Certification Problem
Input (TPDB TRS_Standard/Transformed_CSR_04/Ex3_3_25_Bor03_L)
The rewrite relation of the following TRS is considered.
app(nil,YS) |
→ |
YS |
(1) |
app(cons(X),YS) |
→ |
cons(X) |
(2) |
from(X) |
→ |
cons(X) |
(3) |
zWadr(nil,YS) |
→ |
nil |
(4) |
zWadr(XS,nil) |
→ |
nil |
(5) |
zWadr(cons(X),cons(Y)) |
→ |
cons(app(Y,cons(X))) |
(6) |
prefix(L) |
→ |
cons(nil) |
(7) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(prefix) |
= |
1 |
|
weight(prefix) |
= |
6 |
|
|
|
prec(zWadr) |
= |
5 |
|
weight(zWadr) |
= |
0 |
|
|
|
prec(from) |
= |
7 |
|
weight(from) |
= |
2 |
|
|
|
prec(cons) |
= |
0 |
|
weight(cons) |
= |
2 |
|
|
|
prec(app) |
= |
6 |
|
weight(app) |
= |
0 |
|
|
|
prec(nil) |
= |
4 |
|
weight(nil) |
= |
5 |
|
|
|
all of the following rules can be deleted.
app(nil,YS) |
→ |
YS |
(1) |
app(cons(X),YS) |
→ |
cons(X) |
(2) |
from(X) |
→ |
cons(X) |
(3) |
zWadr(nil,YS) |
→ |
nil |
(4) |
zWadr(XS,nil) |
→ |
nil |
(5) |
zWadr(cons(X),cons(Y)) |
→ |
cons(app(Y,cons(X))) |
(6) |
prefix(L) |
→ |
cons(nil) |
(7) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.