Certification Problem
Input (TPDB SRS_Standard/Zantema_04/z015)
The rewrite relation of the following TRS is considered.
a(b(x1)) |
→ |
b(a(a(a(x1)))) |
(1) |
b(a(x1)) |
→ |
a(a(x1)) |
(2) |
a(a(x1)) |
→ |
a(c(b(x1))) |
(3) |
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
b(a(x1)) |
→ |
a(a(a(b(x1)))) |
(4) |
a(b(x1)) |
→ |
a(a(x1)) |
(5) |
a(a(x1)) |
→ |
b(c(a(x1))) |
(6) |
1.1 Rule Removal
Using the
linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1
over the naturals
[a(x1)] |
= |
|
1 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
|
|
· x1 +
|
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
|
|
[b(x1)] |
= |
|
1 |
1 |
1 |
1 |
1 |
0 |
1 |
1 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
1 |
1 |
1 |
0 |
1 |
1 |
1 |
1 |
|
|
· x1 +
|
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
|
|
[c(x1)] |
= |
|
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
|
· x1 +
|
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
|
|
all of the following rules can be deleted.
b(a(x1)) |
→ |
a(a(a(b(x1)))) |
(4) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
[a(x1)] |
= |
· x1 +
|
[b(x1)] |
= |
· x1 +
|
[c(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
a(a(x1)) |
→ |
b(c(a(x1))) |
(6) |
1.1.1.1 Rule Removal
Using the
Weighted Path Order with the following precedence and status
prec(a) |
= |
0 |
|
status(a) |
= |
[1] |
|
list-extension(a) |
= |
Lex |
prec(b) |
= |
0 |
|
status(b) |
= |
[1] |
|
list-extension(b) |
= |
Lex |
and the following
Max-polynomial interpretation
[a(x1)] |
=
|
max(2, 2 + 1 · x1) |
[b(x1)] |
=
|
max(5, 5 + 1 · x1) |
all of the following rules can be deleted.
1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.