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:

f1(0) -> 1
f1(s1(x)) -> g2(x, s1(x))
g2(0, y) -> y
g2(s1(x), y) -> g2(x, +2(y, s1(x)))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
g2(s1(x), y) -> g2(x, s1(+2(y, x)))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f1(0) -> 1
f1(s1(x)) -> g2(x, s1(x))
g2(0, y) -> y
g2(s1(x), y) -> g2(x, +2(y, s1(x)))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
g2(s1(x), y) -> g2(x, s1(+2(y, x)))

Q is empty.

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

+12(x, s1(y)) -> +12(x, y)
G2(s1(x), y) -> +12(y, s1(x))
G2(s1(x), y) -> G2(x, +2(y, s1(x)))
G2(s1(x), y) -> G2(x, s1(+2(y, x)))
F1(s1(x)) -> G2(x, s1(x))
G2(s1(x), y) -> +12(y, x)

The TRS R consists of the following rules:

f1(0) -> 1
f1(s1(x)) -> g2(x, s1(x))
g2(0, y) -> y
g2(s1(x), y) -> g2(x, +2(y, s1(x)))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
g2(s1(x), y) -> g2(x, s1(+2(y, 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:

+12(x, s1(y)) -> +12(x, y)
G2(s1(x), y) -> +12(y, s1(x))
G2(s1(x), y) -> G2(x, +2(y, s1(x)))
G2(s1(x), y) -> G2(x, s1(+2(y, x)))
F1(s1(x)) -> G2(x, s1(x))
G2(s1(x), y) -> +12(y, x)

The TRS R consists of the following rules:

f1(0) -> 1
f1(s1(x)) -> g2(x, s1(x))
g2(0, y) -> y
g2(s1(x), y) -> g2(x, +2(y, s1(x)))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
g2(s1(x), y) -> g2(x, s1(+2(y, x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPAfsSolverProof
          ↳ QDP

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

+12(x, s1(y)) -> +12(x, y)

The TRS R consists of the following rules:

f1(0) -> 1
f1(s1(x)) -> g2(x, s1(x))
g2(0, y) -> y
g2(s1(x), y) -> g2(x, +2(y, s1(x)))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
g2(s1(x), y) -> g2(x, s1(+2(y, x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

+12(x, s1(y)) -> +12(x, y)
Used argument filtering: +12(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ PisEmptyProof
          ↳ QDP

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

f1(0) -> 1
f1(s1(x)) -> g2(x, s1(x))
g2(0, y) -> y
g2(s1(x), y) -> g2(x, +2(y, s1(x)))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
g2(s1(x), y) -> g2(x, s1(+2(y, 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
            ↳ QDPAfsSolverProof

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

G2(s1(x), y) -> G2(x, +2(y, s1(x)))
G2(s1(x), y) -> G2(x, s1(+2(y, x)))

The TRS R consists of the following rules:

f1(0) -> 1
f1(s1(x)) -> g2(x, s1(x))
g2(0, y) -> y
g2(s1(x), y) -> g2(x, +2(y, s1(x)))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
g2(s1(x), y) -> g2(x, s1(+2(y, x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

G2(s1(x), y) -> G2(x, +2(y, s1(x)))
G2(s1(x), y) -> G2(x, s1(+2(y, x)))
Used argument filtering: G2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
+2(x1, x2)  =  +2(x1, x2)
0  =  0
Used ordering: Quasi Precedence: +_2 > s_1


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ PisEmptyProof

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

f1(0) -> 1
f1(s1(x)) -> g2(x, s1(x))
g2(0, y) -> y
g2(s1(x), y) -> g2(x, +2(y, s1(x)))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
g2(s1(x), y) -> g2(x, s1(+2(y, 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.