Certification Problem
Input (TPDB TRS_Standard/Transformed_CSR_04/Ex25_Luc06_FR)
The rewrite relation of the following TRS is considered.
|
f(f(X)) |
→ |
c(n__f(n__g(n__f(X)))) |
(1) |
|
c(X) |
→ |
d(activate(X)) |
(2) |
|
h(X) |
→ |
c(n__d(X)) |
(3) |
|
f(X) |
→ |
n__f(X) |
(4) |
|
g(X) |
→ |
n__g(X) |
(5) |
|
d(X) |
→ |
n__d(X) |
(6) |
|
activate(n__f(X)) |
→ |
f(activate(X)) |
(7) |
|
activate(n__g(X)) |
→ |
g(X) |
(8) |
|
activate(n__d(X)) |
→ |
d(X) |
(9) |
|
activate(X) |
→ |
X |
(10) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [f(x1)] |
= |
1 · x1
|
| [c(x1)] |
= |
1 · x1
|
| [n__f(x1)] |
= |
1 · x1
|
| [n__g(x1)] |
= |
1 · x1
|
| [d(x1)] |
= |
1 · x1
|
| [activate(x1)] |
= |
1 · x1
|
| [h(x1)] |
= |
1 · x1 + 1 |
| [n__d(x1)] |
= |
1 · x1
|
| [g(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
f#(f(X)) |
→ |
c#(n__f(n__g(n__f(X)))) |
(11) |
|
c#(X) |
→ |
d#(activate(X)) |
(12) |
|
c#(X) |
→ |
activate#(X) |
(13) |
|
activate#(n__f(X)) |
→ |
f#(activate(X)) |
(14) |
|
activate#(n__f(X)) |
→ |
activate#(X) |
(15) |
|
activate#(n__g(X)) |
→ |
g#(X) |
(16) |
|
activate#(n__d(X)) |
→ |
d#(X) |
(17) |
1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.