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:

+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
f1(g1(f1(x))) -> f1(h2(s1(0), x))
f1(g1(h2(x, y))) -> f1(h2(s1(x), y))
f1(h2(x, h2(y, z))) -> f1(h2(+2(x, y), z))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
f1(g1(f1(x))) -> f1(h2(s1(0), x))
f1(g1(h2(x, y))) -> f1(h2(s1(x), y))
f1(h2(x, h2(y, z))) -> f1(h2(+2(x, y), z))

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:

F1(g1(h2(x, y))) -> F1(h2(s1(x), y))
+12(x, s1(y)) -> +12(x, y)
F1(h2(x, h2(y, z))) -> +12(x, y)
F1(h2(x, h2(y, z))) -> F1(h2(+2(x, y), z))
F1(g1(f1(x))) -> F1(h2(s1(0), x))
+12(s1(x), y) -> +12(x, y)
+12(x, +2(y, z)) -> +12(x, y)
+12(x, +2(y, z)) -> +12(+2(x, y), z)

The TRS R consists of the following rules:

+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
f1(g1(f1(x))) -> f1(h2(s1(0), x))
f1(g1(h2(x, y))) -> f1(h2(s1(x), y))
f1(h2(x, h2(y, z))) -> f1(h2(+2(x, y), z))

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:

F1(g1(h2(x, y))) -> F1(h2(s1(x), y))
+12(x, s1(y)) -> +12(x, y)
F1(h2(x, h2(y, z))) -> +12(x, y)
F1(h2(x, h2(y, z))) -> F1(h2(+2(x, y), z))
F1(g1(f1(x))) -> F1(h2(s1(0), x))
+12(s1(x), y) -> +12(x, y)
+12(x, +2(y, z)) -> +12(x, y)
+12(x, +2(y, z)) -> +12(+2(x, y), z)

The TRS R consists of the following rules:

+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
f1(g1(f1(x))) -> f1(h2(s1(0), x))
f1(g1(h2(x, y))) -> f1(h2(s1(x), y))
f1(h2(x, h2(y, z))) -> f1(h2(+2(x, y), z))

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 3 less nodes.

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

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

+12(x, s1(y)) -> +12(x, y)
+12(s1(x), y) -> +12(x, y)
+12(x, +2(y, z)) -> +12(x, y)
+12(x, +2(y, z)) -> +12(+2(x, y), z)

The TRS R consists of the following rules:

+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
f1(g1(f1(x))) -> f1(h2(s1(0), x))
f1(g1(h2(x, y))) -> f1(h2(s1(x), y))
f1(h2(x, h2(y, z))) -> f1(h2(+2(x, y), z))

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.


+12(x, s1(y)) -> +12(x, y)
+12(s1(x), y) -> +12(x, y)
+12(x, +2(y, z)) -> +12(x, y)
+12(x, +2(y, z)) -> +12(+2(x, y), z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(+2(x1, x2)) = 2 + x1 + 2·x2   
POL(+12(x1, x2)) = x1 + 2·x2   
POL(0) = 0   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented:

+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(x, s1(y)) -> s1(+2(x, y))
+2(s1(x), y) -> s1(+2(x, y))
+2(x, 0) -> x
+2(0, y) -> y



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

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

+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
f1(g1(f1(x))) -> f1(h2(s1(0), x))
f1(g1(h2(x, y))) -> f1(h2(s1(x), y))
f1(h2(x, h2(y, z))) -> f1(h2(+2(x, y), z))

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:

F1(h2(x, h2(y, z))) -> F1(h2(+2(x, y), z))

The TRS R consists of the following rules:

+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
f1(g1(f1(x))) -> f1(h2(s1(0), x))
f1(g1(h2(x, y))) -> f1(h2(s1(x), y))
f1(h2(x, h2(y, z))) -> f1(h2(+2(x, y), z))

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.


F1(h2(x, h2(y, z))) -> F1(h2(+2(x, y), z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(+2(x1, x2)) = 1 + x1 + 2·x2   
POL(0) = 0   
POL(F1(x1)) = x1   
POL(h2(x1, x2)) = 1 + x1 + 2·x2   
POL(s1(x1)) = 0   

The following usable rules [14] were oriented:

+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(x, s1(y)) -> s1(+2(x, y))
+2(s1(x), y) -> s1(+2(x, y))
+2(x, 0) -> x
+2(0, y) -> 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:

+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
f1(g1(f1(x))) -> f1(h2(s1(0), x))
f1(g1(h2(x, y))) -> f1(h2(s1(x), y))
f1(h2(x, h2(y, z))) -> f1(h2(+2(x, y), z))

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.