Certification Problem
Input (TPDB SRS_Standard/Waldmann_19/random-234)
The rewrite relation of the following TRS is considered.
|
a(b(b(a(x1)))) |
→ |
b(b(a(a(x1)))) |
(1) |
|
b(b(b(a(x1)))) |
→ |
a(a(a(a(x1)))) |
(2) |
|
a(b(a(a(x1)))) |
→ |
b(a(b(b(x1)))) |
(3) |
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
|
a(b(b(a(x1)))) |
→ |
a(a(b(b(x1)))) |
(4) |
|
a(b(b(b(x1)))) |
→ |
a(a(a(a(x1)))) |
(5) |
|
a(a(b(a(x1)))) |
→ |
b(b(a(b(x1)))) |
(6) |
1.1 Closure Under Flat Contexts
Using the flat contexts
{a(☐), b(☐)}
We obtain the transformed TRS
|
a(b(b(a(x1)))) |
→ |
a(a(b(b(x1)))) |
(4) |
|
a(b(b(b(x1)))) |
→ |
a(a(a(a(x1)))) |
(5) |
|
a(a(a(b(a(x1))))) |
→ |
a(b(b(a(b(x1))))) |
(7) |
|
b(a(a(b(a(x1))))) |
→ |
b(b(b(a(b(x1))))) |
(8) |
1.1.1 Semantic Labeling
Root-labeling is applied.
We obtain the labeled TRS
|
ab(bb(ba(aa(x1)))) |
→ |
aa(ab(bb(ba(x1)))) |
(9) |
|
ab(bb(ba(ab(x1)))) |
→ |
aa(ab(bb(bb(x1)))) |
(10) |
|
ab(bb(bb(ba(x1)))) |
→ |
aa(aa(aa(aa(x1)))) |
(11) |
|
ab(bb(bb(bb(x1)))) |
→ |
aa(aa(aa(ab(x1)))) |
(12) |
|
aa(aa(ab(ba(aa(x1))))) |
→ |
ab(bb(ba(ab(ba(x1))))) |
(13) |
|
aa(aa(ab(ba(ab(x1))))) |
→ |
ab(bb(ba(ab(bb(x1))))) |
(14) |
|
ba(aa(ab(ba(aa(x1))))) |
→ |
bb(bb(ba(ab(ba(x1))))) |
(15) |
|
ba(aa(ab(ba(ab(x1))))) |
→ |
bb(bb(ba(ab(bb(x1))))) |
(16) |
1.1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
ab#(bb(ba(aa(x1)))) |
→ |
aa#(ab(bb(ba(x1)))) |
(17) |
|
ab#(bb(ba(aa(x1)))) |
→ |
ab#(bb(ba(x1))) |
(18) |
|
ab#(bb(ba(aa(x1)))) |
→ |
ba#(x1) |
(19) |
|
ab#(bb(ba(ab(x1)))) |
→ |
aa#(ab(bb(bb(x1)))) |
(20) |
|
ab#(bb(ba(ab(x1)))) |
→ |
ab#(bb(bb(x1))) |
(21) |
|
ab#(bb(bb(ba(x1)))) |
→ |
aa#(aa(aa(aa(x1)))) |
(22) |
|
ab#(bb(bb(ba(x1)))) |
→ |
aa#(aa(aa(x1))) |
(23) |
|
ab#(bb(bb(ba(x1)))) |
→ |
aa#(aa(x1)) |
(24) |
|
ab#(bb(bb(ba(x1)))) |
→ |
aa#(x1) |
(25) |
|
ab#(bb(bb(bb(x1)))) |
→ |
aa#(aa(aa(ab(x1)))) |
(26) |
|
ab#(bb(bb(bb(x1)))) |
→ |
aa#(aa(ab(x1))) |
(27) |
|
ab#(bb(bb(bb(x1)))) |
→ |
aa#(ab(x1)) |
(28) |
|
ab#(bb(bb(bb(x1)))) |
→ |
ab#(x1) |
(29) |
|
aa#(aa(ab(ba(aa(x1))))) |
→ |
ab#(bb(ba(ab(ba(x1))))) |
(30) |
|
aa#(aa(ab(ba(aa(x1))))) |
→ |
ba#(ab(ba(x1))) |
(31) |
|
aa#(aa(ab(ba(aa(x1))))) |
→ |
ab#(ba(x1)) |
(32) |
|
aa#(aa(ab(ba(aa(x1))))) |
→ |
ba#(x1) |
(33) |
|
aa#(aa(ab(ba(ab(x1))))) |
→ |
ab#(bb(ba(ab(bb(x1))))) |
(34) |
|
aa#(aa(ab(ba(ab(x1))))) |
→ |
ba#(ab(bb(x1))) |
(35) |
|
aa#(aa(ab(ba(ab(x1))))) |
→ |
ab#(bb(x1)) |
(36) |
|
ba#(aa(ab(ba(aa(x1))))) |
→ |
ba#(ab(ba(x1))) |
(37) |
|
ba#(aa(ab(ba(aa(x1))))) |
→ |
ab#(ba(x1)) |
(38) |
|
ba#(aa(ab(ba(aa(x1))))) |
→ |
ba#(x1) |
(39) |
|
ba#(aa(ab(ba(ab(x1))))) |
→ |
ba#(ab(bb(x1))) |
(40) |
|
ba#(aa(ab(ba(ab(x1))))) |
→ |
ab#(bb(x1)) |
(41) |
1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.