Certification Problem
Input (TPDB TRS_Standard/SK90/2.49)
The rewrite relation of the following TRS is considered.
|
a(c(d(x))) |
→ |
c(x) |
(1) |
|
u(b(d(d(x)))) |
→ |
b(x) |
(2) |
|
v(a(a(x))) |
→ |
u(v(x)) |
(3) |
|
v(a(c(x))) |
→ |
u(b(d(x))) |
(4) |
|
v(c(x)) |
→ |
b(x) |
(5) |
|
w(a(a(x))) |
→ |
u(w(x)) |
(6) |
|
w(a(c(x))) |
→ |
u(b(d(x))) |
(7) |
|
w(c(x)) |
→ |
b(x) |
(8) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
d(c(a(x))) |
→ |
c(x) |
(9) |
|
d(d(b(u(x)))) |
→ |
b(x) |
(10) |
|
a(a(v(x))) |
→ |
v(u(x)) |
(11) |
|
c(a(v(x))) |
→ |
d(b(u(x))) |
(12) |
|
c(v(x)) |
→ |
b(x) |
(13) |
|
a(a(w(x))) |
→ |
w(u(x)) |
(14) |
|
c(a(w(x))) |
→ |
d(b(u(x))) |
(15) |
|
c(w(x)) |
→ |
b(x) |
(16) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(w) |
= |
2 |
|
weight(w) |
= |
1 |
|
|
|
| prec(v) |
= |
4 |
|
weight(v) |
= |
1 |
|
|
|
| prec(u) |
= |
3 |
|
weight(u) |
= |
6 |
|
|
|
| prec(b) |
= |
6 |
|
weight(b) |
= |
2 |
|
|
|
| prec(a) |
= |
7 |
|
weight(a) |
= |
3 |
|
|
|
| prec(c) |
= |
1 |
|
weight(c) |
= |
6 |
|
|
|
| prec(d) |
= |
0 |
|
weight(d) |
= |
2 |
|
|
|
all of the following rules can be deleted.
|
d(c(a(x))) |
→ |
c(x) |
(9) |
|
d(d(b(u(x)))) |
→ |
b(x) |
(10) |
|
a(a(v(x))) |
→ |
v(u(x)) |
(11) |
|
c(a(v(x))) |
→ |
d(b(u(x))) |
(12) |
|
c(v(x)) |
→ |
b(x) |
(13) |
|
a(a(w(x))) |
→ |
w(u(x)) |
(14) |
|
c(a(w(x))) |
→ |
d(b(u(x))) |
(15) |
|
c(w(x)) |
→ |
b(x) |
(16) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.