Certification Problem
Input (TPDB TRS_Standard/AG01/#3.2)
The rewrite relation of the following TRS is considered.
|
pred(s(x)) |
→ |
x |
(1) |
|
minus(x,0) |
→ |
x |
(2) |
|
minus(x,s(y)) |
→ |
pred(minus(x,y)) |
(3) |
|
quot(0,s(y)) |
→ |
0 |
(4) |
|
quot(s(x),s(y)) |
→ |
s(quot(minus(x,y),s(y))) |
(5) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
minus#(x,s(y)) |
→ |
minus#(x,y) |
(6) |
|
minus#(x,s(y)) |
→ |
pred#(minus(x,y)) |
(7) |
|
quot#(s(x),s(y)) |
→ |
minus#(x,y) |
(8) |
|
quot#(s(x),s(y)) |
→ |
quot#(minus(x,y),s(y)) |
(9) |
1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.
-
The
1st
component contains the
pair
|
quot#(s(x),s(y)) |
→ |
quot#(minus(x,y),s(y)) |
(9) |
1.1.1 Subterm Criterion Processor
We use the projection to multisets
| π(quot#)
|
= |
{
1
}
|
| π(minus)
|
= |
{
1
}
|
| π(pred)
|
= |
{
1
}
|
to remove the pairs:
|
quot#(s(x),s(y)) |
→ |
quot#(minus(x,y),s(y)) |
(9) |
1.1.1.1 P is empty
There are no pairs anymore.
-
The
2nd
component contains the
pair
|
minus#(x,s(y)) |
→ |
minus#(x,y) |
(6) |
1.1.2 Size-Change Termination
Using size-change termination in combination with
the subterm criterion
one obtains the following initial size-change graphs.
|
minus#(x,s(y)) |
→ |
minus#(x,y) |
(6) |
|
|
| 2 |
> |
2 |
| 1 |
≥ |
1 |
As there is no critical graph in the transitive closure, there are no infinite chains.