Term Rewriting System R:
[x, y]
f(g(x), s(0), y) -> f(y, y, g(x))
g(s(x)) -> s(g(x))
g(0) -> 0
Innermost Termination of R to be shown.
R
↳Dependency Pair Analysis
R contains the following Dependency Pairs:
F(g(x), s(0), y) -> F(y, y, g(x))
G(s(x)) -> G(x)
Furthermore, R contains one SCC.
R
↳DPs
→DP Problem 1
↳Forward Instantiation Transformation
Dependency Pair:
G(s(x)) -> G(x)
Rules:
f(g(x), s(0), y) -> f(y, y, g(x))
g(s(x)) -> s(g(x))
g(0) -> 0
Strategy:
innermost
On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule
G(s(x)) -> G(x)
one new Dependency Pair
is created:
G(s(s(x''))) -> G(s(x''))
The transformation is resulting in one new DP problem:
R
↳DPs
→DP Problem 1
↳FwdInst
→DP Problem 2
↳Forward Instantiation Transformation
Dependency Pair:
G(s(s(x''))) -> G(s(x''))
Rules:
f(g(x), s(0), y) -> f(y, y, g(x))
g(s(x)) -> s(g(x))
g(0) -> 0
Strategy:
innermost
On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule
G(s(s(x''))) -> G(s(x''))
one new Dependency Pair
is created:
G(s(s(s(x'''')))) -> G(s(s(x'''')))
The transformation is resulting in one new DP problem:
R
↳DPs
→DP Problem 1
↳FwdInst
→DP Problem 2
↳FwdInst
...
→DP Problem 3
↳Polynomial Ordering
Dependency Pair:
G(s(s(s(x'''')))) -> G(s(s(x'''')))
Rules:
f(g(x), s(0), y) -> f(y, y, g(x))
g(s(x)) -> s(g(x))
g(0) -> 0
Strategy:
innermost
The following dependency pair can be strictly oriented:
G(s(s(s(x'''')))) -> G(s(s(x'''')))
There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(G(x1)) | = 1 + x1 |
POL(s(x1)) | = 1 + x1 |
resulting in one new DP problem.
R
↳DPs
→DP Problem 1
↳FwdInst
→DP Problem 2
↳FwdInst
...
→DP Problem 4
↳Dependency Graph
Dependency Pair:
Rules:
f(g(x), s(0), y) -> f(y, y, g(x))
g(s(x)) -> s(g(x))
g(0) -> 0
Strategy:
innermost
Using the Dependency Graph resulted in no new DP problems.
Innermost Termination of R successfully shown.
Duration:
0:00 minutes