Certification Problem
Input (TPDB TRS_Standard/SK90/2.14)
The rewrite relation of the following TRS is considered.
double(0) |
→ |
0 |
(1) |
double(s(x)) |
→ |
s(s(double(x))) |
(2) |
half(0) |
→ |
0 |
(3) |
half(s(0)) |
→ |
0 |
(4) |
half(s(s(x))) |
→ |
s(half(x)) |
(5) |
-(x,0) |
→ |
x |
(6) |
-(s(x),s(y)) |
→ |
-(x,y) |
(7) |
if(0,y,z) |
→ |
y |
(8) |
if(s(x),y,z) |
→ |
z |
(9) |
half(double(x)) |
→ |
x |
(10) |
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.
-#(s(x),s(y)) |
→ |
-#(x,y) |
(11) |
half#(s(s(x))) |
→ |
half#(x) |
(12) |
double#(s(x)) |
→ |
double#(x) |
(13) |
1.1 Dependency Graph Processor
The dependency pairs are split into 3
components.
-
The
1st
component contains the
pair
double#(s(x)) |
→ |
double#(x) |
(13) |
1.1.1 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation
[s(x1)] |
=
|
x1 + 1 |
[half#(x1)] |
=
|
0 |
[half(x1)] |
=
|
0 |
[0] |
=
|
0 |
[if(x1, x2, x3)] |
=
|
0 |
[double#(x1)] |
=
|
x1 + 0 |
[double(x1)] |
=
|
0 |
[-(x1, x2)] |
=
|
0 |
[-#(x1, x2)] |
=
|
0 |
[if#(x1, x2, x3)] |
=
|
0 |
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pair
double#(s(x)) |
→ |
double#(x) |
(13) |
could be deleted.
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.
-
The
2nd
component contains the
pair
half#(s(s(x))) |
→ |
half#(x) |
(12) |
1.1.2 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation
[s(x1)] |
=
|
x1 + 1 |
[half#(x1)] |
=
|
x1 + 0 |
[half(x1)] |
=
|
0 |
[0] |
=
|
0 |
[if(x1, x2, x3)] |
=
|
0 |
[double#(x1)] |
=
|
0 |
[double(x1)] |
=
|
0 |
[-(x1, x2)] |
=
|
0 |
[-#(x1, x2)] |
=
|
0 |
[if#(x1, x2, x3)] |
=
|
0 |
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pair
half#(s(s(x))) |
→ |
half#(x) |
(12) |
could be deleted.
1.1.2.1 Dependency Graph Processor
The dependency pairs are split into 0
components.
-
The
3rd
component contains the
pair
-#(s(x),s(y)) |
→ |
-#(x,y) |
(11) |
1.1.3 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation
[s(x1)] |
=
|
x1 + 1 |
[half#(x1)] |
=
|
0 |
[half(x1)] |
=
|
0 |
[0] |
=
|
0 |
[if(x1, x2, x3)] |
=
|
0 |
[double#(x1)] |
=
|
0 |
[double(x1)] |
=
|
0 |
[-(x1, x2)] |
=
|
0 |
[-#(x1, x2)] |
=
|
x2 + 0 |
[if#(x1, x2, x3)] |
=
|
0 |
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pair
-#(s(x),s(y)) |
→ |
-#(x,y) |
(11) |
could be deleted.
1.1.3.1 Dependency Graph Processor
The dependency pairs are split into 0
components.