Certification Problem
Input (TPDB SRS_Standard/ICFP_2010/213407)
The rewrite relation of the following TRS is considered.
|
0(1(1(x1))) |
→ |
1(0(2(1(x1)))) |
(1) |
|
0(1(1(x1))) |
→ |
1(1(0(3(2(x1))))) |
(2) |
|
0(1(1(x1))) |
→ |
1(4(0(0(2(1(x1)))))) |
(3) |
|
0(1(4(x1))) |
→ |
1(4(0(3(x1)))) |
(4) |
|
0(1(4(x1))) |
→ |
4(0(2(1(x1)))) |
(5) |
|
0(1(4(x1))) |
→ |
4(0(2(1(3(x1))))) |
(6) |
|
0(4(1(x1))) |
→ |
0(2(4(1(x1)))) |
(7) |
|
0(4(1(x1))) |
→ |
4(0(2(1(x1)))) |
(8) |
|
0(4(1(x1))) |
→ |
4(0(3(1(x1)))) |
(9) |
|
0(4(1(x1))) |
→ |
1(4(4(0(2(x1))))) |
(10) |
|
0(4(1(x1))) |
→ |
2(1(4(0(2(x1))))) |
(11) |
|
0(5(4(x1))) |
→ |
1(4(0(0(2(5(x1)))))) |
(12) |
|
0(5(4(x1))) |
→ |
4(0(2(5(2(5(x1)))))) |
(13) |
|
0(5(4(x1))) |
→ |
5(0(4(0(3(3(x1)))))) |
(14) |
|
4(2(1(x1))) |
→ |
4(0(2(1(x1)))) |
(15) |
|
4(2(1(x1))) |
→ |
1(2(4(0(2(x1))))) |
(16) |
|
0(1(0(4(x1)))) |
→ |
0(0(2(2(1(4(x1)))))) |
(17) |
|
0(1(1(2(x1)))) |
→ |
5(1(0(2(1(x1))))) |
(18) |
|
0(1(1(3(x1)))) |
→ |
1(1(5(0(3(x1))))) |
(19) |
|
0(1(4(1(x1)))) |
→ |
1(0(2(4(1(x1))))) |
(20) |
|
0(1(4(3(x1)))) |
→ |
1(0(2(4(3(x1))))) |
(21) |
|
0(1(4(3(x1)))) |
→ |
4(0(2(1(3(x1))))) |
(22) |
|
0(4(1(2(x1)))) |
→ |
2(0(3(1(4(x1))))) |
(23) |
|
0(4(2(1(x1)))) |
→ |
0(4(3(0(2(1(x1)))))) |
(24) |
|
0(5(0(4(x1)))) |
→ |
1(5(4(0(0(2(x1)))))) |
(25) |
|
0(5(1(3(x1)))) |
→ |
1(1(5(0(3(x1))))) |
(26) |
|
0(5(1(3(x1)))) |
→ |
5(3(1(0(3(x1))))) |
(27) |
|
0(5(4(1(x1)))) |
→ |
4(1(5(4(0(2(x1)))))) |
(28) |
|
0(5(4(3(x1)))) |
→ |
0(2(5(0(3(4(x1)))))) |
(29) |
|
0(5(4(3(x1)))) |
→ |
1(5(3(4(0(3(x1)))))) |
(30) |
|
1(0(5(4(x1)))) |
→ |
1(4(5(0(3(3(x1)))))) |
(31) |
|
1(0(5(4(x1)))) |
→ |
5(5(1(0(2(4(x1)))))) |
(32) |
|
1(4(2(1(x1)))) |
→ |
4(0(2(1(2(1(x1)))))) |
(33) |
|
4(1(2(1(x1)))) |
→ |
4(1(0(2(1(x1))))) |
(34) |
|
4(1(2(1(x1)))) |
→ |
3(4(0(2(1(1(x1)))))) |
(35) |
|
4(3(2(1(x1)))) |
→ |
0(3(4(0(2(1(x1)))))) |
(36) |
|
0(0(1(2(3(x1))))) |
→ |
1(0(3(0(0(2(x1)))))) |
(37) |
|
0(0(5(1(2(x1))))) |
→ |
0(0(2(5(0(1(x1)))))) |
(38) |
|
0(0(5(1(3(x1))))) |
→ |
4(5(0(0(3(1(x1)))))) |
(39) |
|
0(1(3(4(2(x1))))) |
→ |
1(3(4(3(0(2(x1)))))) |
(40) |
|
0(4(5(3(4(x1))))) |
→ |
0(3(2(5(4(4(x1)))))) |
(41) |
|
0(5(0(2(2(x1))))) |
→ |
0(0(2(5(2(4(x1)))))) |
(42) |
|
0(5(1(1(3(x1))))) |
→ |
0(3(2(1(1(5(x1)))))) |
(43) |
|
0(5(1(4(3(x1))))) |
→ |
3(0(2(1(5(4(x1)))))) |
(44) |
|
0(5(5(1(2(x1))))) |
→ |
1(5(3(0(2(5(x1)))))) |
(45) |
|
1(0(1(2(4(x1))))) |
→ |
1(1(0(2(3(4(x1)))))) |
(46) |
|
1(4(2(1(2(x1))))) |
→ |
0(2(2(1(1(4(x1)))))) |
(47) |
|
4(0(0(5(4(x1))))) |
→ |
5(0(0(4(4(5(x1)))))) |
(48) |
|
4(2(5(4(1(x1))))) |
→ |
4(4(1(5(3(2(x1)))))) |
(49) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
There are 140 ruless (increase limit for explicit display).
1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.