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 AProVE @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
[1(x1)] |
= |
1 · x1 + 1 |
[0(x1)] |
= |
1 · 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)) |
→ |
0#(0(0(1(x1)))) |
(5) |
1#(0(x1)) |
→ |
0#(0(1(x1))) |
(6) |
1#(0(x1)) |
→ |
0#(1(x1)) |
(7) |
1#(0(x1)) |
→ |
1#(x1) |
(8) |
1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.
-
The
1st
component contains the
pair
1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
[0(x1)] |
= |
1 · x1
|
[1#(x1)] |
= |
1 · x1
|
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
rule
could be deleted.
1.1.1.1.1 Size-Change Termination
Using size-change termination in combination with
the subterm criterion
one obtains the following initial size-change graphs.
1#(0(x1)) |
→ |
1#(x1) |
(8) |
|
1 |
> |
1 |
As there is no critical graph in the transitive closure, there are no infinite chains.