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 AProVE @ termCOMP 2023)
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
1(1(x1)) |
→ |
3(4(x1)) |
(10) |
|
2(1(x1)) |
→ |
1(2(x1)) |
(11) |
|
2(2(x1)) |
→ |
1(1(1(x1))) |
(3) |
|
3(3(x1)) |
→ |
6(5(x1)) |
(12) |
|
4(3(x1)) |
→ |
1(1(x1)) |
(13) |
|
4(4(x1)) |
→ |
3(x1) |
(6) |
|
5(5(x1)) |
→ |
2(6(x1)) |
(14) |
|
6(5(x1)) |
→ |
2(1(x1)) |
(15) |
|
6(6(x1)) |
→ |
1(2(x1)) |
(16) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [1(x1)] |
= |
1 · x1 + 85 |
| [3(x1)] |
= |
1 · x1 + 113 |
| [4(x1)] |
= |
1 · x1 + 57 |
| [2(x1)] |
= |
1 · x1 + 128 |
| [6(x1)] |
= |
1 · x1 + 107 |
| [5(x1)] |
= |
1 · x1 + 118 |
all of the following rules can be deleted.
|
2(2(x1)) |
→ |
1(1(1(x1))) |
(3) |
|
3(3(x1)) |
→ |
6(5(x1)) |
(12) |
|
4(4(x1)) |
→ |
3(x1) |
(6) |
|
5(5(x1)) |
→ |
2(6(x1)) |
(14) |
|
6(5(x1)) |
→ |
2(1(x1)) |
(15) |
|
6(6(x1)) |
→ |
1(2(x1)) |
(16) |
1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
1#(1(x1)) |
→ |
4#(x1) |
(17) |
|
2#(1(x1)) |
→ |
1#(2(x1)) |
(18) |
|
2#(1(x1)) |
→ |
2#(x1) |
(19) |
|
4#(3(x1)) |
→ |
1#(1(x1)) |
(20) |
|
4#(3(x1)) |
→ |
1#(x1) |
(21) |
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.
-
The
1st
component contains the
pair
1.1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
| [1(x1)] |
= |
1 · x1
|
| [2#(x1)] |
= |
1 · x1
|
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
rule
could be deleted.
1.1.1.1.1.1 Size-Change Termination
Using size-change termination in combination with
the subterm criterion
one obtains the following initial size-change graphs.
|
2#(1(x1)) |
→ |
2#(x1) |
(19) |
|
| 1 |
> |
1 |
As there is no critical graph in the transitive closure, there are no infinite chains.
-
The
2nd
component contains the
pair
|
4#(3(x1)) |
→ |
1#(1(x1)) |
(20) |
|
1#(1(x1)) |
→ |
4#(x1) |
(17) |
|
4#(3(x1)) |
→ |
1#(x1) |
(21) |
1.1.1.1.2 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
| [1(x1)] |
= |
1 · x1
|
| [3(x1)] |
= |
1 · x1
|
| [4(x1)] |
= |
1 · x1
|
| [1#(x1)] |
= |
1 · x1
|
| [4#(x1)] |
= |
1 · x1
|
together with the usable
rules
|
1(1(x1)) |
→ |
3(4(x1)) |
(10) |
|
4(3(x1)) |
→ |
1(1(x1)) |
(13) |
(w.r.t. the implicit argument filter of the reduction pair),
the
rule
could be deleted.
1.1.1.1.2.1 Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [4#(x1)] |
= |
1 · x1
|
| [3(x1)] |
= |
1 + 1 · x1
|
| [1#(x1)] |
= |
1 · x1
|
| [1(x1)] |
= |
1 + 1 · x1
|
| [4(x1)] |
= |
1 + 1 · x1
|
the
pairs
|
1#(1(x1)) |
→ |
4#(x1) |
(17) |
|
4#(3(x1)) |
→ |
1#(x1) |
(21) |
could be deleted.
1.1.1.1.2.1.1 Monotonic Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [1(x1)] |
= |
2 + 2 · x1
|
| [3(x1)] |
= |
3 + 2 · x1
|
| [4(x1)] |
= |
2 · x1
|
| [4#(x1)] |
= |
2 + 3 · x1
|
| [1#(x1)] |
= |
2 · x1
|
the
pair
|
4#(3(x1)) |
→ |
1#(1(x1)) |
(20) |
and
the
rule
could be deleted.
1.1.1.1.2.1.1.1 P is empty
There are no pairs anymore.