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 ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
Weighted Path Order with the following precedence and status
prec(half) |
= |
0 |
|
status(half) |
= |
[1] |
|
list-extension(half) |
= |
Lex |
prec(nil) |
= |
0 |
|
status(nil) |
= |
[] |
|
list-extension(nil) |
= |
Lex |
prec(first) |
= |
1 |
|
status(first) |
= |
[2, 1] |
|
list-extension(first) |
= |
Lex |
prec(add) |
= |
5 |
|
status(add) |
= |
[2, 1] |
|
list-extension(add) |
= |
Lex |
prec(dbl) |
= |
5 |
|
status(dbl) |
= |
[1] |
|
list-extension(dbl) |
= |
Lex |
prec(s) |
= |
4 |
|
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) |
= |
8 |
|
status(sqr) |
= |
[1] |
|
list-extension(sqr) |
= |
Lex |
prec(terms) |
= |
9 |
|
status(terms) |
= |
[1] |
|
list-extension(terms) |
= |
Lex |
and the following
Max-polynomial interpretation
[half(x1)] |
=
|
0 + 1 · x1
|
[nil] |
=
|
max(4) |
[first(x1, x2)] |
=
|
max(3, 1 + 1 · x1, 6 + 1 · x2) |
[add(x1, x2)] |
=
|
max(0, 0 + 1 · x1, 0 + 1 · x2) |
[dbl(x1)] |
=
|
max(2, 0 + 1 · x1) |
[s(x1)] |
=
|
max(4, 0 + 1 · x1) |
[0] |
=
|
max(2) |
[cons(x1)] |
=
|
max(2, 1 + 1 · x1) |
[recip(x1)] |
=
|
max(0, 0 + 1 · x1) |
[sqr(x1)] |
=
|
0 + 1 · x1
|
[terms(x1)] |
=
|
max(0, 2 + 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) |
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.