Certification Problem
Input (TPDB SRS_Standard/Waldmann_19/random-402)
The rewrite relation of the following TRS is considered.
a(b(b(b(x1)))) |
→ |
b(b(b(b(x1)))) |
(1) |
b(b(b(a(x1)))) |
→ |
b(a(b(a(x1)))) |
(2) |
b(b(a(a(x1)))) |
→ |
b(a(b(a(x1)))) |
(3) |
a(a(a(b(x1)))) |
→ |
a(b(a(a(x1)))) |
(4) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
b(b(b(a(x1)))) |
→ |
b(b(b(b(x1)))) |
(5) |
a(b(b(b(x1)))) |
→ |
a(b(a(b(x1)))) |
(6) |
a(a(b(b(x1)))) |
→ |
a(b(a(b(x1)))) |
(7) |
b(a(a(a(x1)))) |
→ |
a(a(b(a(x1)))) |
(8) |
1.1 Closure Under Flat Contexts
Using the flat contexts
{b(☐), a(☐)}
We obtain the transformed TRS
b(b(b(a(x1)))) |
→ |
b(b(b(b(x1)))) |
(5) |
a(b(b(b(x1)))) |
→ |
a(b(a(b(x1)))) |
(6) |
a(a(b(b(x1)))) |
→ |
a(b(a(b(x1)))) |
(7) |
b(b(a(a(a(x1))))) |
→ |
b(a(a(b(a(x1))))) |
(9) |
a(b(a(a(a(x1))))) |
→ |
a(a(a(b(a(x1))))) |
(10) |
1.1.1 Semantic Labeling
Root-labeling is applied.
We obtain the labeled TRS
bb(bb(ba(ab(x1)))) |
→ |
bb(bb(bb(bb(x1)))) |
(11) |
bb(bb(ba(aa(x1)))) |
→ |
bb(bb(bb(ba(x1)))) |
(12) |
ab(bb(bb(bb(x1)))) |
→ |
ab(ba(ab(bb(x1)))) |
(13) |
ab(bb(bb(ba(x1)))) |
→ |
ab(ba(ab(ba(x1)))) |
(14) |
aa(ab(bb(bb(x1)))) |
→ |
ab(ba(ab(bb(x1)))) |
(15) |
aa(ab(bb(ba(x1)))) |
→ |
ab(ba(ab(ba(x1)))) |
(16) |
bb(ba(aa(aa(ab(x1))))) |
→ |
ba(aa(ab(ba(ab(x1))))) |
(17) |
bb(ba(aa(aa(aa(x1))))) |
→ |
ba(aa(ab(ba(aa(x1))))) |
(18) |
ab(ba(aa(aa(ab(x1))))) |
→ |
aa(aa(ab(ba(ab(x1))))) |
(19) |
ab(ba(aa(aa(aa(x1))))) |
→ |
aa(aa(ab(ba(aa(x1))))) |
(20) |
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
[bb(x1)] |
= |
1 · x1
|
[ba(x1)] |
= |
1 · x1
|
[ab(x1)] |
= |
1 · x1
|
[aa(x1)] |
= |
1 · x1 + 1 |
all of the following rules can be deleted.
bb(bb(ba(aa(x1)))) |
→ |
bb(bb(bb(ba(x1)))) |
(12) |
aa(ab(bb(bb(x1)))) |
→ |
ab(ba(ab(bb(x1)))) |
(15) |
aa(ab(bb(ba(x1)))) |
→ |
ab(ba(ab(ba(x1)))) |
(16) |
bb(ba(aa(aa(ab(x1))))) |
→ |
ba(aa(ab(ba(ab(x1))))) |
(17) |
bb(ba(aa(aa(aa(x1))))) |
→ |
ba(aa(ab(ba(aa(x1))))) |
(18) |
1.1.1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
bb#(bb(ba(ab(x1)))) |
→ |
bb#(bb(bb(bb(x1)))) |
(21) |
bb#(bb(ba(ab(x1)))) |
→ |
bb#(bb(bb(x1))) |
(22) |
bb#(bb(ba(ab(x1)))) |
→ |
bb#(bb(x1)) |
(23) |
bb#(bb(ba(ab(x1)))) |
→ |
bb#(x1) |
(24) |
ab#(bb(bb(bb(x1)))) |
→ |
ab#(ba(ab(bb(x1)))) |
(25) |
ab#(bb(bb(bb(x1)))) |
→ |
ab#(bb(x1)) |
(26) |
ab#(bb(bb(ba(x1)))) |
→ |
ab#(ba(ab(ba(x1)))) |
(27) |
ab#(bb(bb(ba(x1)))) |
→ |
ab#(ba(x1)) |
(28) |
ab#(ba(aa(aa(ab(x1))))) |
→ |
ab#(ba(ab(x1))) |
(29) |
ab#(ba(aa(aa(aa(x1))))) |
→ |
ab#(ba(aa(x1))) |
(30) |
1.1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 3
components.
-
The
1st
component contains the
pair
ab#(bb(bb(bb(x1)))) |
→ |
ab#(bb(x1)) |
(26) |
1.1.1.1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
[bb(x1)] |
= |
1 · x1
|
[ba(x1)] |
= |
1 · x1
|
[ab(x1)] |
= |
1 · x1
|
[ab#(x1)] |
= |
1 · x1
|
together with the usable
rule
bb(bb(ba(ab(x1)))) |
→ |
bb(bb(bb(bb(x1)))) |
(11) |
(w.r.t. the implicit argument filter of the reduction pair),
the
rule
could be deleted.
1.1.1.1.1.1.1.1 Size-Change Termination
Using size-change termination in combination with
the subterm criterion
one obtains the following initial size-change graphs.
ab#(bb(bb(bb(x1)))) |
→ |
ab#(bb(x1)) |
(26) |
|
1 |
> |
1 |
As there is no critical graph in the transitive closure, there are no infinite chains.
-
The
2nd
component contains the
pair
bb#(bb(ba(ab(x1)))) |
→ |
bb#(x1) |
(24) |
bb#(bb(ba(ab(x1)))) |
→ |
bb#(bb(x1)) |
(23) |
1.1.1.1.1.1.2 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
[bb(x1)] |
= |
1 · x1
|
[ba(x1)] |
= |
1 · x1
|
[ab(x1)] |
= |
1 · x1
|
[bb#(x1)] |
= |
1 · x1
|
together with the usable
rule
bb(bb(ba(ab(x1)))) |
→ |
bb(bb(bb(bb(x1)))) |
(11) |
(w.r.t. the implicit argument filter of the reduction pair),
the
rule
could be deleted.
1.1.1.1.1.1.2.1 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.
1.1.1.1.1.1.2.1.1 Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
[bb#(x1)] |
= |
1 · x1
|
[bb(x1)] |
= |
1 · x1
|
[ba(x1)] |
= |
1 + 1 · x1
|
[ab(x1)] |
= |
1 + 1 · x1
|
the
pairs
bb#(bb(ba(ab(x1)))) |
→ |
bb#(x1) |
(24) |
bb#(bb(ba(ab(x1)))) |
→ |
bb#(bb(x1)) |
(23) |
could be deleted.
1.1.1.1.1.1.2.1.1.1 P is empty
There are no pairs anymore.
-
The
3rd
component contains the
pair
ab#(ba(aa(aa(aa(x1))))) |
→ |
ab#(ba(aa(x1))) |
(30) |
ab#(ba(aa(aa(ab(x1))))) |
→ |
ab#(ba(ab(x1))) |
(29) |
1.1.1.1.1.1.3 Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
[ab#(x1)] |
= |
1 · x1
|
[ba(x1)] |
= |
1 + 1 · x1
|
[aa(x1)] |
= |
1 + 1 · x1
|
[ab(x1)] |
= |
1 + 1 · x1
|
[bb(x1)] |
= |
1 + 1 · x1
|
the
pairs
ab#(ba(aa(aa(aa(x1))))) |
→ |
ab#(ba(aa(x1))) |
(30) |
ab#(ba(aa(aa(ab(x1))))) |
→ |
ab#(ba(ab(x1))) |
(29) |
could be deleted.
1.1.1.1.1.1.3.1 P is empty
There are no pairs anymore.