Certification Problem
Input (TPDB SRS_Standard/Waldmann_19/random-126)
The rewrite relation of the following TRS is considered.
|
b(a(b(b(x1)))) |
→ |
a(b(a(b(x1)))) |
(1) |
|
b(a(a(b(x1)))) |
→ |
a(b(b(b(x1)))) |
(2) |
|
a(b(b(a(x1)))) |
→ |
a(b(a(a(x1)))) |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by matchbox @ termCOMP 2023)
1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
b#(a(b(b(x1)))) |
→ |
b#(a(b(x1))) |
(4) |
|
b#(a(b(b(x1)))) |
→ |
a#(b(x1)) |
(5) |
|
b#(a(b(b(x1)))) |
→ |
a#(b(a(b(x1)))) |
(6) |
|
b#(a(a(b(x1)))) |
→ |
b#(b(x1)) |
(7) |
|
b#(a(a(b(x1)))) |
→ |
b#(b(b(x1))) |
(8) |
|
b#(a(a(b(x1)))) |
→ |
a#(b(b(b(x1)))) |
(9) |
|
a#(b(b(a(x1)))) |
→ |
b#(a(a(x1))) |
(10) |
|
a#(b(b(a(x1)))) |
→ |
a#(b(a(a(x1)))) |
(11) |
|
a#(b(b(a(x1)))) |
→ |
a#(a(x1)) |
(12) |
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 +
|
| [a#(x1)] |
= |
x1 +
|
together with the usable
rules
|
b(a(b(b(x1)))) |
→ |
a(b(a(b(x1)))) |
(1) |
|
b(a(a(b(x1)))) |
→ |
a(b(b(b(x1)))) |
(2) |
|
a(b(b(a(x1)))) |
→ |
a(b(a(a(x1)))) |
(3) |
(w.r.t. the implicit argument filter of the reduction pair),
the
pairs
|
b#(a(b(b(x1)))) |
→ |
b#(a(b(x1))) |
(4) |
|
b#(a(b(b(x1)))) |
→ |
a#(b(x1)) |
(5) |
|
b#(a(b(b(x1)))) |
→ |
a#(b(a(b(x1)))) |
(6) |
|
b#(a(a(b(x1)))) |
→ |
b#(b(x1)) |
(7) |
|
b#(a(a(b(x1)))) |
→ |
b#(b(b(x1))) |
(8) |
|
b#(a(a(b(x1)))) |
→ |
a#(b(b(b(x1)))) |
(9) |
|
a#(b(b(a(x1)))) |
→ |
b#(a(a(x1))) |
(10) |
|
a#(b(b(a(x1)))) |
→ |
a#(a(x1)) |
(12) |
and
no rules
could be deleted.
1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.