Certification Problem
Input (TPDB SRS_Standard/Secret_06_SRS/aprove09)
The rewrite relation of the following TRS is considered.
|
q(0(x1)) |
→ |
p(p(s(s(0(s(s(s(s(x1))))))))) |
(1) |
|
q(s(x1)) |
→ |
p(p(s(s(s(s(s(s(r(p(p(s(s(x1))))))))))))) |
(2) |
|
r(0(x1)) |
→ |
p(s(p(s(0(p(p(p(s(s(s(x1))))))))))) |
(3) |
|
r(s(x1)) |
→ |
p(s(p(s(s(q(p(s(p(s(x1)))))))))) |
(4) |
|
p(p(s(x1))) |
→ |
p(x1) |
(5) |
|
p(s(x1)) |
→ |
x1 |
(6) |
|
p(0(x1)) |
→ |
0(s(s(s(x1)))) |
(7) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [q(x1)] |
= |
1 · x1 + 1 |
| [0(x1)] |
= |
1 · x1
|
| [p(x1)] |
= |
1 · x1
|
| [s(x1)] |
= |
1 · x1
|
| [r(x1)] |
= |
1 · x1 + 1 |
all of the following rules can be deleted.
|
q(0(x1)) |
→ |
p(p(s(s(0(s(s(s(s(x1))))))))) |
(1) |
|
r(0(x1)) |
→ |
p(s(p(s(0(p(p(p(s(s(s(x1))))))))))) |
(3) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
q#(s(x1)) |
→ |
p#(p(s(s(s(s(s(s(r(p(p(s(s(x1))))))))))))) |
(8) |
|
q#(s(x1)) |
→ |
p#(s(s(s(s(s(s(r(p(p(s(s(x1)))))))))))) |
(9) |
|
q#(s(x1)) |
→ |
r#(p(p(s(s(x1))))) |
(10) |
|
q#(s(x1)) |
→ |
p#(p(s(s(x1)))) |
(11) |
|
q#(s(x1)) |
→ |
p#(s(s(x1))) |
(12) |
|
r#(s(x1)) |
→ |
p#(s(p(s(s(q(p(s(p(s(x1)))))))))) |
(13) |
|
r#(s(x1)) |
→ |
p#(s(s(q(p(s(p(s(x1)))))))) |
(14) |
|
r#(s(x1)) |
→ |
q#(p(s(p(s(x1))))) |
(15) |
|
r#(s(x1)) |
→ |
p#(s(p(s(x1)))) |
(16) |
|
r#(s(x1)) |
→ |
p#(s(x1)) |
(17) |
|
p#(p(s(x1))) |
→ |
p#(x1) |
(18) |
1.1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.
-
The
1st
component contains the
pair
|
q#(s(x1)) |
→ |
r#(p(p(s(s(x1))))) |
(10) |
|
r#(s(x1)) |
→ |
q#(p(s(p(s(x1))))) |
(15) |
1.1.1.1 Switch to Innermost Termination
The TRS does not have overlaps with the pairs and is locally confluent:
20
Hence, it suffices to show innermost termination in the following.
1.1.1.1.1 Usable Rules Processor
We restrict the rewrite rules to the following usable rules of the DP problem.
|
p(s(x1)) |
→ |
x1 |
(6) |
|
p(0(x1)) |
→ |
0(s(s(s(x1)))) |
(7) |
1.1.1.1.1.1 Innermost Lhss Removal Processor
We restrict the innermost strategy to the following left hand sides.
1.1.1.1.1.1.1 Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [q#(x1)] |
= |
-2 + 2 · x1
|
| [r#(x1)] |
= |
-2 + 2 · x1
|
| [p(x1)] |
= |
-2 + x1
|
| [s(x1)] |
= |
2 + x1
|
| [0(x1)] |
= |
-2 |
the
pairs
|
q#(s(x1)) |
→ |
r#(p(p(s(s(x1))))) |
(10) |
|
r#(s(x1)) |
→ |
q#(p(s(p(s(x1))))) |
(15) |
could be deleted.
1.1.1.1.1.1.1.1 P is empty
There are no pairs anymore.
-
The
2nd
component contains the
pair
|
p#(p(s(x1))) |
→ |
p#(x1) |
(18) |
1.1.1.2 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
| [p(x1)] |
= |
1 · x1
|
| [s(x1)] |
= |
1 · x1
|
| [p#(x1)] |
= |
1 · x1
|
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
rule
could be deleted.
1.1.1.2.1 Size-Change Termination
Using size-change termination in combination with
the subterm criterion
one obtains the following initial size-change graphs.
|
p#(p(s(x1))) |
→ |
p#(x1) |
(18) |
|
| 1 |
> |
1 |
As there is no critical graph in the transitive closure, there are no infinite chains.