Certification Problem
Input (TPDB TRS_Standard/Secret_07_TRS/secret3)
The rewrite relation of the following TRS is considered.
|
app(nil,k) |
→ |
k |
(1) |
|
app(l,nil) |
→ |
l |
(2) |
|
app(cons(x,l),k) |
→ |
cons(x,app(l,k)) |
(3) |
|
sum(cons(x,nil)) |
→ |
cons(x,nil) |
(4) |
|
sum(cons(x,cons(y,l))) |
→ |
sum(cons(a(x,y,h),l)) |
(5) |
|
a(h,h,x) |
→ |
s(x) |
(6) |
|
a(x,s(y),h) |
→ |
a(x,y,s(h)) |
(7) |
|
a(x,s(y),s(z)) |
→ |
a(x,y,a(x,s(y),z)) |
(8) |
|
a(s(x),h,z) |
→ |
a(x,z,z) |
(9) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(app) |
= |
4 |
|
stat(app) |
= |
lex
|
| prec(nil) |
= |
0 |
|
stat(nil) |
= |
lex
|
| prec(cons) |
= |
3 |
|
stat(cons) |
= |
lex
|
| prec(sum) |
= |
5 |
|
stat(sum) |
= |
lex
|
| prec(a) |
= |
2 |
|
stat(a) |
= |
lex
|
| prec(h) |
= |
2 |
|
stat(h) |
= |
lex
|
| prec(s) |
= |
1 |
|
stat(s) |
= |
lex
|
| π(app) |
= |
[2,1] |
| π(nil) |
= |
[] |
| π(cons) |
= |
[2,1] |
| π(sum) |
= |
[1] |
| π(a) |
= |
[1,2,3] |
| π(h) |
= |
[] |
| π(s) |
= |
[1] |
all of the following rules can be deleted.
|
app(nil,k) |
→ |
k |
(1) |
|
app(l,nil) |
→ |
l |
(2) |
|
app(cons(x,l),k) |
→ |
cons(x,app(l,k)) |
(3) |
|
sum(cons(x,nil)) |
→ |
cons(x,nil) |
(4) |
|
sum(cons(x,cons(y,l))) |
→ |
sum(cons(a(x,y,h),l)) |
(5) |
|
a(h,h,x) |
→ |
s(x) |
(6) |
|
a(x,s(y),h) |
→ |
a(x,y,s(h)) |
(7) |
|
a(x,s(y),s(z)) |
→ |
a(x,y,a(x,s(y),z)) |
(8) |
|
a(s(x),h,z) |
→ |
a(x,z,z) |
(9) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.