Certification Problem
Input (TPDB TRS_Standard/Secret_06_TRS/gen-17)
The rewrite relation of the following TRS is considered.
|
f(c(c(a,y,a),b(x,z),a)) |
→ |
b(y,f(c(f(a),z,z))) |
(1) |
|
f(b(b(x,f(y)),z)) |
→ |
c(z,x,f(b(b(f(a),y),y))) |
(2) |
|
c(b(a,a),b(y,z),x) |
→ |
b(a,b(z,z)) |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(f) |
= |
1 |
|
stat(f) |
= |
lex
|
| prec(c) |
= |
1 |
|
stat(c) |
= |
lex
|
| prec(a) |
= |
0 |
|
stat(a) |
= |
lex
|
| prec(b) |
= |
0 |
|
stat(b) |
= |
lex
|
| π(f) |
= |
[1] |
| π(c) |
= |
[1,3,2] |
| π(a) |
= |
[] |
| π(b) |
= |
[2,1] |
all of the following rules can be deleted.
|
f(c(c(a,y,a),b(x,z),a)) |
→ |
b(y,f(c(f(a),z,z))) |
(1) |
|
f(b(b(x,f(y)),z)) |
→ |
c(z,x,f(b(b(f(a),y),y))) |
(2) |
|
c(b(a,a),b(y,z),x) |
→ |
b(a,b(z,z)) |
(3) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.