Certification Problem
Input (TPDB SRS_Standard/Zantema_04/z113)
The rewrite relation of the following TRS is considered.
|
1(1(x1)) |
→ |
4(3(x1)) |
(1) |
|
1(2(x1)) |
→ |
2(1(x1)) |
(2) |
|
2(2(x1)) |
→ |
1(1(1(x1))) |
(3) |
|
3(3(x1)) |
→ |
5(6(x1)) |
(4) |
|
3(4(x1)) |
→ |
1(1(x1)) |
(5) |
|
4(4(x1)) |
→ |
3(x1) |
(6) |
|
5(5(x1)) |
→ |
6(2(x1)) |
(7) |
|
5(6(x1)) |
→ |
1(2(x1)) |
(8) |
|
6(6(x1)) |
→ |
2(1(x1)) |
(9) |
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
| [6(x1)] |
= |
x1 +
|
| [5(x1)] |
= |
x1 +
|
| [4(x1)] |
= |
x1 +
|
| [3(x1)] |
= |
x1 +
|
| [2(x1)] |
= |
x1 +
|
| [1(x1)] |
= |
x1 +
|
all of the following rules can be deleted.
|
2(2(x1)) |
→ |
1(1(1(x1))) |
(3) |
|
3(3(x1)) |
→ |
5(6(x1)) |
(4) |
|
4(4(x1)) |
→ |
3(x1) |
(6) |
|
5(5(x1)) |
→ |
6(2(x1)) |
(7) |
|
5(6(x1)) |
→ |
1(2(x1)) |
(8) |
|
6(6(x1)) |
→ |
2(1(x1)) |
(9) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
3#(4(x1)) |
→ |
1#(x1) |
(10) |
|
3#(4(x1)) |
→ |
1#(1(x1)) |
(11) |
|
1#(2(x1)) |
→ |
1#(x1) |
(12) |
|
1#(1(x1)) |
→ |
3#(x1) |
(13) |
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
| [4(x1)] |
= |
x1 +
|
| [3(x1)] |
= |
x1 +
|
| [2(x1)] |
= |
x1 +
|
| [1(x1)] |
= |
x1 +
|
| [3#(x1)] |
= |
x1 +
|
| [1#(x1)] |
= |
x1 +
|
together with the usable
rules
|
1(1(x1)) |
→ |
4(3(x1)) |
(1) |
|
1(2(x1)) |
→ |
2(1(x1)) |
(2) |
|
3(4(x1)) |
→ |
1(1(x1)) |
(5) |
(w.r.t. the implicit argument filter of the reduction pair),
the
pairs
|
3#(4(x1)) |
→ |
1#(x1) |
(10) |
|
3#(4(x1)) |
→ |
1#(1(x1)) |
(11) |
|
1#(2(x1)) |
→ |
1#(x1) |
(12) |
|
1#(1(x1)) |
→ |
3#(x1) |
(13) |
and
no rules
could be deleted.
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.