Certification Problem
Input (TPDB SRS_Standard/Waldmann_19/random-529)
The rewrite relation of the following TRS is considered.
|
a(b(b(a(x1)))) |
→ |
a(a(a(a(x1)))) |
(1) |
|
b(a(a(a(x1)))) |
→ |
a(a(b(b(x1)))) |
(2) |
|
b(b(a(a(x1)))) |
→ |
b(a(b(a(x1)))) |
(3) |
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(b(b(a(x1)))) |
→ |
a(a(a(a(x1)))) |
(1) |
|
b(b(a(a(x1)))) |
→ |
b(a(b(a(x1)))) |
(3) |
|
a(b(a(a(a(x1))))) |
→ |
a(a(a(b(b(x1))))) |
(4) |
|
b(b(a(a(a(x1))))) |
→ |
b(a(a(b(b(x1))))) |
(5) |
1.1 Semantic Labeling
Root-labeling is applied.
We obtain the labeled TRS
|
ab(bb(ba(aa(x1)))) |
→ |
aa(aa(aa(aa(x1)))) |
(6) |
|
ab(bb(ba(ab(x1)))) |
→ |
aa(aa(aa(ab(x1)))) |
(7) |
|
bb(ba(aa(aa(x1)))) |
→ |
ba(ab(ba(aa(x1)))) |
(8) |
|
bb(ba(aa(ab(x1)))) |
→ |
ba(ab(ba(ab(x1)))) |
(9) |
|
ab(ba(aa(aa(aa(x1))))) |
→ |
aa(aa(ab(bb(ba(x1))))) |
(10) |
|
ab(ba(aa(aa(ab(x1))))) |
→ |
aa(aa(ab(bb(bb(x1))))) |
(11) |
|
bb(ba(aa(aa(aa(x1))))) |
→ |
ba(aa(ab(bb(ba(x1))))) |
(12) |
|
bb(ba(aa(aa(ab(x1))))) |
→ |
ba(aa(ab(bb(bb(x1))))) |
(13) |
1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
bb#(ba(aa(aa(x1)))) |
→ |
ab#(ba(aa(x1))) |
(14) |
|
bb#(ba(aa(ab(x1)))) |
→ |
ab#(ba(ab(x1))) |
(15) |
|
ab#(ba(aa(aa(aa(x1))))) |
→ |
ab#(bb(ba(x1))) |
(16) |
|
ab#(ba(aa(aa(aa(x1))))) |
→ |
bb#(ba(x1)) |
(17) |
|
ab#(ba(aa(aa(ab(x1))))) |
→ |
ab#(bb(bb(x1))) |
(18) |
|
ab#(ba(aa(aa(ab(x1))))) |
→ |
bb#(bb(x1)) |
(19) |
|
ab#(ba(aa(aa(ab(x1))))) |
→ |
bb#(x1) |
(20) |
|
bb#(ba(aa(aa(aa(x1))))) |
→ |
ab#(bb(ba(x1))) |
(21) |
|
bb#(ba(aa(aa(aa(x1))))) |
→ |
bb#(ba(x1)) |
(22) |
|
bb#(ba(aa(aa(ab(x1))))) |
→ |
ab#(bb(bb(x1))) |
(23) |
|
bb#(ba(aa(aa(ab(x1))))) |
→ |
bb#(bb(x1)) |
(24) |
|
bb#(ba(aa(aa(ab(x1))))) |
→ |
bb#(x1) |
(25) |
1.1.1.1 Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [bb#(x1)] |
= |
1 · x1
|
| [ba(x1)] |
= |
1 + 1 · x1
|
| [aa(x1)] |
= |
1 + 1 · x1
|
| [ab#(x1)] |
= |
1 · x1
|
| [ab(x1)] |
= |
1 + 1 · x1
|
| [bb(x1)] |
= |
1 + 1 · x1
|
the
pairs
|
bb#(ba(aa(aa(x1)))) |
→ |
ab#(ba(aa(x1))) |
(14) |
|
bb#(ba(aa(ab(x1)))) |
→ |
ab#(ba(ab(x1))) |
(15) |
|
ab#(ba(aa(aa(aa(x1))))) |
→ |
ab#(bb(ba(x1))) |
(16) |
|
ab#(ba(aa(aa(aa(x1))))) |
→ |
bb#(ba(x1)) |
(17) |
|
ab#(ba(aa(aa(ab(x1))))) |
→ |
ab#(bb(bb(x1))) |
(18) |
|
ab#(ba(aa(aa(ab(x1))))) |
→ |
bb#(bb(x1)) |
(19) |
|
ab#(ba(aa(aa(ab(x1))))) |
→ |
bb#(x1) |
(20) |
|
bb#(ba(aa(aa(aa(x1))))) |
→ |
ab#(bb(ba(x1))) |
(21) |
|
bb#(ba(aa(aa(aa(x1))))) |
→ |
bb#(ba(x1)) |
(22) |
|
bb#(ba(aa(aa(ab(x1))))) |
→ |
ab#(bb(bb(x1))) |
(23) |
|
bb#(ba(aa(aa(ab(x1))))) |
→ |
bb#(bb(x1)) |
(24) |
|
bb#(ba(aa(aa(ab(x1))))) |
→ |
bb#(x1) |
(25) |
could be deleted.
1.1.1.1.1 P is empty
There are no pairs anymore.