Certification Problem
Input (TPDB SRS_Standard/Zantema_06/beans6)
The rewrite relation of the following TRS is considered.
|
b(a(a(x1))) |
→ |
a(b(c(x1))) |
(1) |
|
c(a(x1)) |
→ |
a(c(x1)) |
(2) |
|
c(b(x1)) |
→ |
b(a(x1)) |
(3) |
|
a(a(b(x1))) |
→ |
d(b(a(x1))) |
(4) |
|
a(d(x1)) |
→ |
d(a(x1)) |
(5) |
|
b(d(x1)) |
→ |
a(b(x1)) |
(6) |
|
a(a(x1)) |
→ |
a(b(a(x1))) |
(7) |
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(a(b(x1))) |
→ |
c(b(a(x1))) |
(8) |
|
a(c(x1)) |
→ |
c(a(x1)) |
(9) |
|
b(c(x1)) |
→ |
a(b(x1)) |
(10) |
|
b(a(a(x1))) |
→ |
a(b(d(x1))) |
(11) |
|
d(a(x1)) |
→ |
a(d(x1)) |
(12) |
|
d(b(x1)) |
→ |
b(a(x1)) |
(13) |
|
a(a(x1)) |
→ |
a(b(a(x1))) |
(7) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
a#(a(b(x1))) |
→ |
a#(x1) |
(14) |
|
a#(a(b(x1))) |
→ |
b#(a(x1)) |
(15) |
|
a#(c(x1)) |
→ |
a#(x1) |
(16) |
|
b#(c(x1)) |
→ |
b#(x1) |
(17) |
|
b#(c(x1)) |
→ |
a#(b(x1)) |
(18) |
|
b#(a(a(x1))) |
→ |
d#(x1) |
(19) |
|
b#(a(a(x1))) |
→ |
b#(d(x1)) |
(20) |
|
b#(a(a(x1))) |
→ |
a#(b(d(x1))) |
(21) |
|
d#(a(x1)) |
→ |
d#(x1) |
(22) |
|
d#(a(x1)) |
→ |
a#(d(x1)) |
(23) |
|
d#(b(x1)) |
→ |
a#(x1) |
(24) |
|
d#(b(x1)) |
→ |
b#(a(x1)) |
(25) |
|
a#(a(x1)) |
→ |
b#(a(x1)) |
(26) |
|
a#(a(x1)) |
→ |
a#(b(a(x1))) |
(27) |
1.1.1 Subterm Criterion Processor
We use the projection to multisets
| π(d#)
|
= |
{
1, 1
}
|
| π(b#)
|
= |
{
1
}
|
| π(a#)
|
= |
{
1, 1
}
|
| π(d)
|
= |
{
1, 1
}
|
| π(c)
|
= |
{
1, 1
}
|
| π(b)
|
= |
{
1
}
|
| π(a)
|
= |
{
1, 1
}
|
to remove the pairs:
|
a#(a(b(x1))) |
→ |
a#(x1) |
(14) |
|
a#(a(b(x1))) |
→ |
b#(a(x1)) |
(15) |
|
a#(c(x1)) |
→ |
a#(x1) |
(16) |
|
b#(c(x1)) |
→ |
b#(x1) |
(17) |
|
b#(a(a(x1))) |
→ |
d#(x1) |
(19) |
|
b#(a(a(x1))) |
→ |
b#(d(x1)) |
(20) |
|
d#(a(x1)) |
→ |
d#(x1) |
(22) |
|
a#(a(x1)) |
→ |
b#(a(x1)) |
(26) |
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.