Certification Problem
Input (TPDB SRS_Standard/Waldmann_19/random-158)
The rewrite relation of the following TRS is considered.
|
a(a(a(a(x1)))) |
→ |
a(b(a(b(x1)))) |
(1) |
|
a(a(b(a(x1)))) |
→ |
a(b(a(a(x1)))) |
(2) |
|
b(a(b(b(x1)))) |
→ |
b(b(a(a(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(aa(aa(aa(x1)))) |
→ |
ab(ba(ab(ba(x1)))) |
(4) |
|
aa(aa(aa(ab(x1)))) |
→ |
ab(ba(ab(bb(x1)))) |
(5) |
|
aa(ab(ba(aa(x1)))) |
→ |
ab(ba(aa(aa(x1)))) |
(6) |
|
aa(ab(ba(ab(x1)))) |
→ |
ab(ba(aa(ab(x1)))) |
(7) |
|
ba(ab(bb(ba(x1)))) |
→ |
bb(ba(aa(aa(x1)))) |
(8) |
|
ba(ab(bb(bb(x1)))) |
→ |
bb(ba(aa(ab(x1)))) |
(9) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
aa#(aa(aa(aa(x1)))) |
→ |
ba#(ab(ba(x1))) |
(10) |
|
aa#(aa(aa(aa(x1)))) |
→ |
ba#(x1) |
(11) |
|
aa#(aa(aa(ab(x1)))) |
→ |
ba#(ab(bb(x1))) |
(12) |
|
aa#(ab(ba(aa(x1)))) |
→ |
ba#(aa(aa(x1))) |
(13) |
|
aa#(ab(ba(aa(x1)))) |
→ |
aa#(aa(x1)) |
(14) |
|
aa#(ab(ba(ab(x1)))) |
→ |
ba#(aa(ab(x1))) |
(15) |
|
aa#(ab(ba(ab(x1)))) |
→ |
aa#(ab(x1)) |
(16) |
|
ba#(ab(bb(ba(x1)))) |
→ |
ba#(aa(aa(x1))) |
(17) |
|
ba#(ab(bb(ba(x1)))) |
→ |
aa#(aa(x1)) |
(18) |
|
ba#(ab(bb(ba(x1)))) |
→ |
aa#(x1) |
(19) |
|
ba#(ab(bb(bb(x1)))) |
→ |
ba#(aa(ab(x1))) |
(20) |
|
ba#(ab(bb(bb(x1)))) |
→ |
aa#(ab(x1)) |
(21) |
1.1.1 Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [aa#(x1)] |
= |
1 · x1
|
| [aa(x1)] |
= |
1 + 1 · x1
|
| [ba#(x1)] |
= |
1 + 1 · x1
|
| [ab(x1)] |
= |
1 + 1 · x1
|
| [ba(x1)] |
= |
1 + 1 · x1
|
| [bb(x1)] |
= |
1 + 1 · x1
|
the
pairs
|
aa#(aa(aa(aa(x1)))) |
→ |
ba#(x1) |
(11) |
|
aa#(ab(ba(aa(x1)))) |
→ |
aa#(aa(x1)) |
(14) |
|
aa#(ab(ba(ab(x1)))) |
→ |
aa#(ab(x1)) |
(16) |
|
ba#(ab(bb(ba(x1)))) |
→ |
ba#(aa(aa(x1))) |
(17) |
|
ba#(ab(bb(ba(x1)))) |
→ |
aa#(aa(x1)) |
(18) |
|
ba#(ab(bb(ba(x1)))) |
→ |
aa#(x1) |
(19) |
|
ba#(ab(bb(bb(x1)))) |
→ |
ba#(aa(ab(x1))) |
(20) |
|
ba#(ab(bb(bb(x1)))) |
→ |
aa#(ab(x1)) |
(21) |
could be deleted.
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.