Certification Problem
Input (TPDB TRS_Standard/Secret_06_TRS/7)
The rewrite relation of the following TRS is considered.
|
c(c(c(a(x,y)))) |
→ |
b(c(c(c(c(y)))),x) |
(1) |
|
c(c(b(c(y),0))) |
→ |
a(0,c(c(a(y,0)))) |
(2) |
|
c(c(a(a(y,0),x))) |
→ |
c(y) |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [c(x1)] |
= |
2 · x1 + 0 |
| [0] |
= |
6 |
| [a(x1, x2)] |
= |
1 · x1 + 2 · x2 + 10 |
| [b(x1, x2)] |
= |
1 · x1 + 8 · x2 + 0 |
all of the following rules can be deleted.
|
c(c(c(a(x,y)))) |
→ |
b(c(c(c(c(y)))),x) |
(1) |
|
c(c(a(a(y,0),x))) |
→ |
c(y) |
(3) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(0) |
= |
1 |
|
weight(0) |
= |
1 |
|
|
|
| prec(b) |
= |
2 |
|
weight(b) |
= |
1 |
|
|
|
| prec(c) |
= |
3 |
|
weight(c) |
= |
2 |
|
|
|
| prec(a) |
= |
0 |
|
weight(a) |
= |
1 |
|
|
|
all of the following rules can be deleted.
|
c(c(b(c(y),0))) |
→ |
a(0,c(c(a(y,0)))) |
(2) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.