Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/Ste92SLASHperfect2.trs

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:

minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
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:

minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
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:

minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
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:

minus₂(0, x0)
minus₂(s₁(x0), 0)
minus₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
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)

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), 0, z, u) -> MINUS₂(z, s₁(x))
F₄(s₁(x), s₁(y), z, u) -> F₄(s₁(x), minus₂(y, x), z, u)
F₄(s₁(x), s₁(y), z, u) -> MINUS₂(y, x)
F₄(s₁(x), s₁(y), z, u) -> F₄(x, u, z, u)
F₄(s₁(x), s₁(y), z, u) -> LE₂(x, y)
F₄(s₁(x), s₁(y), z, u) -> IF₃(le₂(x, y), f₄(s₁(x), minus₂(y, x), z, u), f₄(x, u, z, u))
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
PERFECTP₁(s₁(x)) -> F₄(x, s₁(0), s₁(x), s₁(x))
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
F₄(s₁(x), 0, z, u) -> F₄(x, u, minus₂(z, s₁(x)), u)

The TRS R consists of the following rules:

minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
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:

minus₂(0, x0)
minus₂(s₁(x0), 0)
minus₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
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), 0, z, u) -> MINUS₂(z, s₁(x))
F₄(s₁(x), s₁(y), z, u) -> F₄(s₁(x), minus₂(y, x), z, u)
F₄(s₁(x), s₁(y), z, u) -> MINUS₂(y, x)
F₄(s₁(x), s₁(y), z, u) -> F₄(x, u, z, u)
F₄(s₁(x), s₁(y), z, u) -> LE₂(x, y)
F₄(s₁(x), s₁(y), z, u) -> IF₃(le₂(x, y), f₄(s₁(x), minus₂(y, x), z, u), f₄(x, u, z, u))
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
PERFECTP₁(s₁(x)) -> F₄(x, s₁(0), s₁(x), s₁(x))
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
F₄(s₁(x), 0, z, u) -> F₄(x, u, minus₂(z, s₁(x)), u)

The TRS R consists of the following rules:

minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
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:

minus₂(0, x0)
minus₂(s₁(x0), 0)
minus₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
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 [13,14,18] contains 3 SCCs with 5 less nodes.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

LE₂(s₁(x), s₁(y)) -> LE₂(x, y)

The TRS R consists of the following rules:

minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
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:

minus₂(0, x0)
minus₂(s₁(x0), 0)
minus₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

LE₂(s₁(x), s₁(y)) -> LE₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
LE₂(x1, x2) = LE₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
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:

minus₂(0, x0)
minus₂(s₁(x0), 0)
minus₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
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.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)

The TRS R consists of the following rules:

minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
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:

minus₂(0, x0)
minus₂(s₁(x0), 0)
minus₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MINUS₂(x1, x2) = MINUS₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
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:

minus₂(0, x0)
minus₂(s₁(x0), 0)
minus₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
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.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

The TRS R consists of the following rules:

minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
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:

minus₂(0, x0)
minus₂(s₁(x0), 0)
minus₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented 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 by weakly be oriented.

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

Used ordering: Combined order from the following AFS and order.
F₄(x1, x2, x3, x4) = x1
s₁(x1) = s₁(x1)
minus₂(x1, x2) = x1
0 = 0

Lexicographic Path Order [19].
Precedence:

0 > s₁

The following usable rules [14] were oriented:

minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

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)

The TRS R consists of the following rules:

minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
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:

minus₂(0, x0)
minus₂(s₁(x0), 0)
minus₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F₄(x1, x2, x3, x4) = x2
s₁(x1) = s₁(x1)
minus₂(x1, x2) = x1
0 = 0

Lexicographic Path Order [19].
Precedence:

0 > s₁

The following usable rules [14] were oriented:

minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
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:

minus₂(0, x0)
minus₂(s₁(x0), 0)
minus₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
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.