Certification Problem
Input (TPDB SRS_Standard/Zantema_04/z124)
The rewrite relation of the following TRS is considered.
|
q0(a(x1)) |
→ |
x(q1(x1)) |
(1) |
|
q1(a(x1)) |
→ |
a(q1(x1)) |
(2) |
|
q1(y(x1)) |
→ |
y(q1(x1)) |
(3) |
|
a(q1(b(x1))) |
→ |
q2(a(y(x1))) |
(4) |
|
a(q2(a(x1))) |
→ |
q2(a(a(x1))) |
(5) |
|
a(q2(y(x1))) |
→ |
q2(a(y(x1))) |
(6) |
|
y(q1(b(x1))) |
→ |
q2(y(y(x1))) |
(7) |
|
y(q2(a(x1))) |
→ |
q2(y(a(x1))) |
(8) |
|
y(q2(y(x1))) |
→ |
q2(y(y(x1))) |
(9) |
|
q2(x(x1)) |
→ |
x(q0(x1)) |
(10) |
|
q0(y(x1)) |
→ |
y(q3(x1)) |
(11) |
|
q3(y(x1)) |
→ |
y(q3(x1)) |
(12) |
|
q3(bl(x1)) |
→ |
bl(q4(x1)) |
(13) |
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
|
a(q0(x1)) |
→ |
q1(x(x1)) |
(14) |
|
a(q1(x1)) |
→ |
q1(a(x1)) |
(15) |
|
y(q1(x1)) |
→ |
q1(y(x1)) |
(16) |
|
b(q1(a(x1))) |
→ |
y(a(q2(x1))) |
(17) |
|
a(q2(a(x1))) |
→ |
a(a(q2(x1))) |
(18) |
|
y(q2(a(x1))) |
→ |
y(a(q2(x1))) |
(19) |
|
b(q1(y(x1))) |
→ |
y(y(q2(x1))) |
(20) |
|
a(q2(y(x1))) |
→ |
a(y(q2(x1))) |
(21) |
|
y(q2(y(x1))) |
→ |
y(y(q2(x1))) |
(22) |
|
x(q2(x1)) |
→ |
q0(x(x1)) |
(23) |
|
y(q0(x1)) |
→ |
q3(y(x1)) |
(24) |
|
y(q3(x1)) |
→ |
q3(y(x1)) |
(25) |
|
bl(q3(x1)) |
→ |
q4(bl(x1)) |
(26) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(q4) |
= |
2 |
|
weight(q4) |
= |
2 |
|
|
|
| prec(bl) |
= |
11 |
|
weight(bl) |
= |
2 |
|
|
|
| prec(q3) |
= |
4 |
|
weight(q3) |
= |
2 |
|
|
|
| prec(q2) |
= |
7 |
|
weight(q2) |
= |
5 |
|
|
|
| prec(b) |
= |
14 |
|
weight(b) |
= |
2 |
|
|
|
| prec(y) |
= |
5 |
|
weight(y) |
= |
2 |
|
|
|
| prec(x) |
= |
10 |
|
weight(x) |
= |
1 |
|
|
|
| prec(q1) |
= |
0 |
|
weight(q1) |
= |
5 |
|
|
|
| prec(q0) |
= |
15 |
|
weight(q0) |
= |
3 |
|
|
|
| prec(a) |
= |
1 |
|
weight(a) |
= |
3 |
|
|
|
all of the following rules can be deleted.
|
a(q0(x1)) |
→ |
q1(x(x1)) |
(14) |
|
a(q1(x1)) |
→ |
q1(a(x1)) |
(15) |
|
y(q1(x1)) |
→ |
q1(y(x1)) |
(16) |
|
b(q1(a(x1))) |
→ |
y(a(q2(x1))) |
(17) |
|
a(q2(a(x1))) |
→ |
a(a(q2(x1))) |
(18) |
|
y(q2(a(x1))) |
→ |
y(a(q2(x1))) |
(19) |
|
b(q1(y(x1))) |
→ |
y(y(q2(x1))) |
(20) |
|
a(q2(y(x1))) |
→ |
a(y(q2(x1))) |
(21) |
|
y(q2(y(x1))) |
→ |
y(y(q2(x1))) |
(22) |
|
x(q2(x1)) |
→ |
q0(x(x1)) |
(23) |
|
y(q0(x1)) |
→ |
q3(y(x1)) |
(24) |
|
y(q3(x1)) |
→ |
q3(y(x1)) |
(25) |
|
bl(q3(x1)) |
→ |
q4(bl(x1)) |
(26) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.