Certification Problem
Input (TPDB SRS_Standard/Zantema_04/z117)
The rewrite relation of the following TRS is considered.
d(a(x1)) |
→ |
b(d(x1)) |
(1) |
b(x1) |
→ |
a(a(a(x1))) |
(2) |
c(d(c(x1))) |
→ |
a(d(x1)) |
(3) |
b(d(d(x1))) |
→ |
c(c(d(d(c(x1))))) |
(4) |
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
[d(x1)] |
= |
4 · x1 + 0 |
[c(x1)] |
= |
1 · x1 + 1 |
[a(x1)] |
= |
1 · x1 + 5 |
[b(x1)] |
= |
1 · x1 + 20 |
all of the following rules can be deleted.
b(x1) |
→ |
a(a(a(x1))) |
(2) |
b(d(d(x1))) |
→ |
c(c(d(d(c(x1))))) |
(4) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(c) |
= |
1 |
|
weight(c) |
= |
2 |
|
|
|
prec(b) |
= |
2 |
|
weight(b) |
= |
4 |
|
|
|
prec(d) |
= |
3 |
|
weight(d) |
= |
2 |
|
|
|
prec(a) |
= |
0 |
|
weight(a) |
= |
4 |
|
|
|
all of the following rules can be deleted.
d(a(x1)) |
→ |
b(d(x1)) |
(1) |
c(d(c(x1))) |
→ |
a(d(x1)) |
(3) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.