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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 2 SCCs with 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


H23(x, j2(y, h12(z, u)), h12(z, u)) -> H23(s1(x), y, h12(s1(z), u))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
H23(x1, x2, x3)  =  x2
j2(x1, x2)  =  j1(x1)
h12(x1, x2)  =  h1
s1(x1)  =  s

Lexicographic Path Order [19].
Precedence:
j1 > s > h1


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:

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
            ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


F2(j2(x, y), y) -> F2(x, k1(y))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F2(x1, x2)  =  x1
j2(x1, x2)  =  j1(x1)
k1(x1)  =  k1(x1)
h1(x1)  =  x1
h12(x1, x2)  =  x2
0  =  0
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
[j1, k1]


The following usable rules [14] were oriented:

k1(h1(x)) -> h12(0, x)
k1(h12(x, y)) -> h12(s1(x), y)



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