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
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳AFS


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))


The following usable rules for innermost w.r.t. to the 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: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
1 > g

resulting in one new DP problem.
Used Argument Filtering System:
F(x1) -> F(x1)
g(x1) -> g
f(x1) -> f(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 3
             ↳Dependency Graph
       →DP Problem 2
         ↳AFS


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
         ↳AFS
       →DP Problem 2
         ↳Argument Filtering and 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))


The following usable rules for innermost w.r.t. to the 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: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
0 > f

resulting in one new DP problem.
Used Argument Filtering System:
G(x1) -> G(x1)
f(x1) -> f
g(x1) -> g(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
           →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