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
  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
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))

The set Q consists of the following terms:

perfectp1(0)
perfectp1(s1(x0))
f4(0, x0, 0, x1)
f4(0, x0, s1(x1), x2)
f4(s1(x0), 0, x1, x2)
f4(s1(x0), s1(x1), x2, x3)


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))

The set Q consists of the following terms:

perfectp1(0)
perfectp1(s1(x0))
f4(0, x0, 0, x1)
f4(0, x0, s1(x1), x2)
f4(s1(x0), 0, x1, x2)
f4(s1(x0), s1(x1), x2, x3)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

perfectp1(0)
perfectp1(s1(x0))
f4(0, x0, 0, x1)
f4(0, x0, s1(x1), x2)
f4(s1(x0), 0, x1, x2)
f4(s1(x0), s1(x1), x2, x3)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
QDP
              ↳ QDPAfsSolverProof

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))

The set Q consists of the following terms:

perfectp1(0)
perfectp1(s1(x0))
f4(0, x0, 0, x1)
f4(0, x0, s1(x1), x2)
f4(s1(x0), 0, x1, x2)
f4(s1(x0), s1(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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)
Used argument filtering: F4(x1, x2, x3, x4)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPAfsSolverProof
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))

The set Q consists of the following terms:

perfectp1(0)
perfectp1(s1(x0))
f4(0, x0, 0, x1)
f4(0, x0, s1(x1), x2)
f4(s1(x0), 0, x1, x2)
f4(s1(x0), s1(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.