) |
= |
[1] |
|
list-extension(terms) |
= |
Lex |
and the following
Max-polynomial interpretation
| [nil] |
=
|
max(0) |
| [first(x1, x2)] |
=
|
max(4, 3 + 1 · x1, 0 + 1 · x2) |
| [add(x1, x2)] |
=
|
max(2, 0 + 1 · x1, 0 + 1 · x2) |
| [dbl(x1)] |
=
|
max(0, 0 + 1 · x1) |
| [s(x1)] |
=
|
0 + 1 · x1
|
| [0] |
=
|
max(0) |
| [cons(x1)] |
=
|
0 + 1 · x1
|
| [recip(x1)] |
=
|
max(2, 0 + 1 · x1) |
| [sqr(x1)] |
=
|
max(2, 0 + 1 · x1) |
| [terms(x1)] |
=
|
max(0, 3 + 1 · x1) |
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.