Certification Problem
Input (TPDB SRS_Standard/Trafo_06/dup06)
The rewrite relation of the following TRS is considered.
|
a(a(b(b(x1)))) |
→ |
b(b(c(c(a(a(x1)))))) |
(1) |
|
b(b(c(c(x1)))) |
→ |
c(c(b(b(b(b(x1)))))) |
(2) |
|
b(b(a(a(x1)))) |
→ |
a(a(c(c(b(b(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#(b(c(c(x1)))) |
→ |
b#(x1) |
(4) |
|
b#(b(c(c(x1)))) |
→ |
b#(b(x1)) |
(5) |
|
b#(b(c(c(x1)))) |
→ |
b#(b(b(x1))) |
(6) |
|
b#(b(c(c(x1)))) |
→ |
b#(b(b(b(x1)))) |
(7) |
|
b#(b(a(a(x1)))) |
→ |
b#(x1) |
(8) |
|
b#(b(a(a(x1)))) |
→ |
b#(b(x1)) |
(9) |
|
b#(b(a(a(x1)))) |
→ |
a#(c(c(b(b(x1))))) |
(10) |
|
b#(b(a(a(x1)))) |
→ |
a#(a(c(c(b(b(x1)))))) |
(11) |
|
a#(a(b(b(x1)))) |
→ |
b#(c(c(a(a(x1))))) |
(12) |
|
a#(a(b(b(x1)))) |
→ |
b#(b(c(c(a(a(x1)))))) |
(13) |
|
a#(a(b(b(x1)))) |
→ |
a#(x1) |
(14) |
|
a#(a(b(b(x1)))) |
→ |
a#(a(x1)) |
(15) |
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
| [c(x1)] |
= |
x1 +
|
| [b(x1)] |
= |
x1 +
|
| [a(x1)] |
= |
x1 +
|
| [b#(x1)] |
= |
x1 +
|
| [a#(x1)] |
= |
x1 +
|
together with the usable
rules
|
a(a(b(b(x1)))) |
→ |
b(b(c(c(a(a(x1)))))) |
(1) |
|
b(b(c(c(x1)))) |
→ |
c(c(b(b(b(b(x1)))))) |
(2) |
|
b(b(a(a(x1)))) |
→ |
a(a(c(c(b(b(x1)))))) |
(3) |
(w.r.t. the implicit argument filter of the reduction pair),
the
pairs
|
b#(b(a(a(x1)))) |
→ |
b#(x1) |
(8) |
|
b#(b(a(a(x1)))) |
→ |
b#(b(x1)) |
(9) |
|
b#(b(a(a(x1)))) |
→ |
a#(c(c(b(b(x1))))) |
(10) |
|
a#(a(b(b(x1)))) |
→ |
a#(x1) |
(14) |
and
no rules
could be deleted.
1.1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.