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:

f(a) → b
f(c) → d
f(g(x, y)) → g(f(x), f(y))
f(h(x, y)) → g(h(y, f(x)), h(x, f(y)))
g(x, x) → h(e, x)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(a) → b
f(c) → d
f(g(x, y)) → g(f(x), f(y))
f(h(x, y)) → g(h(y, f(x)), h(x, f(y)))
g(x, x) → h(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:

F(h(x, y)) → F(y)
F(g(x, y)) → F(y)
F(g(x, y)) → G(f(x), f(y))
F(h(x, y)) → G(h(y, f(x)), h(x, f(y)))
F(g(x, y)) → F(x)
F(h(x, y)) → F(x)

The TRS R consists of the following rules:

f(a) → b
f(c) → d
f(g(x, y)) → g(f(x), f(y))
f(h(x, y)) → g(h(y, f(x)), h(x, f(y)))
g(x, x) → h(e, x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

F(h(x, y)) → F(y)
F(g(x, y)) → F(y)
F(g(x, y)) → G(f(x), f(y))
F(h(x, y)) → G(h(y, f(x)), h(x, f(y)))
F(g(x, y)) → F(x)
F(h(x, y)) → F(x)

The TRS R consists of the following rules:

f(a) → b
f(c) → d
f(g(x, y)) → g(f(x), f(y))
f(h(x, y)) → g(h(y, f(x)), h(x, f(y)))
g(x, x) → h(e, x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

F(h(x, y)) → F(y)
F(g(x, y)) → F(y)
F(g(x, y)) → G(f(x), f(y))
F(h(x, y)) → G(h(y, f(x)), h(x, f(y)))
F(g(x, y)) → F(x)
F(h(x, y)) → F(x)

The TRS R consists of the following rules:

f(a) → b
f(c) → d
f(g(x, y)) → g(f(x), f(y))
f(h(x, y)) → g(h(y, f(x)), h(x, f(y)))
g(x, x) → h(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof

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

F(h(x, y)) → F(y)
F(g(x, y)) → F(y)
F(g(x, y)) → F(x)
F(h(x, y)) → F(x)

The TRS R consists of the following rules:

f(a) → b
f(c) → d
f(g(x, y)) → g(f(x), f(y))
f(h(x, y)) → g(h(y, f(x)), h(x, f(y)))
g(x, x) → h(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.


F(h(x, y)) → F(y)
F(g(x, y)) → F(y)
F(g(x, y)) → F(x)
F(h(x, y)) → F(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1)  =  F(x1)
h(x1, x2)  =  h(x1, x2)
g(x1, x2)  =  g(x1, x2)

Recursive path order with status [2].
Precedence:
h2 > F1
g2 > F1

Status:
g2: multiset
F1: multiset
h2: multiset

The following usable rules [14] were oriented: none



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

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

f(a) → b
f(c) → d
f(g(x, y)) → g(f(x), f(y))
f(h(x, y)) → g(h(y, f(x)), h(x, f(y)))
g(x, x) → h(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.