Certification Problem
Input (TPDB SRS_Standard/Waldmann_19/random-320)
The rewrite relation of the following TRS is considered.
a(a(a(a(x1)))) |
→ |
a(b(b(b(x1)))) |
(1) |
a(b(b(b(x1)))) |
→ |
b(a(b(a(x1)))) |
(2) |
b(a(a(b(x1)))) |
→ |
b(a(b(b(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(a(a(a(x1)))) |
→ |
a(b(b(b(x1)))) |
(1) |
b(a(a(b(x1)))) |
→ |
b(a(b(b(x1)))) |
(3) |
a(a(b(b(b(x1))))) |
→ |
a(b(a(b(a(x1))))) |
(4) |
b(a(b(b(b(x1))))) |
→ |
b(b(a(b(a(x1))))) |
(5) |
1.1 Semantic Labeling
Root-labeling is applied.
We obtain the labeled TRS
aa(aa(aa(aa(x1)))) |
→ |
ab(bb(bb(ba(x1)))) |
(6) |
aa(aa(aa(ab(x1)))) |
→ |
ab(bb(bb(bb(x1)))) |
(7) |
ba(aa(ab(ba(x1)))) |
→ |
ba(ab(bb(ba(x1)))) |
(8) |
ba(aa(ab(bb(x1)))) |
→ |
ba(ab(bb(bb(x1)))) |
(9) |
aa(ab(bb(bb(ba(x1))))) |
→ |
ab(ba(ab(ba(aa(x1))))) |
(10) |
aa(ab(bb(bb(bb(x1))))) |
→ |
ab(ba(ab(ba(ab(x1))))) |
(11) |
ba(ab(bb(bb(ba(x1))))) |
→ |
bb(ba(ab(ba(aa(x1))))) |
(12) |
ba(ab(bb(bb(bb(x1))))) |
→ |
bb(ba(ab(ba(ab(x1))))) |
(13) |
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.
aa(aa(aa(aa(x1)))) |
→ |
ab(bb(bb(ba(x1)))) |
(6) |
aa(ab(bb(bb(ba(x1))))) |
→ |
ab(ba(ab(ba(aa(x1))))) |
(10) |
aa(ab(bb(bb(bb(x1))))) |
→ |
ab(ba(ab(ba(ab(x1))))) |
(11) |
ba(ab(bb(bb(bb(x1))))) |
→ |
bb(ba(ab(ba(ab(x1))))) |
(13) |
1.1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
ba#(aa(ab(ba(x1)))) |
→ |
ba#(ab(bb(ba(x1)))) |
(14) |
ba#(aa(ab(bb(x1)))) |
→ |
ba#(ab(bb(bb(x1)))) |
(15) |
ba#(ab(bb(bb(ba(x1))))) |
→ |
ba#(ab(ba(aa(x1)))) |
(16) |
ba#(ab(bb(bb(ba(x1))))) |
→ |
ba#(aa(x1)) |
(17) |
ba#(ab(bb(bb(ba(x1))))) |
→ |
aa#(x1) |
(18) |
1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.