Term Rewriting System R:
[x, y]
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

MINUS(s(x), s(y)) -> MINUS(x, y)
F(s(x)) -> MINUS(s(x), g(f(x)))
F(s(x)) -> G(f(x))
F(s(x)) -> F(x)
G(s(x)) -> MINUS(s(x), f(g(x)))
G(s(x)) -> F(g(x))
G(s(x)) -> G(x)

Furthermore, R contains two SCCs.


   R
DPs
       →DP Problem 1
Argument Filtering and Ordering
       →DP Problem 2
Remaining


Dependency Pair:

MINUS(s(x), s(y)) -> MINUS(x, y)


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

MINUS(s(x), s(y)) -> MINUS(x, y)


There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
MINUS(x1, x2) -> MINUS(x1, x2)
s(x1) -> s(x1)


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


Dependency Pair:


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

G(s(x)) -> G(x)
F(s(x)) -> F(x)
G(s(x)) -> F(g(x))
F(s(x)) -> G(f(x))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
f(0) -> s(0)
f(s(x)) -> minus(s(x), g(f(x)))
g(0) -> 0
g(s(x)) -> minus(s(x), f(g(x)))


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:00 minutes