Certification Problem
Input (TPDB TRS_Standard/Transformed_CSR_04/Ex8_BLR02_FR)
The rewrite relation of the following TRS is considered.
|
fib(N) |
→ |
sel(N,fib1(s(0),s(0))) |
(1) |
|
fib1(X,Y) |
→ |
cons(X,n__fib1(Y,n__add(X,Y))) |
(2) |
|
add(0,X) |
→ |
X |
(3) |
|
add(s(X),Y) |
→ |
s(add(X,Y)) |
(4) |
|
sel(0,cons(X,XS)) |
→ |
X |
(5) |
|
sel(s(N),cons(X,XS)) |
→ |
sel(N,activate(XS)) |
(6) |
|
fib1(X1,X2) |
→ |
n__fib1(X1,X2) |
(7) |
|
add(X1,X2) |
→ |
n__add(X1,X2) |
(8) |
|
activate(n__fib1(X1,X2)) |
→ |
fib1(activate(X1),activate(X2)) |
(9) |
|
activate(n__add(X1,X2)) |
→ |
add(activate(X1),activate(X2)) |
(10) |
|
activate(X) |
→ |
X |
(11) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(fib) |
= |
7 |
|
stat(fib) |
= |
mul
|
| prec(sel) |
= |
5 |
|
stat(sel) |
= |
lex
|
| prec(fib1) |
= |
2 |
|
stat(fib1) |
= |
mul
|
| prec(s) |
= |
0 |
|
stat(s) |
= |
mul
|
| prec(0) |
= |
6 |
|
stat(0) |
= |
mul
|
| prec(cons) |
= |
0 |
|
stat(cons) |
= |
mul
|
| prec(n__fib1) |
= |
0 |
|
stat(n__fib1) |
= |
mul
|
| prec(n__add) |
= |
1 |
|
stat(n__add) |
= |
mul
|
| prec(add) |
= |
3 |
|
stat(add) |
= |
mul
|
| prec(activate) |
= |
4 |
|
stat(activate) |
= |
mul
|
| π(fib) |
= |
[1] |
| π(sel) |
= |
[1,2] |
| π(fib1) |
= |
[1,2] |
| π(s) |
= |
[1] |
| π(0) |
= |
[] |
| π(cons) |
= |
[1,2] |
| π(n__fib1) |
= |
[1,2] |
| π(n__add) |
= |
[1,2] |
| π(add) |
= |
[1,2] |
| π(activate) |
= |
[1] |
all of the following rules can be deleted.
|
fib(N) |
→ |
sel(N,fib1(s(0),s(0))) |
(1) |
|
fib1(X,Y) |
→ |
cons(X,n__fib1(Y,n__add(X,Y))) |
(2) |
|
add(0,X) |
→ |
X |
(3) |
|
add(s(X),Y) |
→ |
s(add(X,Y)) |
(4) |
|
sel(0,cons(X,XS)) |
→ |
X |
(5) |
|
sel(s(N),cons(X,XS)) |
→ |
sel(N,activate(XS)) |
(6) |
|
fib1(X1,X2) |
→ |
n__fib1(X1,X2) |
(7) |
|
add(X1,X2) |
→ |
n__add(X1,X2) |
(8) |
|
activate(n__fib1(X1,X2)) |
→ |
fib1(activate(X1),activate(X2)) |
(9) |
|
activate(n__add(X1,X2)) |
→ |
add(activate(X1),activate(X2)) |
(10) |
|
activate(X) |
→ |
X |
(11) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.