Certification Problem
Input (TPDB SRS_Standard/Waldmann_19/random-408)
The rewrite relation of the following TRS is considered.
a(b(b(b(x1)))) |
→ |
b(b(a(b(x1)))) |
(1) |
a(b(a(a(x1)))) |
→ |
b(a(b(a(x1)))) |
(2) |
b(b(a(a(x1)))) |
→ |
b(a(a(a(x1)))) |
(3) |
a(b(b(b(x1)))) |
→ |
a(b(a(a(x1)))) |
(4) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
b(b(b(a(x1)))) |
→ |
b(a(b(b(x1)))) |
(5) |
a(a(b(a(x1)))) |
→ |
a(b(a(b(x1)))) |
(6) |
a(a(b(b(x1)))) |
→ |
a(a(a(b(x1)))) |
(7) |
b(b(b(a(x1)))) |
→ |
a(a(b(a(x1)))) |
(8) |
1.1 Rule Removal
Using the
linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1
over the naturals
[a(x1)] |
= |
|
1 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
|
· x1 +
|
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
|
|
[b(x1)] |
= |
|
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
1 |
1 |
|
|
· x1 +
|
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
|
|
all of the following rules can be deleted.
a(a(b(b(x1)))) |
→ |
a(a(a(b(x1)))) |
(7) |
1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
b#(b(b(a(x1)))) |
→ |
b#(x1) |
(9) |
b#(b(b(a(x1)))) |
→ |
b#(b(x1)) |
(10) |
b#(b(b(a(x1)))) |
→ |
a#(b(b(x1))) |
(11) |
b#(b(b(a(x1)))) |
→ |
b#(a(b(b(x1)))) |
(12) |
a#(a(b(a(x1)))) |
→ |
b#(x1) |
(13) |
a#(a(b(a(x1)))) |
→ |
a#(b(x1)) |
(14) |
a#(a(b(a(x1)))) |
→ |
b#(a(b(x1))) |
(15) |
a#(a(b(a(x1)))) |
→ |
a#(b(a(b(x1)))) |
(16) |
b#(b(b(a(x1)))) |
→ |
a#(b(a(x1))) |
(17) |
b#(b(b(a(x1)))) |
→ |
a#(a(b(a(x1)))) |
(18) |
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.