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 oriented strictly 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 be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(H23(x1, x2, x3)) = x2   
POL(h12(x1, x2)) = 1   
POL(j2(x1, x2)) = x1 + x2   
POL(s1(x1)) = 0   

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 oriented strictly and are deleted.


F2(j2(x, y), y) -> F2(x, k1(y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(F2(x1, x2)) = x1   
POL(h1(x1)) = 0   
POL(h12(x1, x2)) = 0   
POL(j2(x1, x2)) = 1 + x1   
POL(k1(x1)) = 0   
POL(s1(x1)) = 0   

The following usable rules [14] were oriented: none



↳ 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.