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(a) -> b
f1(c) -> d
f1(g2(x, y)) -> g2(f1(x), f1(y))
f1(h2(x, y)) -> g2(h2(y, f1(x)), h2(x, f1(y)))
g2(x, x) -> h2(e, x)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f1(a) -> b
f1(c) -> d
f1(g2(x, y)) -> g2(f1(x), f1(y))
f1(h2(x, y)) -> g2(h2(y, f1(x)), h2(x, f1(y)))
g2(x, x) -> h2(e, x)

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(h2(x, y)) -> F1(y)
F1(g2(x, y)) -> F1(y)
F1(g2(x, y)) -> G2(f1(x), f1(y))
F1(h2(x, y)) -> G2(h2(y, f1(x)), h2(x, f1(y)))
F1(g2(x, y)) -> F1(x)
F1(h2(x, y)) -> F1(x)

The TRS R consists of the following rules:

f1(a) -> b
f1(c) -> d
f1(g2(x, y)) -> g2(f1(x), f1(y))
f1(h2(x, y)) -> g2(h2(y, f1(x)), h2(x, f1(y)))
g2(x, x) -> h2(e, 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:

F1(h2(x, y)) -> F1(y)
F1(g2(x, y)) -> F1(y)
F1(g2(x, y)) -> G2(f1(x), f1(y))
F1(h2(x, y)) -> G2(h2(y, f1(x)), h2(x, f1(y)))
F1(g2(x, y)) -> F1(x)
F1(h2(x, y)) -> F1(x)

The TRS R consists of the following rules:

f1(a) -> b
f1(c) -> d
f1(g2(x, y)) -> g2(f1(x), f1(y))
f1(h2(x, y)) -> g2(h2(y, f1(x)), h2(x, f1(y)))
g2(x, x) -> h2(e, x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.

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

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

F1(h2(x, y)) -> F1(y)
F1(g2(x, y)) -> F1(y)
F1(g2(x, y)) -> F1(x)
F1(h2(x, y)) -> F1(x)

The TRS R consists of the following rules:

f1(a) -> b
f1(c) -> d
f1(g2(x, y)) -> g2(f1(x), f1(y))
f1(h2(x, y)) -> g2(h2(y, f1(x)), h2(x, f1(y)))
g2(x, x) -> h2(e, x)

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(g2(x, y)) -> F1(y)
F1(g2(x, y)) -> F1(x)
The remaining pairs can at least be oriented weakly.

F1(h2(x, y)) -> F1(y)
F1(h2(x, y)) -> F1(x)
Used ordering: Polynomial interpretation [21]:

POL(F1(x1)) = x1   
POL(g2(x1, x2)) = 1 + x1 + x2   
POL(h2(x1, x2)) = x1 + x2   

The following usable rules [14] were oriented: none



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

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

F1(h2(x, y)) -> F1(y)
F1(h2(x, y)) -> F1(x)

The TRS R consists of the following rules:

f1(a) -> b
f1(c) -> d
f1(g2(x, y)) -> g2(f1(x), f1(y))
f1(h2(x, y)) -> g2(h2(y, f1(x)), h2(x, f1(y)))
g2(x, x) -> h2(e, x)

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

POL(F1(x1)) = x1   
POL(h2(x1, x2)) = 1 + x1 + x2   

The following usable rules [14] were oriented: none



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

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

f1(a) -> b
f1(c) -> d
f1(g2(x, y)) -> g2(f1(x), f1(y))
f1(h2(x, y)) -> g2(h2(y, f1(x)), h2(x, f1(y)))
g2(x, x) -> h2(e, 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.