Certification Problem
Input (TPDB SRS_Standard/Zantema_06/beans1)
The rewrite relation of the following TRS is considered.
|
1(2(1(x1))) |
→ |
2(0(2(x1))) |
(1) |
|
0(2(1(x1))) |
→ |
1(0(2(x1))) |
(2) |
|
L(2(1(x1))) |
→ |
L(1(0(2(x1)))) |
(3) |
|
1(2(0(x1))) |
→ |
2(0(1(x1))) |
(4) |
|
1(2(R(x1))) |
→ |
2(0(1(R(x1)))) |
(5) |
|
0(2(0(x1))) |
→ |
1(0(1(x1))) |
(6) |
|
L(2(0(x1))) |
→ |
L(1(0(1(x1)))) |
(7) |
|
0(2(R(x1))) |
→ |
1(0(1(R(x1)))) |
(8) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by NaTT @ termCOMP 2023)
1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
0#(2(1(x1))) |
→ |
0#(2(x1)) |
(9) |
|
L#(2(0(x1))) |
→ |
L#(1(0(1(x1)))) |
(10) |
|
L#(2(1(x1))) |
→ |
1#(0(2(x1))) |
(11) |
|
L#(2(0(x1))) |
→ |
1#(x1) |
(12) |
|
1#(2(R(x1))) |
→ |
1#(R(x1)) |
(13) |
|
L#(2(1(x1))) |
→ |
L#(1(0(2(x1)))) |
(14) |
|
L#(2(1(x1))) |
→ |
0#(2(x1)) |
(15) |
|
1#(2(0(x1))) |
→ |
1#(x1) |
(16) |
|
L#(2(0(x1))) |
→ |
1#(0(1(x1))) |
(17) |
|
1#(2(R(x1))) |
→ |
0#(1(R(x1))) |
(18) |
|
0#(2(0(x1))) |
→ |
1#(x1) |
(19) |
|
0#(2(R(x1))) |
→ |
0#(1(R(x1))) |
(20) |
|
1#(2(0(x1))) |
→ |
0#(1(x1)) |
(21) |
|
0#(2(R(x1))) |
→ |
1#(0(1(R(x1)))) |
(22) |
|
0#(2(0(x1))) |
→ |
1#(0(1(x1))) |
(23) |
|
0#(2(1(x1))) |
→ |
1#(0(2(x1))) |
(24) |
|
L#(2(0(x1))) |
→ |
0#(1(x1)) |
(25) |
|
1#(2(1(x1))) |
→ |
0#(2(x1)) |
(26) |
|
0#(2(0(x1))) |
→ |
0#(1(x1)) |
(27) |
|
0#(2(R(x1))) |
→ |
1#(R(x1)) |
(28) |
1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.