Certification Problem
Input (TPDB SRS_Standard/Waldmann_19/random-141)
The rewrite relation of the following TRS is considered.
|
a(a(a(a(x1)))) |
→ |
b(a(b(b(x1)))) |
(1) |
|
b(b(a(b(x1)))) |
→ |
a(b(b(b(x1)))) |
(2) |
|
a(a(b(a(x1)))) |
→ |
a(a(a(a(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(a(a(a(x1)))) |
→ |
b(b(a(b(x1)))) |
(4) |
|
b(a(b(b(x1)))) |
→ |
b(b(b(a(x1)))) |
(5) |
|
a(b(a(a(x1)))) |
→ |
a(a(a(a(x1)))) |
(6) |
1.1 Closure Under Flat Contexts
Using the flat contexts
{a(☐), b(☐)}
We obtain the transformed TRS
|
b(a(b(b(x1)))) |
→ |
b(b(b(a(x1)))) |
(5) |
|
a(b(a(a(x1)))) |
→ |
a(a(a(a(x1)))) |
(6) |
|
a(a(a(a(a(x1))))) |
→ |
a(b(b(a(b(x1))))) |
(7) |
|
b(a(a(a(a(x1))))) |
→ |
b(b(b(a(b(x1))))) |
(8) |
1.1.1 Semantic Labeling
Root-labeling is applied.
We obtain the labeled TRS
|
ba(ab(bb(bb(x1)))) |
→ |
bb(bb(ba(ab(x1)))) |
(9) |
|
ba(ab(bb(ba(x1)))) |
→ |
bb(bb(ba(aa(x1)))) |
(10) |
|
ab(ba(aa(ab(x1)))) |
→ |
aa(aa(aa(ab(x1)))) |
(11) |
|
ab(ba(aa(aa(x1)))) |
→ |
aa(aa(aa(aa(x1)))) |
(12) |
|
aa(aa(aa(aa(ab(x1))))) |
→ |
ab(bb(ba(ab(bb(x1))))) |
(13) |
|
aa(aa(aa(aa(aa(x1))))) |
→ |
ab(bb(ba(ab(ba(x1))))) |
(14) |
|
ba(aa(aa(aa(ab(x1))))) |
→ |
bb(bb(ba(ab(bb(x1))))) |
(15) |
|
ba(aa(aa(aa(aa(x1))))) |
→ |
bb(bb(ba(ab(ba(x1))))) |
(16) |
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [ba(x1)] |
= |
1 · x1 + 2 |
| [ab(x1)] |
= |
1 · x1
|
| [bb(x1)] |
= |
1 · x1
|
| [aa(x1)] |
= |
1 · x1 + 1 |
all of the following rules can be deleted.
|
ba(ab(bb(ba(x1)))) |
→ |
bb(bb(ba(aa(x1)))) |
(10) |
|
aa(aa(aa(aa(ab(x1))))) |
→ |
ab(bb(ba(ab(bb(x1))))) |
(13) |
|
aa(aa(aa(aa(aa(x1))))) |
→ |
ab(bb(ba(ab(ba(x1))))) |
(14) |
|
ba(aa(aa(aa(ab(x1))))) |
→ |
bb(bb(ba(ab(bb(x1))))) |
(15) |
|
ba(aa(aa(aa(aa(x1))))) |
→ |
bb(bb(ba(ab(ba(x1))))) |
(16) |
1.1.1.1.1 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.
|
ab(ba(aa(ab(x1)))) |
→ |
aa(aa(aa(ab(x1)))) |
(11) |
|
ab(ba(aa(aa(x1)))) |
→ |
aa(aa(aa(aa(x1)))) |
(12) |
1.1.1.1.1.1 Switch to Innermost Termination
The TRS is overlay and locally confluent:
10Hence, it suffices to show innermost termination in the following.
1.1.1.1.1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
ba#(ab(bb(bb(x1)))) |
→ |
ba#(ab(x1)) |
(17) |
1.1.1.1.1.1.1.1 Usable Rules Processor
We restrict the rewrite rules to the following usable rules of the DP problem.
There are no rules.
1.1.1.1.1.1.1.1.1 Innermost Lhss Removal Processor
We restrict the innermost strategy to the following left hand sides.
There are no lhss.
1.1.1.1.1.1.1.1.1.1 Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [ba#(x1)] |
= |
-2 + 2 · x1
|
| [ab(x1)] |
= |
x1 |
| [bb(x1)] |
= |
1 + 2 · x1
|
the
pair
|
ba#(ab(bb(bb(x1)))) |
→ |
ba#(ab(x1)) |
(17) |
could be deleted.
1.1.1.1.1.1.1.1.1.1.1 P is empty
There are no pairs anymore.