Certification Problem
Input (TPDB SRS_Standard/Secret_06_SRS/secr1)
The rewrite relation of the following TRS is considered.
|
a(b(x1)) |
→ |
b(b(b(b(x1)))) |
(1) |
|
b(a(x1)) |
→ |
a(a(a(a(x1)))) |
(2) |
|
a(x1) |
→ |
x1 |
(3) |
|
b(x1) |
→ |
x1 |
(4) |
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
|
b(a(x1)) |
→ |
b(b(b(b(x1)))) |
(5) |
|
a(b(x1)) |
→ |
a(a(a(a(x1)))) |
(6) |
|
a(x1) |
→ |
x1 |
(3) |
|
b(x1) |
→ |
x1 |
(4) |
1.1 Closure Under Flat Contexts
Using the flat contexts
{b(☐), a(☐)}
We obtain the transformed TRS
|
b(a(x1)) |
→ |
b(b(b(b(x1)))) |
(5) |
|
a(b(x1)) |
→ |
a(a(a(a(x1)))) |
(6) |
|
b(a(x1)) |
→ |
b(x1) |
(7) |
|
a(a(x1)) |
→ |
a(x1) |
(8) |
|
b(b(x1)) |
→ |
b(x1) |
(9) |
|
a(b(x1)) |
→ |
a(x1) |
(10) |
1.1.1 Semantic Labeling
Root-labeling is applied.
We obtain the labeled TRS
|
ba(ab(x1)) |
→ |
bb(bb(bb(bb(x1)))) |
(11) |
|
ba(aa(x1)) |
→ |
bb(bb(bb(ba(x1)))) |
(12) |
|
ab(bb(x1)) |
→ |
aa(aa(aa(ab(x1)))) |
(13) |
|
ab(ba(x1)) |
→ |
aa(aa(aa(aa(x1)))) |
(14) |
|
ba(ab(x1)) |
→ |
bb(x1) |
(15) |
|
ba(aa(x1)) |
→ |
ba(x1) |
(16) |
|
aa(ab(x1)) |
→ |
ab(x1) |
(17) |
|
aa(aa(x1)) |
→ |
aa(x1) |
(18) |
|
bb(bb(x1)) |
→ |
bb(x1) |
(19) |
|
bb(ba(x1)) |
→ |
ba(x1) |
(20) |
|
ab(bb(x1)) |
→ |
ab(x1) |
(21) |
|
ab(ba(x1)) |
→ |
aa(x1) |
(22) |
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [ba(x1)] |
= |
1 · x1
|
| [ab(x1)] |
= |
1 · x1 + 1 |
| [bb(x1)] |
= |
1 · x1
|
| [aa(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
ba(ab(x1)) |
→ |
bb(bb(bb(bb(x1)))) |
(11) |
|
ab(ba(x1)) |
→ |
aa(aa(aa(aa(x1)))) |
(14) |
|
ba(ab(x1)) |
→ |
bb(x1) |
(15) |
|
ab(ba(x1)) |
→ |
aa(x1) |
(22) |
1.1.1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
ba#(aa(x1)) |
→ |
bb#(bb(bb(ba(x1)))) |
(23) |
|
ba#(aa(x1)) |
→ |
bb#(bb(ba(x1))) |
(24) |
|
ba#(aa(x1)) |
→ |
bb#(ba(x1)) |
(25) |
|
ba#(aa(x1)) |
→ |
ba#(x1) |
(26) |
|
ab#(bb(x1)) |
→ |
aa#(aa(aa(ab(x1)))) |
(27) |
|
ab#(bb(x1)) |
→ |
aa#(aa(ab(x1))) |
(28) |
|
ab#(bb(x1)) |
→ |
aa#(ab(x1)) |
(29) |
|
ab#(bb(x1)) |
→ |
ab#(x1) |
(30) |
1.1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.