Certification Problem
Input (TPDB TRS_Innermost/Transformed_CSR_innermost_04/Ex6_GM04_iGM)
The rewrite relation of the following TRS is considered.
active(c) |
→ |
mark(f(g(c))) |
(1) |
active(f(g(X))) |
→ |
mark(g(X)) |
(2) |
mark(c) |
→ |
active(c) |
(3) |
mark(f(X)) |
→ |
active(f(X)) |
(4) |
mark(g(X)) |
→ |
active(g(X)) |
(5) |
f(mark(X)) |
→ |
f(X) |
(6) |
f(active(X)) |
→ |
f(X) |
(7) |
g(mark(X)) |
→ |
g(X) |
(8) |
g(active(X)) |
→ |
g(X) |
(9) |
The evaluation strategy is innermost.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
[active(x1)] |
= |
1 · x1 + 1 |
[c] |
= |
0 |
[mark(x1)] |
= |
1 · x1 + 1 |
[f(x1)] |
= |
1 · x1
|
[g(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
f(mark(X)) |
→ |
f(X) |
(6) |
f(active(X)) |
→ |
f(X) |
(7) |
g(mark(X)) |
→ |
g(X) |
(8) |
g(active(X)) |
→ |
g(X) |
(9) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
active#(c) |
→ |
mark#(f(g(c))) |
(10) |
active#(f(g(X))) |
→ |
mark#(g(X)) |
(11) |
mark#(c) |
→ |
active#(c) |
(12) |
mark#(f(X)) |
→ |
active#(f(X)) |
(13) |
mark#(g(X)) |
→ |
active#(g(X)) |
(14) |
1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.