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:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
f1(0) -> s1(0)
f1(s1(x)) -> minus2(s1(x), g1(f1(x)))
g1(0) -> 0
g1(s1(x)) -> minus2(s1(x), f1(g1(x)))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
f1(0) -> s1(0)
f1(s1(x)) -> minus2(s1(x), g1(f1(x)))
g1(0) -> 0
g1(s1(x)) -> minus2(s1(x), f1(g1(x)))

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:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
f1(0) -> s1(0)
f1(s1(x)) -> minus2(s1(x), g1(f1(x)))
g1(0) -> 0
g1(s1(x)) -> minus2(s1(x), f1(g1(x)))

The set Q consists of the following terms:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
f1(0)
f1(s1(x0))
g1(0)
g1(s1(x0))


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)) -> G1(x)
G1(s1(x)) -> F1(g1(x))
F1(s1(x)) -> G1(f1(x))
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
G1(s1(x)) -> MINUS2(s1(x), f1(g1(x)))
F1(s1(x)) -> MINUS2(s1(x), g1(f1(x)))
F1(s1(x)) -> F1(x)

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
f1(0) -> s1(0)
f1(s1(x)) -> minus2(s1(x), g1(f1(x)))
g1(0) -> 0
g1(s1(x)) -> minus2(s1(x), f1(g1(x)))

The set Q consists of the following terms:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
f1(0)
f1(s1(x0))
g1(0)
g1(s1(x0))

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

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

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

G1(s1(x)) -> G1(x)
G1(s1(x)) -> F1(g1(x))
F1(s1(x)) -> G1(f1(x))
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
G1(s1(x)) -> MINUS2(s1(x), f1(g1(x)))
F1(s1(x)) -> MINUS2(s1(x), g1(f1(x)))
F1(s1(x)) -> F1(x)

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
f1(0) -> s1(0)
f1(s1(x)) -> minus2(s1(x), g1(f1(x)))
g1(0) -> 0
g1(s1(x)) -> minus2(s1(x), f1(g1(x)))

The set Q consists of the following terms:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
f1(0)
f1(s1(x0))
g1(0)
g1(s1(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 2 less nodes.

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

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

MINUS2(s1(x), s1(y)) -> MINUS2(x, y)

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
f1(0) -> s1(0)
f1(s1(x)) -> minus2(s1(x), g1(f1(x)))
g1(0) -> 0
g1(s1(x)) -> minus2(s1(x), f1(g1(x)))

The set Q consists of the following terms:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
f1(0)
f1(s1(x0))
g1(0)
g1(s1(x0))

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.


MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MINUS2(x1, x2)  =  MINUS1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
f1(0) -> s1(0)
f1(s1(x)) -> minus2(s1(x), g1(f1(x)))
g1(0) -> 0
g1(s1(x)) -> minus2(s1(x), f1(g1(x)))

The set Q consists of the following terms:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
f1(0)
f1(s1(x0))
g1(0)
g1(s1(x0))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

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

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

G1(s1(x)) -> G1(x)
G1(s1(x)) -> F1(g1(x))
F1(s1(x)) -> G1(f1(x))
F1(s1(x)) -> F1(x)

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
f1(0) -> s1(0)
f1(s1(x)) -> minus2(s1(x), g1(f1(x)))
g1(0) -> 0
g1(s1(x)) -> minus2(s1(x), f1(g1(x)))

The set Q consists of the following terms:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
f1(0)
f1(s1(x0))
g1(0)
g1(s1(x0))

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.


G1(s1(x)) -> G1(x)
F1(s1(x)) -> G1(f1(x))
F1(s1(x)) -> F1(x)
The remaining pairs can at least by weakly be oriented.

G1(s1(x)) -> F1(g1(x))
Used ordering: Combined order from the following AFS and order.
G1(x1)  =  x1
s1(x1)  =  s1(x1)
F1(x1)  =  F1(x1)
g1(x1)  =  x1
f1(x1)  =  f1(x1)
0  =  0
minus2(x1, x2)  =  x1

Lexicographic Path Order [19].
Precedence:
0 > [s1, F1, f1]


The following usable rules [14] were oriented:

f1(0) -> s1(0)
f1(s1(x)) -> minus2(s1(x), g1(f1(x)))
g1(0) -> 0
g1(s1(x)) -> minus2(s1(x), f1(g1(x)))
minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)



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

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

G1(s1(x)) -> F1(g1(x))

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
f1(0) -> s1(0)
f1(s1(x)) -> minus2(s1(x), g1(f1(x)))
g1(0) -> 0
g1(s1(x)) -> minus2(s1(x), f1(g1(x)))

The set Q consists of the following terms:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
f1(0)
f1(s1(x0))
g1(0)
g1(s1(x0))

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.