Certification Problem
Input (TPDB TRS_Standard/Der95/17)
The rewrite relation of the following TRS is considered.
|
.(1,x) |
→ |
x |
(1) |
|
.(x,1) |
→ |
x |
(2) |
|
.(i(x),x) |
→ |
1 |
(3) |
|
.(x,i(x)) |
→ |
1 |
(4) |
|
i(1) |
→ |
1 |
(5) |
|
i(i(x)) |
→ |
x |
(6) |
|
.(i(y),.(y,z)) |
→ |
z |
(7) |
|
.(y,.(i(y),z)) |
→ |
z |
(8) |
|
.(.(x,y),z) |
→ |
.(x,.(y,z)) |
(9) |
|
i(.(x,y)) |
→ |
.(i(y),i(x)) |
(10) |
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.
|
.#(.(x,y),z) |
→ |
.#(y,z) |
(11) |
|
i#(.(x,y)) |
→ |
i#(x) |
(12) |
|
i#(.(x,y)) |
→ |
.#(i(y),i(x)) |
(13) |
|
.#(.(x,y),z) |
→ |
.#(x,.(y,z)) |
(14) |
|
i#(.(x,y)) |
→ |
i#(y) |
(15) |
1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.