Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

g1(s1(x)) -> f1(x)
f1(0) -> s1(0)
f1(s1(x)) -> s1(s1(g1(x)))
g1(0) -> 0

Q is empty.


QTRS
  ↳ Non-Overlap Check

Q restricted rewrite system:
The TRS R consists of the following rules:

g1(s1(x)) -> f1(x)
f1(0) -> s1(0)
f1(s1(x)) -> s1(s1(g1(x)))
g1(0) -> 0

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
QTRS
      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

g1(s1(x)) -> f1(x)
f1(0) -> s1(0)
f1(s1(x)) -> s1(s1(g1(x)))
g1(0) -> 0

The set Q consists of the following terms:

g1(s1(x0))
f1(0)
f1(s1(x0))
g1(0)


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

G1(s1(x)) -> F1(x)
F1(s1(x)) -> G1(x)

The TRS R consists of the following rules:

g1(s1(x)) -> f1(x)
f1(0) -> s1(0)
f1(s1(x)) -> s1(s1(g1(x)))
g1(0) -> 0

The set Q consists of the following terms:

g1(s1(x0))
f1(0)
f1(s1(x0))
g1(0)

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

G1(s1(x)) -> F1(x)
F1(s1(x)) -> G1(x)

The TRS R consists of the following rules:

g1(s1(x)) -> f1(x)
f1(0) -> s1(0)
f1(s1(x)) -> s1(s1(g1(x)))
g1(0) -> 0

The set Q consists of the following terms:

g1(s1(x0))
f1(0)
f1(s1(x0))
g1(0)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


F1(s1(x)) -> G1(x)
The remaining pairs can at least by weakly be oriented.

G1(s1(x)) -> F1(x)
Used ordering: Combined order from the following AFS and order.
G1(x1)  =  x1
s1(x1)  =  s1(x1)
F1(x1)  =  F1(x1)

Lexicographic Path Order [19].
Precedence:
[s1, F1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof
QDP
              ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

G1(s1(x)) -> F1(x)

The TRS R consists of the following rules:

g1(s1(x)) -> f1(x)
f1(0) -> s1(0)
f1(s1(x)) -> s1(s1(g1(x)))
g1(0) -> 0

The set Q consists of the following terms:

g1(s1(x0))
f1(0)
f1(s1(x0))
g1(0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.