Certification Problem
Input (TPDB TRS_Standard/SK90/2.19)
The rewrite relation of the following TRS is considered.
|
sqr(0) |
→ |
0 |
(1) |
|
sqr(s(x)) |
→ |
+(sqr(x),s(double(x))) |
(2) |
|
double(0) |
→ |
0 |
(3) |
|
double(s(x)) |
→ |
s(s(double(x))) |
(4) |
|
+(x,0) |
→ |
x |
(5) |
|
+(x,s(y)) |
→ |
s(+(x,y)) |
(6) |
|
sqr(s(x)) |
→ |
s(+(sqr(x),double(x))) |
(7) |
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.
|
sqr#(s(x)) |
→ |
sqr#(x) |
(8) |
|
sqr#(s(x)) |
→ |
sqr#(x) |
(8) |
|
sqr#(s(x)) |
→ |
double#(x) |
(9) |
|
sqr#(s(x)) |
→ |
+#(sqr(x),double(x)) |
(10) |
|
sqr#(s(x)) |
→ |
double#(x) |
(9) |
|
sqr#(s(x)) |
→ |
+#(sqr(x),s(double(x))) |
(11) |
|
double#(s(x)) |
→ |
double#(x) |
(12) |
|
+#(x,s(y)) |
→ |
+#(x,y) |
(13) |
1.1 Dependency Graph Processor
The dependency pairs are split into 3
components.
-
The
1st
component contains the
pair
|
sqr#(s(x)) |
→ |
sqr#(x) |
(8) |
|
sqr#(s(x)) |
→ |
sqr#(x) |
(8) |
1.1.1 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation
| [s(x1)] |
=
|
x1 + 1 |
| [sqr#(x1)] |
=
|
x1 + 0 |
| [0] |
=
|
0 |
| [double#(x1)] |
=
|
0 |
| [double(x1)] |
=
|
0 |
| [sqr(x1)] |
=
|
0 |
| [+(x1, x2)] |
=
|
0 |
| [+#(x1, x2)] |
=
|
0 |
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pairs
|
sqr#(s(x)) |
→ |
sqr#(x) |
(8) |
|
sqr#(s(x)) |
→ |
sqr#(x) |
(8) |
could be deleted.
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.
-
The
2nd
component contains the
pair
|
+#(x,s(y)) |
→ |
+#(x,y) |
(13) |
1.1.2 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation
| [s(x1)] |
=
|
x1 + 1 |
| [sqr#(x1)] |
=
|
0 |
| [0] |
=
|
0 |
| [double#(x1)] |
=
|
0 |
| [double(x1)] |
=
|
0 |
| [sqr(x1)] |
=
|
0 |
| [+(x1, x2)] |
=
|
0 |
| [+#(x1, x2)] |
=
|
x2 + 0 |
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pair
|
+#(x,s(y)) |
→ |
+#(x,y) |
(13) |
could be deleted.
1.1.2.1 Dependency Graph Processor
The dependency pairs are split into 0
components.
-
The
3rd
component contains the
pair
|
double#(s(x)) |
→ |
double#(x) |
(12) |
1.1.3 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation
| [s(x1)] |
=
|
x1 + 1 |
| [sqr#(x1)] |
=
|
0 |
| [0] |
=
|
0 |
| [double#(x1)] |
=
|
x1 + 0 |
| [double(x1)] |
=
|
0 |
| [sqr(x1)] |
=
|
0 |
| [+(x1, x2)] |
=
|
0 |
| [+#(x1, x2)] |
=
|
0 |
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pair
|
double#(s(x)) |
→ |
double#(x) |
(12) |
could be deleted.
1.1.3.1 Dependency Graph Processor
The dependency pairs are split into 0
components.