Certification Problem
Input (TPDB SRS_Standard/Waldmann_19/random-97)
The rewrite relation of the following TRS is considered.
|
a(a(a(b(x1)))) |
→ |
b(a(b(a(x1)))) |
(1) |
|
a(a(b(b(x1)))) |
→ |
a(a(a(b(x1)))) |
(2) |
|
a(a(a(b(x1)))) |
→ |
b(b(a(b(x1)))) |
(3) |
|
b(b(b(a(x1)))) |
→ |
b(a(b(b(x1)))) |
(4) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Closure Under Flat Contexts
Using the flat contexts
{a(☐), b(☐)}
We obtain the transformed TRS
|
a(a(b(b(x1)))) |
→ |
a(a(a(b(x1)))) |
(2) |
|
b(b(b(a(x1)))) |
→ |
b(a(b(b(x1)))) |
(4) |
|
a(a(a(a(b(x1))))) |
→ |
a(b(a(b(a(x1))))) |
(5) |
|
b(a(a(a(b(x1))))) |
→ |
b(b(a(b(a(x1))))) |
(6) |
|
a(a(a(a(b(x1))))) |
→ |
a(b(b(a(b(x1))))) |
(7) |
|
b(a(a(a(b(x1))))) |
→ |
b(b(b(a(b(x1))))) |
(8) |
1.1 Semantic Labeling
Root-labeling is applied.
We obtain the labeled TRS
|
aa(ab(bb(ba(x1)))) |
→ |
aa(aa(ab(ba(x1)))) |
(9) |
|
aa(ab(bb(bb(x1)))) |
→ |
aa(aa(ab(bb(x1)))) |
(10) |
|
bb(bb(ba(aa(x1)))) |
→ |
ba(ab(bb(ba(x1)))) |
(11) |
|
bb(bb(ba(ab(x1)))) |
→ |
ba(ab(bb(bb(x1)))) |
(12) |
|
aa(aa(aa(ab(ba(x1))))) |
→ |
ab(ba(ab(ba(aa(x1))))) |
(13) |
|
aa(aa(aa(ab(bb(x1))))) |
→ |
ab(ba(ab(ba(ab(x1))))) |
(14) |
|
ba(aa(aa(ab(ba(x1))))) |
→ |
bb(ba(ab(ba(aa(x1))))) |
(15) |
|
ba(aa(aa(ab(bb(x1))))) |
→ |
bb(ba(ab(ba(ab(x1))))) |
(16) |
|
aa(aa(aa(ab(ba(x1))))) |
→ |
ab(bb(ba(ab(ba(x1))))) |
(17) |
|
aa(aa(aa(ab(bb(x1))))) |
→ |
ab(bb(ba(ab(bb(x1))))) |
(18) |
|
ba(aa(aa(ab(ba(x1))))) |
→ |
bb(bb(ba(ab(ba(x1))))) |
(19) |
|
ba(aa(aa(ab(bb(x1))))) |
→ |
bb(bb(ba(ab(bb(x1))))) |
(20) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [aa(x1)] |
= |
1 · x1 + 1 |
| [ab(x1)] |
= |
1 · x1
|
| [bb(x1)] |
= |
1 · x1 + 1 |
| [ba(x1)] |
= |
1 · x1 + 1 |
all of the following rules can be deleted.
|
bb(bb(ba(aa(x1)))) |
→ |
ba(ab(bb(ba(x1)))) |
(11) |
|
aa(aa(aa(ab(ba(x1))))) |
→ |
ab(ba(ab(ba(aa(x1))))) |
(13) |
|
aa(aa(aa(ab(bb(x1))))) |
→ |
ab(ba(ab(ba(ab(x1))))) |
(14) |
|
ba(aa(aa(ab(bb(x1))))) |
→ |
bb(ba(ab(ba(ab(x1))))) |
(16) |
|
aa(aa(aa(ab(ba(x1))))) |
→ |
ab(bb(ba(ab(ba(x1))))) |
(17) |
|
aa(aa(aa(ab(bb(x1))))) |
→ |
ab(bb(ba(ab(bb(x1))))) |
(18) |
1.1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
aa#(ab(bb(ba(x1)))) |
→ |
aa#(aa(ab(ba(x1)))) |
(21) |
|
aa#(ab(bb(ba(x1)))) |
→ |
aa#(ab(ba(x1))) |
(22) |
|
aa#(ab(bb(bb(x1)))) |
→ |
aa#(aa(ab(bb(x1)))) |
(23) |
|
aa#(ab(bb(bb(x1)))) |
→ |
aa#(ab(bb(x1))) |
(24) |
|
bb#(bb(ba(ab(x1)))) |
→ |
ba#(ab(bb(bb(x1)))) |
(25) |
|
bb#(bb(ba(ab(x1)))) |
→ |
bb#(bb(x1)) |
(26) |
|
bb#(bb(ba(ab(x1)))) |
→ |
bb#(x1) |
(27) |
|
ba#(aa(aa(ab(ba(x1))))) |
→ |
bb#(ba(ab(ba(aa(x1))))) |
(28) |
|
ba#(aa(aa(ab(ba(x1))))) |
→ |
ba#(ab(ba(aa(x1)))) |
(29) |
|
ba#(aa(aa(ab(ba(x1))))) |
→ |
ba#(aa(x1)) |
(30) |
|
ba#(aa(aa(ab(ba(x1))))) |
→ |
aa#(x1) |
(31) |
|
ba#(aa(aa(ab(ba(x1))))) |
→ |
bb#(bb(ba(ab(ba(x1))))) |
(32) |
|
ba#(aa(aa(ab(ba(x1))))) |
→ |
bb#(ba(ab(ba(x1)))) |
(33) |
|
ba#(aa(aa(ab(ba(x1))))) |
→ |
ba#(ab(ba(x1))) |
(34) |
|
ba#(aa(aa(ab(bb(x1))))) |
→ |
bb#(bb(ba(ab(bb(x1))))) |
(35) |
|
ba#(aa(aa(ab(bb(x1))))) |
→ |
bb#(ba(ab(bb(x1)))) |
(36) |
|
ba#(aa(aa(ab(bb(x1))))) |
→ |
ba#(ab(bb(x1))) |
(37) |
1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 3
components.
-
The
1st
component contains the
pair
|
ba#(aa(aa(ab(ba(x1))))) |
→ |
ba#(aa(x1)) |
(30) |
1.1.1.1.1.1 Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [ba#(x1)] |
= |
1 · x1
|
| [aa(x1)] |
= |
1 · x1
|
| [ab(x1)] |
= |
1 + 1 · x1
|
| [ba(x1)] |
= |
1 · x1
|
| [bb(x1)] |
= |
1 · x1
|
the
pair
|
ba#(aa(aa(ab(ba(x1))))) |
→ |
ba#(aa(x1)) |
(30) |
could be deleted.
1.1.1.1.1.1.1 P is empty
There are no pairs anymore.
-
The
2nd
component contains the
pair
|
bb#(bb(ba(ab(x1)))) |
→ |
bb#(x1) |
(27) |
|
bb#(bb(ba(ab(x1)))) |
→ |
bb#(bb(x1)) |
(26) |
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)))) |
→ |
ba(ab(bb(bb(x1)))) |
(12) |
(w.r.t. the implicit argument filter of the reduction pair),
the
rule
could be deleted.
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.
1.1.1.1.1.2.1.1 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 · x1
|
the
pairs
|
bb#(bb(ba(ab(x1)))) |
→ |
bb#(x1) |
(27) |
|
bb#(bb(ba(ab(x1)))) |
→ |
bb#(bb(x1)) |
(26) |
could be deleted.
1.1.1.1.1.2.1.1.1 P is empty
There are no pairs anymore.
-
The
3rd
component contains the
pair
|
aa#(ab(bb(bb(x1)))) |
→ |
aa#(ab(bb(x1))) |
(24) |
|
aa#(ab(bb(ba(x1)))) |
→ |
aa#(ab(ba(x1))) |
(22) |
1.1.1.1.1.3 Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [aa#(x1)] |
= |
1 · x1
|
| [ab(x1)] |
= |
1 · x1
|
| [bb(x1)] |
= |
1 + 1 · x1
|
| [ba(x1)] |
= |
1 + 1 · x1
|
| [aa(x1)] |
= |
1 + 1 · x1
|
the
pairs
|
aa#(ab(bb(bb(x1)))) |
→ |
aa#(ab(bb(x1))) |
(24) |
|
aa#(ab(bb(ba(x1)))) |
→ |
aa#(ab(ba(x1))) |
(22) |
could be deleted.
1.1.1.1.1.3.1 P is empty
There are no pairs anymore.