Certification Problem
Input (TPDB SRS_Standard/ICFP_2010/54532)
The rewrite relation of the following TRS is considered.
|
0(x1) |
→ |
1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(...display limit reached...))))))))))))))))))))))))))))))))))))))))))))))))))) |
(1) |
|
2(x1) |
→ |
3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(...display limit reached...))))))))))))))))))))))))))))))))))))))))))))))))))) |
(2) |
|
0(1(2(3(4(5(4(5(x1)))))))) |
→ |
0(1(2(3(4(4(5(5(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
|
0(x1) |
→ |
1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(...display limit reached...))))))))))))))))))))))))))))))))))))))))))))))))))) |
(1) |
|
2(x1) |
→ |
3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(...display limit reached...))))))))))))))))))))))))))))))))))))))))))))))))))) |
(2) |
|
5(4(5(4(3(2(1(0(x1)))))))) |
→ |
5(5(4(4(3(2(1(0(x1)))))))) |
(4) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [0(x1)] |
= |
1 · x1 + 1 |
| [1(x1)] |
= |
1 · x1
|
| [2(x1)] |
= |
1 · x1 + 1 |
| [3(x1)] |
= |
1 · x1
|
| [5(x1)] |
= |
1 · x1
|
| [4(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
0(x1) |
→ |
1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(...display limit reached...))))))))))))))))))))))))))))))))))))))))))))))))))) |
(1) |
|
2(x1) |
→ |
3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(...display limit reached...))))))))))))))))))))))))))))))))))))))))))))))))))) |
(2) |
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 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
5#(4(5(4(3(2(1(0(x1)))))))) |
→ |
5#(5(4(4(3(2(1(0(x1)))))))) |
(5) |
|
5#(4(5(4(3(2(1(0(x1)))))))) |
→ |
5#(4(4(3(2(1(0(x1))))))) |
(6) |
1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.