Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/AG01SLASHHASH3.8b.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:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

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:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
if_minus₃(true, s₁(x0), x1)
if_minus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

QUOT₂(s₁(x), s₁(y)) -> QUOT₂(minus₂(x, y), s₁(y))
MINUS₂(s₁(x), y) -> IF_MINUS₃(le₂(s₁(x), y), s₁(x), y)
MINUS₂(s₁(x), y) -> LE₂(s₁(x), y)
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
LOG₁(s₁(s₁(x))) -> QUOT₂(x, s₁(s₁(0)))
LOG₁(s₁(s₁(x))) -> LOG₁(s₁(quot₂(x, s₁(s₁(0)))))
QUOT₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
IF_MINUS₃(false, s₁(x), y) -> MINUS₂(x, y)

The TRS R consists of the following rules:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
if_minus₃(true, s₁(x0), x1)
if_minus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

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:

QUOT₂(s₁(x), s₁(y)) -> QUOT₂(minus₂(x, y), s₁(y))
MINUS₂(s₁(x), y) -> IF_MINUS₃(le₂(s₁(x), y), s₁(x), y)
MINUS₂(s₁(x), y) -> LE₂(s₁(x), y)
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
LOG₁(s₁(s₁(x))) -> QUOT₂(x, s₁(s₁(0)))
LOG₁(s₁(s₁(x))) -> LOG₁(s₁(quot₂(x, s₁(s₁(0)))))
QUOT₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
IF_MINUS₃(false, s₁(x), y) -> MINUS₂(x, y)

The TRS R consists of the following rules:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
if_minus₃(true, s₁(x0), x1)
if_minus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 3 less nodes.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ 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:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
if_minus₃(true, s₁(x0), x1)
if_minus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

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

              ↳ QDP

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

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
if_minus₃(true, s₁(x0), x1)
if_minus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

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

              ↳ QDP

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

MINUS₂(s₁(x), y) -> IF_MINUS₃(le₂(s₁(x), y), s₁(x), y)
IF_MINUS₃(false, s₁(x), y) -> MINUS₂(x, y)

The TRS R consists of the following rules:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
if_minus₃(true, s₁(x0), x1)
if_minus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

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), y) -> IF_MINUS₃(le₂(s₁(x), y), s₁(x), y)

The remaining pairs can at least by weakly be oriented.

IF_MINUS₃(false, s₁(x), y) -> MINUS₂(x, y)

Used ordering: Combined order from the following AFS and order.
MINUS₂(x1, x2) = MINUS₁(x1)
s₁(x1) = s₁(x1)
IF_MINUS₃(x1, x2, x3) = x2
le₂(x1, x2) = le
false = false
0 = 0
true = true

Lexicographic Path Order [19].
Precedence:

[le, false, true] > [MINUS₁, s_1]
0 > [MINUS₁, s_1]

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

              ↳ QDP

              ↳ QDP

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

IF_MINUS₃(false, s₁(x), y) -> MINUS₂(x, y)

The TRS R consists of the following rules:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
if_minus₃(true, s₁(x0), x1)
if_minus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

QUOT₂(s₁(x), s₁(y)) -> QUOT₂(minus₂(x, y), s₁(y))

The TRS R consists of the following rules:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
if_minus₃(true, s₁(x0), x1)
if_minus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

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.

QUOT₂(s₁(x), s₁(y)) -> QUOT₂(minus₂(x, y), s₁(y))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
QUOT₂(x1, x2) = x1
s₁(x1) = s₁(x1)
minus₂(x1, x2) = x1
0 = 0
if_minus₃(x1, x2, x3) = x2
le₂(x1, x2) = x2
false = false
true = true

Lexicographic Path Order [19].
Precedence:

s₁ > [0, false, true]

The following usable rules [14] were oriented:

minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
le₂(0, y) -> true

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
if_minus₃(true, s₁(x0), x1)
if_minus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

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

              ↳ QDP

                ↳ QDPOrderProof

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

LOG₁(s₁(s₁(x))) -> LOG₁(s₁(quot₂(x, s₁(s₁(0)))))

The TRS R consists of the following rules:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
if_minus₃(true, s₁(x0), x1)
if_minus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

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.

LOG₁(s₁(s₁(x))) -> LOG₁(s₁(quot₂(x, s₁(s₁(0)))))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
LOG₁(x1) = LOG₁(x1)
s₁(x1) = s₁(x1)
quot₂(x1, x2) = x1
0 = 0
minus₂(x1, x2) = x1
if_minus₃(x1, x2, x3) = x2
le₂(x1, x2) = le₁(x1)
false = false
true = true

Lexicographic Path Order [19].
Precedence:

s₁ > LOG₁
s₁ > [0, le₁, false] > true

The following usable rules [14] were oriented:

quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
le₂(0, y) -> true

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
quot₂(0, s₁(y)) -> 0
quot₂(s₁(x), s₁(y)) -> s₁(quot₂(minus₂(x, y), s₁(y)))
log₁(s₁(0)) -> 0
log₁(s₁(s₁(x))) -> s₁(log₁(s₁(quot₂(x, s₁(s₁(0))))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
minus₂(0, x0)
minus₂(s₁(x0), x1)
if_minus₃(true, s₁(x0), x1)
if_minus₃(false, s₁(x0), x1)
quot₂(0, s₁(x0))
quot₂(s₁(x0), s₁(x1))
log₁(s₁(0))
log₁(s₁(s₁(x0)))

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