Certification Problem
Input (TPDB SRS_Standard/Waldmann_07_size12/size-12-alpha-3-num-232)
The rewrite relation of the following TRS is considered.
a(x1) |
→ |
x1 |
(1) |
a(b(x1)) |
→ |
b(c(a(x1))) |
(2) |
c(x1) |
→ |
b(x1) |
(3) |
c(c(x1)) |
→ |
a(c(x1)) |
(4) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
a(x1) |
→ |
x1 |
(1) |
b(a(x1)) |
→ |
a(c(b(x1))) |
(5) |
c(x1) |
→ |
b(x1) |
(3) |
c(c(x1)) |
→ |
c(a(x1)) |
(6) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
b#(a(x1)) |
→ |
b#(x1) |
(7) |
b#(a(x1)) |
→ |
c#(b(x1)) |
(8) |
b#(a(x1)) |
→ |
a#(c(b(x1))) |
(9) |
c#(x1) |
→ |
b#(x1) |
(10) |
c#(c(x1)) |
→ |
a#(x1) |
(11) |
c#(c(x1)) |
→ |
c#(a(x1)) |
(12) |
1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.