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:

perfectp1(0) -> false
perfectp1(s1(x)) -> f4(x, s1(0), s1(x), s1(x))
f4(0, y, 0, u) -> true
f4(0, y, s1(z), u) -> false
f4(s1(x), 0, z, u) -> f4(x, u, minus2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(le2(x, y), f4(s1(x), minus2(y, x), z, u), f4(x, u, z, u))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

perfectp1(0) -> false
perfectp1(s1(x)) -> f4(x, s1(0), s1(x), s1(x))
f4(0, y, 0, u) -> true
f4(0, y, s1(z), u) -> false
f4(s1(x), 0, z, u) -> f4(x, u, minus2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(le2(x, y), f4(s1(x), minus2(y, x), z, u), f4(x, u, z, u))

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:

F4(s1(x), s1(y), z, u) -> F4(s1(x), minus2(y, x), z, u)
F4(s1(x), s1(y), z, u) -> F4(x, u, z, u)
PERFECTP1(s1(x)) -> F4(x, s1(0), s1(x), s1(x))
F4(s1(x), 0, z, u) -> F4(x, u, minus2(z, s1(x)), u)

The TRS R consists of the following rules:

perfectp1(0) -> false
perfectp1(s1(x)) -> f4(x, s1(0), s1(x), s1(x))
f4(0, y, 0, u) -> true
f4(0, y, s1(z), u) -> false
f4(s1(x), 0, z, u) -> f4(x, u, minus2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(le2(x, y), f4(s1(x), minus2(y, x), z, u), f4(x, u, z, u))

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:

F4(s1(x), s1(y), z, u) -> F4(s1(x), minus2(y, x), z, u)
F4(s1(x), s1(y), z, u) -> F4(x, u, z, u)
PERFECTP1(s1(x)) -> F4(x, s1(0), s1(x), s1(x))
F4(s1(x), 0, z, u) -> F4(x, u, minus2(z, s1(x)), u)

The TRS R consists of the following rules:

perfectp1(0) -> false
perfectp1(s1(x)) -> f4(x, s1(0), s1(x), s1(x))
f4(0, y, 0, u) -> true
f4(0, y, s1(z), u) -> false
f4(s1(x), 0, z, u) -> f4(x, u, minus2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(le2(x, y), f4(s1(x), minus2(y, x), z, u), f4(x, u, z, u))

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:

F4(s1(x), s1(y), z, u) -> F4(x, u, z, u)
F4(s1(x), 0, z, u) -> F4(x, u, minus2(z, s1(x)), u)

The TRS R consists of the following rules:

perfectp1(0) -> false
perfectp1(s1(x)) -> f4(x, s1(0), s1(x), s1(x))
f4(0, y, 0, u) -> true
f4(0, y, s1(z), u) -> false
f4(s1(x), 0, z, u) -> f4(x, u, minus2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(le2(x, y), f4(s1(x), minus2(y, x), z, u), f4(x, u, z, u))

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.


F4(s1(x), s1(y), z, u) -> F4(x, u, z, u)
F4(s1(x), 0, z, u) -> F4(x, u, minus2(z, s1(x)), u)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 3   
POL(F4(x1, x2, x3, x4)) = 2·x1 + x1·x2 + 2·x1·x4 + 2·x2 + 2·x3   
POL(minus2(x1, x2)) = 2 + x2   
POL(s1(x1)) = 1 + 3·x1 + 3·x12   

The following usable rules [14] were oriented: none



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

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

perfectp1(0) -> false
perfectp1(s1(x)) -> f4(x, s1(0), s1(x), s1(x))
f4(0, y, 0, u) -> true
f4(0, y, s1(z), u) -> false
f4(s1(x), 0, z, u) -> f4(x, u, minus2(z, s1(x)), u)
f4(s1(x), s1(y), z, u) -> if3(le2(x, y), f4(s1(x), minus2(y, x), z, u), f4(x, u, z, u))

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.