Certification Problem
Input (TPDB SRS_Standard/Zantema_04/z053)
The rewrite relation of the following TRS is considered.
|
b(a(a(b(a(b(x1)))))) |
→ |
a(b(a(a(b(b(a(x1))))))) |
(1) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by matchbox @ termCOMP 2023)
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
b(a(b(a(a(b(x1)))))) |
→ |
a(b(b(a(a(b(a(x1))))))) |
(2) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
b#(a(b(a(a(b(x1)))))) |
→ |
b#(b(a(a(b(a(x1)))))) |
(3) |
|
b#(a(b(a(a(b(x1)))))) |
→ |
b#(a(x1)) |
(4) |
|
b#(a(b(a(a(b(x1)))))) |
→ |
b#(a(a(b(a(x1))))) |
(5) |
1.1.1 Monotonic Reduction Pair Processor with Usable Rules
Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
| [b(x1)] |
= |
x1 +
|
| [a(x1)] |
= |
x1 +
|
| [b#(x1)] |
= |
x1 +
|
together with the usable
rule
|
b(a(b(a(a(b(x1)))))) |
→ |
a(b(b(a(a(b(a(x1))))))) |
(2) |
(w.r.t. the implicit argument filter of the reduction pair),
the
pairs
|
b#(a(b(a(a(b(x1)))))) |
→ |
b#(a(x1)) |
(4) |
|
b#(a(b(a(a(b(x1)))))) |
→ |
b#(a(a(b(a(x1))))) |
(5) |
and
no rules
could be deleted.
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.