Certification Problem
Input (TPDB TRS_Standard/SK90/2.42)
The rewrite relation of the following TRS is considered.
|
flatten(nil) |
→ |
nil |
(1) |
|
flatten(unit(x)) |
→ |
flatten(x) |
(2) |
|
flatten(++(x,y)) |
→ |
++(flatten(x),flatten(y)) |
(3) |
|
flatten(++(unit(x),y)) |
→ |
++(flatten(x),flatten(y)) |
(4) |
|
flatten(flatten(x)) |
→ |
flatten(x) |
(5) |
|
rev(nil) |
→ |
nil |
(6) |
|
rev(unit(x)) |
→ |
unit(x) |
(7) |
|
rev(++(x,y)) |
→ |
++(rev(y),rev(x)) |
(8) |
|
rev(rev(x)) |
→ |
x |
(9) |
|
++(x,nil) |
→ |
x |
(10) |
|
++(nil,y) |
→ |
y |
(11) |
|
++(++(x,y),z) |
→ |
++(x,++(y,z)) |
(12) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by NaTT @ termCOMP 2023)
1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
++#(++(x,y),z) |
→ |
++#(x,++(y,z)) |
(13) |
|
flatten#(++(x,y)) |
→ |
flatten#(y) |
(14) |
|
rev#(++(x,y)) |
→ |
rev#(x) |
(15) |
|
flatten#(++(unit(x),y)) |
→ |
flatten#(x) |
(16) |
|
flatten#(++(unit(x),y)) |
→ |
flatten#(y) |
(17) |
|
rev#(++(x,y)) |
→ |
++#(rev(y),rev(x)) |
(18) |
|
flatten#(++(unit(x),y)) |
→ |
++#(flatten(x),flatten(y)) |
(19) |
|
flatten#(++(x,y)) |
→ |
flatten#(x) |
(20) |
|
++#(++(x,y),z) |
→ |
++#(y,z) |
(21) |
|
flatten#(unit(x)) |
→ |
flatten#(x) |
(22) |
|
rev#(++(x,y)) |
→ |
rev#(y) |
(23) |
|
flatten#(++(x,y)) |
→ |
++#(flatten(x),flatten(y)) |
(24) |
1.1 Dependency Graph Processor
The dependency pairs are split into 3
components.
-
The
1st
component contains the
pair
|
rev#(++(x,y)) |
→ |
rev#(y) |
(23) |
|
rev#(++(x,y)) |
→ |
rev#(x) |
(15) |
1.1.1 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation
| [rev#(x1)] |
=
|
x1 + 0 |
| [flatten(x1)] |
=
|
0 |
| [++(x1, x2)] |
=
|
x1 + x2 + 1 |
| [unit(x1)] |
=
|
0 |
| [++#(x1, x2)] |
=
|
0 |
| [nil] |
=
|
0 |
| [rev(x1)] |
=
|
0 |
| [flatten#(x1)] |
=
|
0 |
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pairs
|
rev#(++(x,y)) |
→ |
rev#(y) |
(23) |
|
rev#(++(x,y)) |
→ |
rev#(x) |
(15) |
could be deleted.
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.
-
The
2nd
component contains the
pair
|
flatten#(++(unit(x),y)) |
→ |
flatten#(y) |
(17) |
|
flatten#(unit(x)) |
→ |
flatten#(x) |
(22) |
|
flatten#(++(unit(x),y)) |
→ |
flatten#(x) |
(16) |
|
flatten#(++(x,y)) |
→ |
flatten#(y) |
(14) |
|
flatten#(++(x,y)) |
→ |
flatten#(x) |
(20) |
1.1.2 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation
| [rev#(x1)] |
=
|
0 |
| [flatten(x1)] |
=
|
0 |
| [++(x1, x2)] |
=
|
x1 + x2 + 1 |
| [unit(x1)] |
=
|
x1 + 1 |
| [++#(x1, x2)] |
=
|
0 |
| [nil] |
=
|
0 |
| [rev(x1)] |
=
|
0 |
| [flatten#(x1)] |
=
|
x1 + 0 |
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pairs
|
flatten#(++(unit(x),y)) |
→ |
flatten#(y) |
(17) |
|
flatten#(unit(x)) |
→ |
flatten#(x) |
(22) |
|
flatten#(++(unit(x),y)) |
→ |
flatten#(x) |
(16) |
|
flatten#(++(x,y)) |
→ |
flatten#(y) |
(14) |
|
flatten#(++(x,y)) |
→ |
flatten#(x) |
(20) |
could be deleted.
1.1.2.1 Dependency Graph Processor
The dependency pairs are split into 0
components.
-
The
3rd
component contains the
pair
|
++#(++(x,y),z) |
→ |
++#(y,z) |
(21) |
|
++#(++(x,y),z) |
→ |
++#(x,++(y,z)) |
(13) |
1.1.3 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation
| [rev#(x1)] |
=
|
0 |
| [flatten(x1)] |
=
|
0 |
| [++(x1, x2)] |
=
|
x1 + x2 + 1 |
| [unit(x1)] |
=
|
1 |
| [++#(x1, x2)] |
=
|
x1 + 0 |
| [nil] |
=
|
7758 |
| [rev(x1)] |
=
|
0 |
| [flatten#(x1)] |
=
|
0 |
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pairs
|
++#(++(x,y),z) |
→ |
++#(y,z) |
(21) |
|
++#(++(x,y),z) |
→ |
++#(x,++(y,z)) |
(13) |
could be deleted.
1.1.3.1 Dependency Graph Processor
The dependency pairs are split into 0
components.