Certification Problem
Input (TPDB SRS_Relative/Waldmann_23/size-10-alpha-2-num-85)
The relative rewrite relation R/S is considered where R is the following TRS
and S is the following TRS.
|
a(b(x)) |
→ |
b(b(a(a(b(x))))) |
(2) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
a(b(a(x))) |
→ |
x |
(1) |
|
b(a(x)) |
→ |
b(a(a(b(b(x))))) |
(3) |
1.1 Closure Under Flat Contexts
Using the flat contexts
{a(☐), b(☐)}
We obtain the transformed TRS
|
a(a(b(a(x)))) |
→ |
a(x) |
(4) |
|
b(a(b(a(x)))) |
→ |
b(x) |
(5) |
|
b(a(x)) |
→ |
b(a(a(b(b(x))))) |
(3) |
1.1.1 Semantic Labeling
Root-labeling is applied.
We obtain the labeled TRS
|
aa(ab(ba(aa(x)))) |
→ |
aa(x) |
(6) |
|
aa(ab(ba(ab(x)))) |
→ |
ab(x) |
(7) |
|
ba(ab(ba(aa(x)))) |
→ |
ba(x) |
(8) |
|
ba(ab(ba(ab(x)))) |
→ |
bb(x) |
(9) |
|
ba(aa(x)) |
→ |
ba(aa(ab(bb(ba(x))))) |
(10) |
|
ba(ab(x)) |
→ |
ba(aa(ab(bb(bb(x))))) |
(11) |
1.1.1.1 Rule Removal
Using the
matrix interpretations of dimension 2 with strict dimension 1 over the integers
| [aa(x1)] |
= |
+ · x1
|
| [ab(x1)] |
= |
+ · x1
|
| [ba(x1)] |
= |
+ · x1
|
| [bb(x1)] |
= |
+ · x1
|
all of the following rules can be deleted.
|
aa(ab(ba(aa(x)))) |
→ |
aa(x) |
(6) |
|
aa(ab(ba(ab(x)))) |
→ |
ab(x) |
(7) |
|
ba(ab(ba(aa(x)))) |
→ |
ba(x) |
(8) |
|
ba(ab(ba(ab(x)))) |
→ |
bb(x) |
(9) |
1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.