Certification Problem
Input (TPDB SRS_Standard/ICFP_2010/264405)
The rewrite relation of the following TRS is considered.
|
0(1(2(3(4(x1))))) |
→ |
0(2(3(1(4(x1))))) |
(1) |
|
0(5(1(2(3(4(x1)))))) |
→ |
0(1(2(5(3(4(x1)))))) |
(2) |
|
0(5(1(2(3(4(x1)))))) |
→ |
0(5(2(1(3(4(x1)))))) |
(3) |
|
0(5(1(2(3(4(x1)))))) |
→ |
5(0(2(3(1(4(x1)))))) |
(4) |
|
0(5(2(3(1(4(x1)))))) |
→ |
0(1(5(2(3(4(x1)))))) |
(5) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by matchbox @ termCOMP 2023)
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
4(3(2(1(0(x1))))) |
→ |
4(1(3(2(0(x1))))) |
(6) |
|
4(3(2(1(5(0(x1)))))) |
→ |
4(3(5(2(1(0(x1)))))) |
(7) |
|
4(3(2(1(5(0(x1)))))) |
→ |
4(3(1(2(5(0(x1)))))) |
(8) |
|
4(3(2(1(5(0(x1)))))) |
→ |
4(1(3(2(0(5(x1)))))) |
(9) |
|
4(1(3(2(5(0(x1)))))) |
→ |
4(3(2(5(1(0(x1)))))) |
(10) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
4#(3(2(1(5(0(x1)))))) |
→ |
4#(3(5(2(1(0(x1)))))) |
(11) |
|
4#(3(2(1(5(0(x1)))))) |
→ |
4#(3(1(2(5(0(x1)))))) |
(12) |
|
4#(3(2(1(5(0(x1)))))) |
→ |
4#(1(3(2(0(5(x1)))))) |
(13) |
|
4#(3(2(1(0(x1))))) |
→ |
4#(1(3(2(0(x1))))) |
(14) |
|
4#(1(3(2(5(0(x1)))))) |
→ |
4#(3(2(5(1(0(x1)))))) |
(15) |
1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.