Certification Problem
Input (TPDB TRS_Standard/AProVE_04/IJCAR_12)
The rewrite relation of the following TRS is considered.
|
plus(x,0) |
→ |
x |
(1) |
|
plus(0,y) |
→ |
y |
(2) |
|
plus(s(x),y) |
→ |
s(plus(x,y)) |
(3) |
|
times(0,y) |
→ |
0 |
(4) |
|
times(s(0),y) |
→ |
y |
(5) |
|
times(s(x),y) |
→ |
plus(y,times(x,y)) |
(6) |
|
div(0,y) |
→ |
0 |
(7) |
|
div(x,y) |
→ |
quot(x,y,y) |
(8) |
|
quot(0,s(y),z) |
→ |
0 |
(9) |
|
quot(s(x),s(y),z) |
→ |
quot(x,y,z) |
(10) |
|
quot(x,0,s(z)) |
→ |
s(div(x,s(z))) |
(11) |
|
div(div(x,y),z) |
→ |
div(x,times(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.
|
times#(s(x),y) |
→ |
times#(x,y) |
(13) |
|
quot#(s(x),s(y),z) |
→ |
quot#(x,y,z) |
(14) |
|
plus#(s(x),y) |
→ |
plus#(x,y) |
(15) |
|
div#(div(x,y),z) |
→ |
times#(y,z) |
(16) |
|
quot#(x,0,s(z)) |
→ |
div#(x,s(z)) |
(17) |
|
times#(s(x),y) |
→ |
plus#(y,times(x,y)) |
(18) |
|
div#(x,y) |
→ |
quot#(x,y,y) |
(19) |
|
div#(div(x,y),z) |
→ |
div#(x,times(y,z)) |
(20) |
1.1 Dependency Graph Processor
The dependency pairs are split into 3
components.
-
The
1st
component contains the
pair
|
div#(div(x,y),z) |
→ |
div#(x,times(y,z)) |
(20) |
|
div#(x,y) |
→ |
quot#(x,y,y) |
(19) |
|
quot#(x,0,s(z)) |
→ |
div#(x,s(z)) |
(17) |
|
quot#(s(x),s(y),z) |
→ |
quot#(x,y,z) |
(14) |
1.1.1 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation
| [div#(x1, x2)] |
=
|
x1 + 0 |
| [s(x1)] |
=
|
x1 + 1 |
| [plus#(x1, x2)] |
=
|
0 |
| [div(x1, x2)] |
=
|
x1 + x2 + 1 |
| [times#(x1, x2)] |
=
|
0 |
| [0] |
=
|
58714 |
| [quot(x1, x2, x3)] |
=
|
0 |
| [times(x1, x2)] |
=
|
x2 + 58713 |
| [plus(x1, x2)] |
=
|
x2 + 5854 |
| [quot#(x1, x2, x3)] |
=
|
x1 + 0 |
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pairs
|
div#(div(x,y),z) |
→ |
div#(x,times(y,z)) |
(20) |
|
quot#(s(x),s(y),z) |
→ |
quot#(x,y,z) |
(14) |
could be deleted.
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.
-
The
2nd
component contains the
pair
|
times#(s(x),y) |
→ |
times#(x,y) |
(13) |
1.1.2 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation
| [div#(x1, x2)] |
=
|
0 |
| [s(x1)] |
=
|
x1 + 1 |
| [plus#(x1, x2)] |
=
|
0 |
| [div(x1, x2)] |
=
|
1 |
| [times#(x1, x2)] |
=
|
x1 + 0 |
| [0] |
=
|
2 |
| [quot(x1, x2, x3)] |
=
|
0 |
| [times(x1, x2)] |
=
|
1 |
| [plus(x1, x2)] |
=
|
x1 + 5854 |
| [quot#(x1, x2, x3)] |
=
|
0 |
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pair
|
times#(s(x),y) |
→ |
times#(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
|
plus#(s(x),y) |
→ |
plus#(x,y) |
(15) |
1.1.3 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation
| [div#(x1, x2)] |
=
|
0 |
| [s(x1)] |
=
|
x1 + 1 |
| [plus#(x1, x2)] |
=
|
x1 + 0 |
| [div(x1, x2)] |
=
|
1 |
| [times#(x1, x2)] |
=
|
0 |
| [0] |
=
|
2 |
| [quot(x1, x2, x3)] |
=
|
0 |
| [times(x1, x2)] |
=
|
1 |
| [plus(x1, x2)] |
=
|
x1 + 25908 |
| [quot#(x1, x2, x3)] |
=
|
0 |
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pair
|
plus#(s(x),y) |
→ |
plus#(x,y) |
(15) |
could be deleted.
1.1.3.1 Dependency Graph Processor
The dependency pairs are split into 0
components.