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
  ↳ 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)))

Q is empty.

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)))

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

↳ 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)))

Q is empty.
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
  ↳ 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)))

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


The following pairs can be oriented strictly and are deleted.


MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MINUS2(x1, x2)) = 2·x2   
POL(s1(x1)) = 2 + 2·x1   

The following usable rules [14] were oriented: none



↳ 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)))

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

↳ 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)))

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


The following pairs can be oriented strictly and are deleted.


G1(s1(x)) -> G1(x)
G1(s1(x)) -> F1(g1(x))
F1(s1(x)) -> F1(x)
The remaining pairs can at least be oriented weakly.

F1(s1(x)) -> G1(f1(x))
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(F1(x1)) = 2·x1   
POL(G1(x1)) = 2·x1   
POL(f1(x1)) = 1 + 2·x1   
POL(g1(x1)) = 2·x1   
POL(minus2(x1, x2)) = 2·x1   
POL(s1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ DependencyGraphProof

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

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

Q is empty.
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.