Certification Problem
Input (TPDB SRS_Standard/Waldmann_19/random-87)
The rewrite relation of the following TRS is considered.
|
a(a(b(a(x1)))) |
→ |
a(b(a(a(x1)))) |
(1) |
|
b(a(b(b(x1)))) |
→ |
b(b(a(b(x1)))) |
(2) |
|
a(b(a(b(x1)))) |
→ |
a(b(b(b(x1)))) |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Semantic Labeling
Root-labeling is applied.
We obtain the labeled TRS
|
aa(ab(ba(aa(x1)))) |
→ |
ab(ba(aa(aa(x1)))) |
(4) |
|
aa(ab(ba(ab(x1)))) |
→ |
ab(ba(aa(ab(x1)))) |
(5) |
|
ba(ab(bb(ba(x1)))) |
→ |
bb(ba(ab(ba(x1)))) |
(6) |
|
ba(ab(bb(bb(x1)))) |
→ |
bb(ba(ab(bb(x1)))) |
(7) |
|
ab(ba(ab(ba(x1)))) |
→ |
ab(bb(bb(ba(x1)))) |
(8) |
|
ab(ba(ab(bb(x1)))) |
→ |
ab(bb(bb(bb(x1)))) |
(9) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [aa(x1)] |
= |
1 · x1
|
| [ab(x1)] |
= |
1 · x1
|
| [ba(x1)] |
= |
1 · x1 + 1 |
| [bb(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
ab(ba(ab(ba(x1)))) |
→ |
ab(bb(bb(ba(x1)))) |
(8) |
|
ab(ba(ab(bb(x1)))) |
→ |
ab(bb(bb(bb(x1)))) |
(9) |
1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
aa#(ab(ba(aa(x1)))) |
→ |
ba#(aa(aa(x1))) |
(10) |
|
aa#(ab(ba(aa(x1)))) |
→ |
aa#(aa(x1)) |
(11) |
|
aa#(ab(ba(ab(x1)))) |
→ |
ba#(aa(ab(x1))) |
(12) |
|
aa#(ab(ba(ab(x1)))) |
→ |
aa#(ab(x1)) |
(13) |
|
ba#(ab(bb(ba(x1)))) |
→ |
ba#(ab(ba(x1))) |
(14) |
|
ba#(ab(bb(bb(x1)))) |
→ |
ba#(ab(bb(x1))) |
(15) |
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.
-
The
1st
component contains the
pair
|
aa#(ab(ba(ab(x1)))) |
→ |
aa#(ab(x1)) |
(13) |
|
aa#(ab(ba(aa(x1)))) |
→ |
aa#(aa(x1)) |
(11) |
1.1.1.1.1 Size-Change Termination
Using size-change termination in combination with
the subterm criterion
one obtains the following initial size-change graphs.
|
aa#(ab(ba(ab(x1)))) |
→ |
aa#(ab(x1)) |
(13) |
|
| 1 |
> |
1 |
|
aa#(ab(ba(aa(x1)))) |
→ |
aa#(aa(x1)) |
(11) |
|
| 1 |
> |
1 |
As there is no critical graph in the transitive closure, there are no infinite chains.
-
The
2nd
component contains the
pair
|
ba#(ab(bb(bb(x1)))) |
→ |
ba#(ab(bb(x1))) |
(15) |
|
ba#(ab(bb(ba(x1)))) |
→ |
ba#(ab(ba(x1))) |
(14) |
1.1.1.1.2 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
| [ba(x1)] |
= |
1 · x1
|
| [ab(x1)] |
= |
1 · x1
|
| [bb(x1)] |
= |
1 · x1
|
| [ba#(x1)] |
= |
1 · x1
|
together with the usable
rules
|
ba(ab(bb(ba(x1)))) |
→ |
bb(ba(ab(ba(x1)))) |
(6) |
|
ba(ab(bb(bb(x1)))) |
→ |
bb(ba(ab(bb(x1)))) |
(7) |
(w.r.t. the implicit argument filter of the reduction pair),
the
rule
could be deleted.
1.1.1.1.2.1 Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [ba#(x1)] |
= |
1 · x1
|
| [ab(x1)] |
= |
1 · x1
|
| [bb(x1)] |
= |
1 + 1 · x1
|
| [ba(x1)] |
= |
1 + 1 · x1
|
the
pairs
|
ba#(ab(bb(bb(x1)))) |
→ |
ba#(ab(bb(x1))) |
(15) |
|
ba#(ab(bb(ba(x1)))) |
→ |
ba#(ab(ba(x1))) |
(14) |
could be deleted.
1.1.1.1.2.1.1 P is empty
There are no pairs anymore.