Certification Problem
Input (TPDB SRS_Standard/Secret_07_SRS/dj)
The rewrite relation of the following TRS is considered.
|
1(0(x1)) |
→ |
0(0(0(1(x1)))) |
(1) |
|
0(1(x1)) |
→ |
1(x1) |
(2) |
|
1(1(x1)) |
→ |
0(0(0(0(x1)))) |
(3) |
|
0(0(x1)) |
→ |
0(x1) |
(4) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by matchbox @ termCOMP 2023)
1 Rule Removal
Using the
matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
| [1(x1)] |
= |
x1 +
|
| [0(x1)] |
= |
x1 +
|
all of the following rules can be deleted.
|
1(1(x1)) |
→ |
0(0(0(0(x1)))) |
(3) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
1#(0(x1)) |
→ |
1#(x1) |
(5) |
|
1#(0(x1)) |
→ |
0#(1(x1)) |
(6) |
|
1#(0(x1)) |
→ |
0#(0(1(x1))) |
(7) |
|
1#(0(x1)) |
→ |
0#(0(0(1(x1)))) |
(8) |
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
| [1(x1)] |
= |
x1 +
|
| [0(x1)] |
= |
x1 +
|
| [1#(x1)] |
= |
x1 +
|
| [0#(x1)] |
= |
x1 +
|
together with the usable
rules
|
1(0(x1)) |
→ |
0(0(0(1(x1)))) |
(1) |
|
0(1(x1)) |
→ |
1(x1) |
(2) |
|
0(0(x1)) |
→ |
0(x1) |
(4) |
(w.r.t. the implicit argument filter of the reduction pair),
the
pairs
|
1#(0(x1)) |
→ |
0#(1(x1)) |
(6) |
|
1#(0(x1)) |
→ |
0#(0(1(x1))) |
(7) |
|
1#(0(x1)) |
→ |
0#(0(0(1(x1)))) |
(8) |
and
no rules
could be deleted.
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.