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:

f2(j2(x, y), y) -> g1(f2(x, k1(y)))
f2(x, h12(y, z)) -> h23(0, x, h12(y, z))
g1(h23(x, y, h12(z, u))) -> h23(s1(x), y, h12(z, u))
h23(x, j2(y, h12(z, u)), h12(z, u)) -> h23(s1(x), y, h12(s1(z), u))
i1(f2(x, h1(y))) -> y
i1(h23(s1(x), y, h12(x, z))) -> z
k1(h1(x)) -> h12(0, x)
k1(h12(x, y)) -> h12(s1(x), y)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f2(j2(x, y), y) -> g1(f2(x, k1(y)))
f2(x, h12(y, z)) -> h23(0, x, h12(y, z))
g1(h23(x, y, h12(z, u))) -> h23(s1(x), y, h12(z, u))
h23(x, j2(y, h12(z, u)), h12(z, u)) -> h23(s1(x), y, h12(s1(z), u))
i1(f2(x, h1(y))) -> y
i1(h23(s1(x), y, h12(x, z))) -> z
k1(h1(x)) -> h12(0, x)
k1(h12(x, y)) -> h12(s1(x), y)

Q is empty.

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

F2(j2(x, y), y) -> G1(f2(x, k1(y)))
F2(j2(x, y), y) -> K1(y)
H23(x, j2(y, h12(z, u)), h12(z, u)) -> H23(s1(x), y, h12(s1(z), u))
F2(x, h12(y, z)) -> H23(0, x, h12(y, z))
G1(h23(x, y, h12(z, u))) -> H23(s1(x), y, h12(z, u))
F2(j2(x, y), y) -> F2(x, k1(y))

The TRS R consists of the following rules:

f2(j2(x, y), y) -> g1(f2(x, k1(y)))
f2(x, h12(y, z)) -> h23(0, x, h12(y, z))
g1(h23(x, y, h12(z, u))) -> h23(s1(x), y, h12(z, u))
h23(x, j2(y, h12(z, u)), h12(z, u)) -> h23(s1(x), y, h12(s1(z), u))
i1(f2(x, h1(y))) -> y
i1(h23(s1(x), y, h12(x, z))) -> z
k1(h1(x)) -> h12(0, x)
k1(h12(x, y)) -> h12(s1(x), y)

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:

F2(j2(x, y), y) -> G1(f2(x, k1(y)))
F2(j2(x, y), y) -> K1(y)
H23(x, j2(y, h12(z, u)), h12(z, u)) -> H23(s1(x), y, h12(s1(z), u))
F2(x, h12(y, z)) -> H23(0, x, h12(y, z))
G1(h23(x, y, h12(z, u))) -> H23(s1(x), y, h12(z, u))
F2(j2(x, y), y) -> F2(x, k1(y))

The TRS R consists of the following rules:

f2(j2(x, y), y) -> g1(f2(x, k1(y)))
f2(x, h12(y, z)) -> h23(0, x, h12(y, z))
g1(h23(x, y, h12(z, u))) -> h23(s1(x), y, h12(z, u))
h23(x, j2(y, h12(z, u)), h12(z, u)) -> h23(s1(x), y, h12(s1(z), u))
i1(f2(x, h1(y))) -> y
i1(h23(s1(x), y, h12(x, z))) -> z
k1(h1(x)) -> h12(0, x)
k1(h12(x, y)) -> h12(s1(x), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 4 less nodes.

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

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

H23(x, j2(y, h12(z, u)), h12(z, u)) -> H23(s1(x), y, h12(s1(z), u))

The TRS R consists of the following rules:

f2(j2(x, y), y) -> g1(f2(x, k1(y)))
f2(x, h12(y, z)) -> h23(0, x, h12(y, z))
g1(h23(x, y, h12(z, u))) -> h23(s1(x), y, h12(z, u))
h23(x, j2(y, h12(z, u)), h12(z, u)) -> h23(s1(x), y, h12(s1(z), u))
i1(f2(x, h1(y))) -> y
i1(h23(s1(x), y, h12(x, z))) -> z
k1(h1(x)) -> h12(0, x)
k1(h12(x, y)) -> h12(s1(x), y)

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.

H23(x, j2(y, h12(z, u)), h12(z, u)) -> H23(s1(x), y, h12(s1(z), u))
Used argument filtering: H23(x1, x2, x3)  =  x2
j2(x1, x2)  =  j1(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:

f2(j2(x, y), y) -> g1(f2(x, k1(y)))
f2(x, h12(y, z)) -> h23(0, x, h12(y, z))
g1(h23(x, y, h12(z, u))) -> h23(s1(x), y, h12(z, u))
h23(x, j2(y, h12(z, u)), h12(z, u)) -> h23(s1(x), y, h12(s1(z), u))
i1(f2(x, h1(y))) -> y
i1(h23(s1(x), y, h12(x, z))) -> z
k1(h1(x)) -> h12(0, x)
k1(h12(x, y)) -> h12(s1(x), y)

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:

F2(j2(x, y), y) -> F2(x, k1(y))

The TRS R consists of the following rules:

f2(j2(x, y), y) -> g1(f2(x, k1(y)))
f2(x, h12(y, z)) -> h23(0, x, h12(y, z))
g1(h23(x, y, h12(z, u))) -> h23(s1(x), y, h12(z, u))
h23(x, j2(y, h12(z, u)), h12(z, u)) -> h23(s1(x), y, h12(s1(z), u))
i1(f2(x, h1(y))) -> y
i1(h23(s1(x), y, h12(x, z))) -> z
k1(h1(x)) -> h12(0, x)
k1(h12(x, y)) -> h12(s1(x), y)

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.

F2(j2(x, y), y) -> F2(x, k1(y))
Used argument filtering: F2(x1, x2)  =  x1
j2(x1, x2)  =  j1(x1)
k1(x1)  =  k
h1(x1)  =  h
h12(x1, x2)  =  h1
0  =  0
s1(x1)  =  s
Used ordering: Quasi Precedence: k > h1


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

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

f2(j2(x, y), y) -> g1(f2(x, k1(y)))
f2(x, h12(y, z)) -> h23(0, x, h12(y, z))
g1(h23(x, y, h12(z, u))) -> h23(s1(x), y, h12(z, u))
h23(x, j2(y, h12(z, u)), h12(z, u)) -> h23(s1(x), y, h12(s1(z), u))
i1(f2(x, h1(y))) -> y
i1(h23(s1(x), y, h12(x, z))) -> z
k1(h1(x)) -> h12(0, x)
k1(h12(x, y)) -> h12(s1(x), y)

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.