Term Rewriting System R:
[x]
f(1) -> f(g(1))
f(f(x)) -> f(x)
g(0) -> g(f(0))
g(g(x)) -> g(x)
Innermost Termination of R to be shown.
R
↳Dependency Pair Analysis
R contains the following Dependency Pairs:
F(1) -> F(g(1))
F(1) -> G(1)
G(0) -> G(f(0))
G(0) -> F(0)
Furthermore, R contains two SCCs.
R
↳DPs
→DP Problem 1
↳Polynomial Ordering
→DP Problem 2
↳Polo
Dependency Pair:
F(1) -> F(g(1))
Rules:
f(1) -> f(g(1))
f(f(x)) -> f(x)
g(0) -> g(f(0))
g(g(x)) -> g(x)
Strategy:
innermost
The following dependency pair can be strictly oriented:
F(1) -> F(g(1))
Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:
g(0) -> g(f(0))
g(g(x)) -> g(x)
f(1) -> f(g(1))
f(f(x)) -> f(x)
Used ordering: Polynomial ordering with Polynomial interpretation:
| POL(0) | = 0 |
| POL(g(x1)) | = 0 |
| POL(1) | = 1 |
| POL(f(x1)) | = 0 |
| POL(F(x1)) | = x1 |
resulting in one new DP problem.
R
↳DPs
→DP Problem 1
↳Polo
→DP Problem 3
↳Dependency Graph
→DP Problem 2
↳Polo
Dependency Pair:
Rules:
f(1) -> f(g(1))
f(f(x)) -> f(x)
g(0) -> g(f(0))
g(g(x)) -> g(x)
Strategy:
innermost
Using the Dependency Graph resulted in no new DP problems.
R
↳DPs
→DP Problem 1
↳Polo
→DP Problem 2
↳Polynomial Ordering
Dependency Pair:
G(0) -> G(f(0))
Rules:
f(1) -> f(g(1))
f(f(x)) -> f(x)
g(0) -> g(f(0))
g(g(x)) -> g(x)
Strategy:
innermost
The following dependency pair can be strictly oriented:
G(0) -> G(f(0))
Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:
g(0) -> g(f(0))
g(g(x)) -> g(x)
f(1) -> f(g(1))
f(f(x)) -> f(x)
Used ordering: Polynomial ordering with Polynomial interpretation:
| POL(0) | = 1 |
| POL(g(x1)) | = 0 |
| POL(G(x1)) | = x1 |
| POL(1) | = 0 |
| POL(f(x1)) | = 0 |
resulting in one new DP problem.
R
↳DPs
→DP Problem 1
↳Polo
→DP Problem 2
↳Polo
→DP Problem 4
↳Dependency Graph
Dependency Pair:
Rules:
f(1) -> f(g(1))
f(f(x)) -> f(x)
g(0) -> g(f(0))
g(g(x)) -> g(x)
Strategy:
innermost
Using the Dependency Graph resulted in no new DP problems.
Innermost Termination of R successfully shown.
Duration:
0:00 minutes