Certification Problem
Input (TPDB SRS_Standard/Mixed_SRS/turing_add)
The rewrite relation of the following TRS is considered.
|
1(q0(1(x1))) |
→ |
0(1(q1(x1))) |
(1) |
|
1(q0(0(x1))) |
→ |
0(0(q1(x1))) |
(2) |
|
1(q1(1(x1))) |
→ |
1(1(q1(x1))) |
(3) |
|
1(q1(0(x1))) |
→ |
1(0(q1(x1))) |
(4) |
|
0(q1(x1)) |
→ |
q2(1(x1)) |
(5) |
|
1(q2(x1)) |
→ |
q2(1(x1)) |
(6) |
|
0(q2(x1)) |
→ |
0(q0(x1)) |
(7) |
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
|
1(q0(1(x1))) |
→ |
q1(1(0(x1))) |
(8) |
|
0(q0(1(x1))) |
→ |
q1(0(0(x1))) |
(9) |
|
1(q1(1(x1))) |
→ |
q1(1(1(x1))) |
(10) |
|
0(q1(1(x1))) |
→ |
q1(0(1(x1))) |
(11) |
|
q1(0(x1)) |
→ |
1(q2(x1)) |
(12) |
|
q2(1(x1)) |
→ |
1(q2(x1)) |
(13) |
|
q2(0(x1)) |
→ |
q0(0(x1)) |
(14) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
1#(q0(1(x1))) |
→ |
q1#(1(0(x1))) |
(15) |
|
1#(q0(1(x1))) |
→ |
1#(0(x1)) |
(16) |
|
1#(q0(1(x1))) |
→ |
0#(x1) |
(17) |
|
0#(q0(1(x1))) |
→ |
q1#(0(0(x1))) |
(18) |
|
0#(q0(1(x1))) |
→ |
0#(0(x1)) |
(19) |
|
0#(q0(1(x1))) |
→ |
0#(x1) |
(20) |
|
1#(q1(1(x1))) |
→ |
q1#(1(1(x1))) |
(21) |
|
1#(q1(1(x1))) |
→ |
1#(1(x1)) |
(22) |
|
0#(q1(1(x1))) |
→ |
q1#(0(1(x1))) |
(23) |
|
0#(q1(1(x1))) |
→ |
0#(1(x1)) |
(24) |
|
q1#(0(x1)) |
→ |
1#(q2(x1)) |
(25) |
|
q1#(0(x1)) |
→ |
q2#(x1) |
(26) |
|
q2#(1(x1)) |
→ |
1#(q2(x1)) |
(27) |
|
q2#(1(x1)) |
→ |
q2#(x1) |
(28) |
1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.