Term Rewriting System R:
[X]
f(0) -> cons(0, nf(s(0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
p(s(0)) -> 0
activate(nf(X)) -> f(X)
activate(X) -> X
Innermost Termination of R to be shown.
R
↳Dependency Pair Analysis
R contains the following Dependency Pairs:
F(s(0)) -> F(p(s(0)))
F(s(0)) -> P(s(0))
ACTIVATE(nf(X)) -> F(X)
Furthermore, R contains one SCC.
R
↳DPs
→DP Problem 1
↳Polynomial Ordering
Dependency Pair:
F(s(0)) -> F(p(s(0)))
Rules:
f(0) -> cons(0, nf(s(0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
p(s(0)) -> 0
activate(nf(X)) -> f(X)
activate(X) -> X
Strategy:
innermost
The following dependency pair can be strictly oriented:
F(s(0)) -> F(p(s(0)))
Additionally, the following usable rule for innermost w.r.t. to the implicit AFS can be oriented:
p(s(0)) -> 0
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(0) | = 0 |
POL(s(x1)) | = 1 |
POL(F(x1)) | = x1 |
POL(p(x1)) | = 0 |
resulting in one new DP problem.
R
↳DPs
→DP Problem 1
↳Polo
→DP Problem 2
↳Dependency Graph
Dependency Pair:
Rules:
f(0) -> cons(0, nf(s(0)))
f(s(0)) -> f(p(s(0)))
f(X) -> nf(X)
p(s(0)) -> 0
activate(nf(X)) -> f(X)
activate(X) -> X
Strategy:
innermost
Using the Dependency Graph resulted in no new DP problems.
Innermost Termination of R successfully shown.
Duration:
0:00 minutes