Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar18/AG01SLASHHASH3.10.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

eq₂(0, 0) -> true
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
min₁(add₂(n, nil)) -> n
min₁(add₂(n, add₂(m, x))) -> if_min₂(le₂(n, m), add₂(n, add₂(m, x)))
if_min₂(true, add₂(n, add₂(m, x))) -> min₁(add₂(n, x))
if_min₂(false, add₂(n, add₂(m, x))) -> min₁(add₂(m, x))
rm₂(n, nil) -> nil
rm₂(n, add₂(m, x)) -> if_rm₃(eq₂(n, m), n, add₂(m, x))
if_rm₃(true, n, add₂(m, x)) -> rm₂(n, x)
if_rm₃(false, n, add₂(m, x)) -> add₂(m, rm₂(n, x))
minsort₂(nil, nil) -> nil
minsort₂(add₂(n, x), y) -> if_minsort₃(eq₂(n, min₁(add₂(n, x))), add₂(n, x), y)
if_minsort₃(true, add₂(n, x), y) -> add₂(n, minsort₂(app₂(rm₂(n, x), y), nil))
if_minsort₃(false, add₂(n, x), y) -> minsort₂(x, add₂(n, y))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

eq₂(0, 0) -> true
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
min₁(add₂(n, nil)) -> n
min₁(add₂(n, add₂(m, x))) -> if_min₂(le₂(n, m), add₂(n, add₂(m, x)))
if_min₂(true, add₂(n, add₂(m, x))) -> min₁(add₂(n, x))
if_min₂(false, add₂(n, add₂(m, x))) -> min₁(add₂(m, x))
rm₂(n, nil) -> nil
rm₂(n, add₂(m, x)) -> if_rm₃(eq₂(n, m), n, add₂(m, x))
if_rm₃(true, n, add₂(m, x)) -> rm₂(n, x)
if_rm₃(false, n, add₂(m, x)) -> add₂(m, rm₂(n, x))
minsort₂(nil, nil) -> nil
minsort₂(add₂(n, x), y) -> if_minsort₃(eq₂(n, min₁(add₂(n, x))), add₂(n, x), y)
if_minsort₃(true, add₂(n, x), y) -> add₂(n, minsort₂(app₂(rm₂(n, x), y), nil))
if_minsort₃(false, add₂(n, x), y) -> minsort₂(x, add₂(n, y))

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:

eq₂(0, 0) -> true
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
min₁(add₂(n, nil)) -> n
min₁(add₂(n, add₂(m, x))) -> if_min₂(le₂(n, m), add₂(n, add₂(m, x)))
if_min₂(true, add₂(n, add₂(m, x))) -> min₁(add₂(n, x))
if_min₂(false, add₂(n, add₂(m, x))) -> min₁(add₂(m, x))
rm₂(n, nil) -> nil
rm₂(n, add₂(m, x)) -> if_rm₃(eq₂(n, m), n, add₂(m, x))
if_rm₃(true, n, add₂(m, x)) -> rm₂(n, x)
if_rm₃(false, n, add₂(m, x)) -> add₂(m, rm₂(n, x))
minsort₂(nil, nil) -> nil
minsort₂(add₂(n, x), y) -> if_minsort₃(eq₂(n, min₁(add₂(n, x))), add₂(n, x), y)
if_minsort₃(true, add₂(n, x), y) -> add₂(n, minsort₂(app₂(rm₂(n, x), y), nil))
if_minsort₃(false, add₂(n, x), y) -> minsort₂(x, add₂(n, y))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
min₁(add₂(x0, nil))
min₁(add₂(x0, add₂(x1, x2)))
if_min₂(true, add₂(x0, add₂(x1, x2)))
if_min₂(false, add₂(x0, add₂(x1, x2)))
rm₂(x0, nil)
rm₂(x0, add₂(x1, x2))
if_rm₃(true, x0, add₂(x1, x2))
if_rm₃(false, x0, add₂(x1, x2))
minsort₂(nil, nil)
minsort₂(add₂(x0, x1), x2)
if_minsort₃(true, add₂(x0, x1), x2)
if_minsort₃(false, add₂(x0, x1), x2)

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

MINSORT₂(add₂(n, x), y) -> IF_MINSORT₃(eq₂(n, min₁(add₂(n, x))), add₂(n, x), y)
MINSORT₂(add₂(n, x), y) -> MIN₁(add₂(n, x))
MIN₁(add₂(n, add₂(m, x))) -> IF_MIN₂(le₂(n, m), add₂(n, add₂(m, x)))
IF_MINSORT₃(true, add₂(n, x), y) -> APP₂(rm₂(n, x), y)
IF_RM₃(true, n, add₂(m, x)) -> RM₂(n, x)
IF_MIN₂(false, add₂(n, add₂(m, x))) -> MIN₁(add₂(m, x))
IF_MINSORT₃(false, add₂(n, x), y) -> MINSORT₂(x, add₂(n, y))
IF_MIN₂(true, add₂(n, add₂(m, x))) -> MIN₁(add₂(n, x))
RM₂(n, add₂(m, x)) -> EQ₂(n, m)
EQ₂(s₁(x), s₁(y)) -> EQ₂(x, y)
IF_MINSORT₃(true, add₂(n, x), y) -> MINSORT₂(app₂(rm₂(n, x), y), nil)
MIN₁(add₂(n, add₂(m, x))) -> LE₂(n, m)
RM₂(n, add₂(m, x)) -> IF_RM₃(eq₂(n, m), n, add₂(m, x))
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
MINSORT₂(add₂(n, x), y) -> EQ₂(n, min₁(add₂(n, x)))
IF_RM₃(false, n, add₂(m, x)) -> RM₂(n, x)
IF_MINSORT₃(true, add₂(n, x), y) -> RM₂(n, x)
APP₂(add₂(n, x), y) -> APP₂(x, y)

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
min₁(add₂(n, nil)) -> n
min₁(add₂(n, add₂(m, x))) -> if_min₂(le₂(n, m), add₂(n, add₂(m, x)))
if_min₂(true, add₂(n, add₂(m, x))) -> min₁(add₂(n, x))
if_min₂(false, add₂(n, add₂(m, x))) -> min₁(add₂(m, x))
rm₂(n, nil) -> nil
rm₂(n, add₂(m, x)) -> if_rm₃(eq₂(n, m), n, add₂(m, x))
if_rm₃(true, n, add₂(m, x)) -> rm₂(n, x)
if_rm₃(false, n, add₂(m, x)) -> add₂(m, rm₂(n, x))
minsort₂(nil, nil) -> nil
minsort₂(add₂(n, x), y) -> if_minsort₃(eq₂(n, min₁(add₂(n, x))), add₂(n, x), y)
if_minsort₃(true, add₂(n, x), y) -> add₂(n, minsort₂(app₂(rm₂(n, x), y), nil))
if_minsort₃(false, add₂(n, x), y) -> minsort₂(x, add₂(n, y))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
min₁(add₂(x0, nil))
min₁(add₂(x0, add₂(x1, x2)))
if_min₂(true, add₂(x0, add₂(x1, x2)))
if_min₂(false, add₂(x0, add₂(x1, x2)))
rm₂(x0, nil)
rm₂(x0, add₂(x1, x2))
if_rm₃(true, x0, add₂(x1, x2))
if_rm₃(false, x0, add₂(x1, x2))
minsort₂(nil, nil)
minsort₂(add₂(x0, x1), x2)
if_minsort₃(true, add₂(x0, x1), x2)
if_minsort₃(false, add₂(x0, x1), x2)

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:

MINSORT₂(add₂(n, x), y) -> IF_MINSORT₃(eq₂(n, min₁(add₂(n, x))), add₂(n, x), y)
MINSORT₂(add₂(n, x), y) -> MIN₁(add₂(n, x))
MIN₁(add₂(n, add₂(m, x))) -> IF_MIN₂(le₂(n, m), add₂(n, add₂(m, x)))
IF_MINSORT₃(true, add₂(n, x), y) -> APP₂(rm₂(n, x), y)
IF_RM₃(true, n, add₂(m, x)) -> RM₂(n, x)
IF_MIN₂(false, add₂(n, add₂(m, x))) -> MIN₁(add₂(m, x))
IF_MINSORT₃(false, add₂(n, x), y) -> MINSORT₂(x, add₂(n, y))
IF_MIN₂(true, add₂(n, add₂(m, x))) -> MIN₁(add₂(n, x))
RM₂(n, add₂(m, x)) -> EQ₂(n, m)
EQ₂(s₁(x), s₁(y)) -> EQ₂(x, y)
IF_MINSORT₃(true, add₂(n, x), y) -> MINSORT₂(app₂(rm₂(n, x), y), nil)
MIN₁(add₂(n, add₂(m, x))) -> LE₂(n, m)
RM₂(n, add₂(m, x)) -> IF_RM₃(eq₂(n, m), n, add₂(m, x))
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
MINSORT₂(add₂(n, x), y) -> EQ₂(n, min₁(add₂(n, x)))
IF_RM₃(false, n, add₂(m, x)) -> RM₂(n, x)
IF_MINSORT₃(true, add₂(n, x), y) -> RM₂(n, x)
APP₂(add₂(n, x), y) -> APP₂(x, y)

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
min₁(add₂(n, nil)) -> n
min₁(add₂(n, add₂(m, x))) -> if_min₂(le₂(n, m), add₂(n, add₂(m, x)))
if_min₂(true, add₂(n, add₂(m, x))) -> min₁(add₂(n, x))
if_min₂(false, add₂(n, add₂(m, x))) -> min₁(add₂(m, x))
rm₂(n, nil) -> nil
rm₂(n, add₂(m, x)) -> if_rm₃(eq₂(n, m), n, add₂(m, x))
if_rm₃(true, n, add₂(m, x)) -> rm₂(n, x)
if_rm₃(false, n, add₂(m, x)) -> add₂(m, rm₂(n, x))
minsort₂(nil, nil) -> nil
minsort₂(add₂(n, x), y) -> if_minsort₃(eq₂(n, min₁(add₂(n, x))), add₂(n, x), y)
if_minsort₃(true, add₂(n, x), y) -> add₂(n, minsort₂(app₂(rm₂(n, x), y), nil))
if_minsort₃(false, add₂(n, x), y) -> minsort₂(x, add₂(n, y))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
min₁(add₂(x0, nil))
min₁(add₂(x0, add₂(x1, x2)))
if_min₂(true, add₂(x0, add₂(x1, x2)))
if_min₂(false, add₂(x0, add₂(x1, x2)))
rm₂(x0, nil)
rm₂(x0, add₂(x1, x2))
if_rm₃(true, x0, add₂(x1, x2))
if_rm₃(false, x0, add₂(x1, x2))
minsort₂(nil, nil)
minsort₂(add₂(x0, x1), x2)
if_minsort₃(true, add₂(x0, x1), x2)
if_minsort₃(false, add₂(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 6 SCCs with 6 less nodes.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

APP₂(add₂(n, x), y) -> APP₂(x, y)

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
min₁(add₂(n, nil)) -> n
min₁(add₂(n, add₂(m, x))) -> if_min₂(le₂(n, m), add₂(n, add₂(m, x)))
if_min₂(true, add₂(n, add₂(m, x))) -> min₁(add₂(n, x))
if_min₂(false, add₂(n, add₂(m, x))) -> min₁(add₂(m, x))
rm₂(n, nil) -> nil
rm₂(n, add₂(m, x)) -> if_rm₃(eq₂(n, m), n, add₂(m, x))
if_rm₃(true, n, add₂(m, x)) -> rm₂(n, x)
if_rm₃(false, n, add₂(m, x)) -> add₂(m, rm₂(n, x))
minsort₂(nil, nil) -> nil
minsort₂(add₂(n, x), y) -> if_minsort₃(eq₂(n, min₁(add₂(n, x))), add₂(n, x), y)
if_minsort₃(true, add₂(n, x), y) -> add₂(n, minsort₂(app₂(rm₂(n, x), y), nil))
if_minsort₃(false, add₂(n, x), y) -> minsort₂(x, add₂(n, y))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
min₁(add₂(x0, nil))
min₁(add₂(x0, add₂(x1, x2)))
if_min₂(true, add₂(x0, add₂(x1, x2)))
if_min₂(false, add₂(x0, add₂(x1, x2)))
rm₂(x0, nil)
rm₂(x0, add₂(x1, x2))
if_rm₃(true, x0, add₂(x1, x2))
if_rm₃(false, x0, add₂(x1, x2))
minsort₂(nil, nil)
minsort₂(add₂(x0, x1), x2)
if_minsort₃(true, add₂(x0, x1), x2)
if_minsort₃(false, add₂(x0, x1), x2)

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.

APP₂(add₂(n, x), y) -> APP₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP₂(x1, x2) = APP₁(x1)
add₂(x1, x2) = add₁(x2)

Lexicographic Path Order [19].
Precedence:

add₁ > APP₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

eq₂(0, 0) -> true
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
min₁(add₂(n, nil)) -> n
min₁(add₂(n, add₂(m, x))) -> if_min₂(le₂(n, m), add₂(n, add₂(m, x)))
if_min₂(true, add₂(n, add₂(m, x))) -> min₁(add₂(n, x))
if_min₂(false, add₂(n, add₂(m, x))) -> min₁(add₂(m, x))
rm₂(n, nil) -> nil
rm₂(n, add₂(m, x)) -> if_rm₃(eq₂(n, m), n, add₂(m, x))
if_rm₃(true, n, add₂(m, x)) -> rm₂(n, x)
if_rm₃(false, n, add₂(m, x)) -> add₂(m, rm₂(n, x))
minsort₂(nil, nil) -> nil
minsort₂(add₂(n, x), y) -> if_minsort₃(eq₂(n, min₁(add₂(n, x))), add₂(n, x), y)
if_minsort₃(true, add₂(n, x), y) -> add₂(n, minsort₂(app₂(rm₂(n, x), y), nil))
if_minsort₃(false, add₂(n, x), y) -> minsort₂(x, add₂(n, y))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
min₁(add₂(x0, nil))
min₁(add₂(x0, add₂(x1, x2)))
if_min₂(true, add₂(x0, add₂(x1, x2)))
if_min₂(false, add₂(x0, add₂(x1, x2)))
rm₂(x0, nil)
rm₂(x0, add₂(x1, x2))
if_rm₃(true, x0, add₂(x1, x2))
if_rm₃(false, x0, add₂(x1, x2))
minsort₂(nil, nil)
minsort₂(add₂(x0, x1), x2)
if_minsort₃(true, add₂(x0, x1), x2)
if_minsort₃(false, add₂(x0, x1), x2)

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

              ↳ 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:

eq₂(0, 0) -> true
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
min₁(add₂(n, nil)) -> n
min₁(add₂(n, add₂(m, x))) -> if_min₂(le₂(n, m), add₂(n, add₂(m, x)))
if_min₂(true, add₂(n, add₂(m, x))) -> min₁(add₂(n, x))
if_min₂(false, add₂(n, add₂(m, x))) -> min₁(add₂(m, x))
rm₂(n, nil) -> nil
rm₂(n, add₂(m, x)) -> if_rm₃(eq₂(n, m), n, add₂(m, x))
if_rm₃(true, n, add₂(m, x)) -> rm₂(n, x)
if_rm₃(false, n, add₂(m, x)) -> add₂(m, rm₂(n, x))
minsort₂(nil, nil) -> nil
minsort₂(add₂(n, x), y) -> if_minsort₃(eq₂(n, min₁(add₂(n, x))), add₂(n, x), y)
if_minsort₃(true, add₂(n, x), y) -> add₂(n, minsort₂(app₂(rm₂(n, x), y), nil))
if_minsort₃(false, add₂(n, x), y) -> minsort₂(x, add₂(n, y))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
min₁(add₂(x0, nil))
min₁(add₂(x0, add₂(x1, x2)))
if_min₂(true, add₂(x0, add₂(x1, x2)))
if_min₂(false, add₂(x0, add₂(x1, x2)))
rm₂(x0, nil)
rm₂(x0, add₂(x1, x2))
if_rm₃(true, x0, add₂(x1, x2))
if_rm₃(false, x0, add₂(x1, x2))
minsort₂(nil, nil)
minsort₂(add₂(x0, x1), x2)
if_minsort₃(true, add₂(x0, x1), x2)
if_minsort₃(false, add₂(x0, x1), x2)

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) = x2
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

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

eq₂(0, 0) -> true
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
min₁(add₂(n, nil)) -> n
min₁(add₂(n, add₂(m, x))) -> if_min₂(le₂(n, m), add₂(n, add₂(m, x)))
if_min₂(true, add₂(n, add₂(m, x))) -> min₁(add₂(n, x))
if_min₂(false, add₂(n, add₂(m, x))) -> min₁(add₂(m, x))
rm₂(n, nil) -> nil
rm₂(n, add₂(m, x)) -> if_rm₃(eq₂(n, m), n, add₂(m, x))
if_rm₃(true, n, add₂(m, x)) -> rm₂(n, x)
if_rm₃(false, n, add₂(m, x)) -> add₂(m, rm₂(n, x))
minsort₂(nil, nil) -> nil
minsort₂(add₂(n, x), y) -> if_minsort₃(eq₂(n, min₁(add₂(n, x))), add₂(n, x), y)
if_minsort₃(true, add₂(n, x), y) -> add₂(n, minsort₂(app₂(rm₂(n, x), y), nil))
if_minsort₃(false, add₂(n, x), y) -> minsort₂(x, add₂(n, y))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
min₁(add₂(x0, nil))
min₁(add₂(x0, add₂(x1, x2)))
if_min₂(true, add₂(x0, add₂(x1, x2)))
if_min₂(false, add₂(x0, add₂(x1, x2)))
rm₂(x0, nil)
rm₂(x0, add₂(x1, x2))
if_rm₃(true, x0, add₂(x1, x2))
if_rm₃(false, x0, add₂(x1, x2))
minsort₂(nil, nil)
minsort₂(add₂(x0, x1), x2)
if_minsort₃(true, add₂(x0, x1), x2)
if_minsort₃(false, add₂(x0, x1), x2)

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

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

MIN₁(add₂(n, add₂(m, x))) -> IF_MIN₂(le₂(n, m), add₂(n, add₂(m, x)))
IF_MIN₂(false, add₂(n, add₂(m, x))) -> MIN₁(add₂(m, x))
IF_MIN₂(true, add₂(n, add₂(m, x))) -> MIN₁(add₂(n, x))

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
min₁(add₂(n, nil)) -> n
min₁(add₂(n, add₂(m, x))) -> if_min₂(le₂(n, m), add₂(n, add₂(m, x)))
if_min₂(true, add₂(n, add₂(m, x))) -> min₁(add₂(n, x))
if_min₂(false, add₂(n, add₂(m, x))) -> min₁(add₂(m, x))
rm₂(n, nil) -> nil
rm₂(n, add₂(m, x)) -> if_rm₃(eq₂(n, m), n, add₂(m, x))
if_rm₃(true, n, add₂(m, x)) -> rm₂(n, x)
if_rm₃(false, n, add₂(m, x)) -> add₂(m, rm₂(n, x))
minsort₂(nil, nil) -> nil
minsort₂(add₂(n, x), y) -> if_minsort₃(eq₂(n, min₁(add₂(n, x))), add₂(n, x), y)
if_minsort₃(true, add₂(n, x), y) -> add₂(n, minsort₂(app₂(rm₂(n, x), y), nil))
if_minsort₃(false, add₂(n, x), y) -> minsort₂(x, add₂(n, y))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
min₁(add₂(x0, nil))
min₁(add₂(x0, add₂(x1, x2)))
if_min₂(true, add₂(x0, add₂(x1, x2)))
if_min₂(false, add₂(x0, add₂(x1, x2)))
rm₂(x0, nil)
rm₂(x0, add₂(x1, x2))
if_rm₃(true, x0, add₂(x1, x2))
if_rm₃(false, x0, add₂(x1, x2))
minsort₂(nil, nil)
minsort₂(add₂(x0, x1), x2)
if_minsort₃(true, add₂(x0, x1), x2)
if_minsort₃(false, add₂(x0, x1), x2)

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.

MIN₁(add₂(n, add₂(m, x))) -> IF_MIN₂(le₂(n, m), add₂(n, add₂(m, x)))
IF_MIN₂(false, add₂(n, add₂(m, x))) -> MIN₁(add₂(m, x))
IF_MIN₂(true, add₂(n, add₂(m, x))) -> MIN₁(add₂(n, x))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MIN₁(x1) = MIN₁(x1)
add₂(x1, x2) = add₁(x2)
IF_MIN₂(x1, x2) = IF_MIN₂(x1, x2)
le₂(x1, x2) = le
false = false
true = true
0 = 0
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

add₁ > le > false > MIN₁ > IF_MIN₂ > true
0 > false > MIN₁ > IF_MIN₂ > true
s₁ > le > false > MIN₁ > IF_MIN₂ > true

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

eq₂(0, 0) -> true
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
min₁(add₂(n, nil)) -> n
min₁(add₂(n, add₂(m, x))) -> if_min₂(le₂(n, m), add₂(n, add₂(m, x)))
if_min₂(true, add₂(n, add₂(m, x))) -> min₁(add₂(n, x))
if_min₂(false, add₂(n, add₂(m, x))) -> min₁(add₂(m, x))
rm₂(n, nil) -> nil
rm₂(n, add₂(m, x)) -> if_rm₃(eq₂(n, m), n, add₂(m, x))
if_rm₃(true, n, add₂(m, x)) -> rm₂(n, x)
if_rm₃(false, n, add₂(m, x)) -> add₂(m, rm₂(n, x))
minsort₂(nil, nil) -> nil
minsort₂(add₂(n, x), y) -> if_minsort₃(eq₂(n, min₁(add₂(n, x))), add₂(n, x), y)
if_minsort₃(true, add₂(n, x), y) -> add₂(n, minsort₂(app₂(rm₂(n, x), y), nil))
if_minsort₃(false, add₂(n, x), y) -> minsort₂(x, add₂(n, y))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
min₁(add₂(x0, nil))
min₁(add₂(x0, add₂(x1, x2)))
if_min₂(true, add₂(x0, add₂(x1, x2)))
if_min₂(false, add₂(x0, add₂(x1, x2)))
rm₂(x0, nil)
rm₂(x0, add₂(x1, x2))
if_rm₃(true, x0, add₂(x1, x2))
if_rm₃(false, x0, add₂(x1, x2))
minsort₂(nil, nil)
minsort₂(add₂(x0, x1), x2)
if_minsort₃(true, add₂(x0, x1), x2)
if_minsort₃(false, add₂(x0, x1), x2)

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

              ↳ QDP

              ↳ QDP

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

EQ₂(s₁(x), s₁(y)) -> EQ₂(x, y)

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
min₁(add₂(n, nil)) -> n
min₁(add₂(n, add₂(m, x))) -> if_min₂(le₂(n, m), add₂(n, add₂(m, x)))
if_min₂(true, add₂(n, add₂(m, x))) -> min₁(add₂(n, x))
if_min₂(false, add₂(n, add₂(m, x))) -> min₁(add₂(m, x))
rm₂(n, nil) -> nil
rm₂(n, add₂(m, x)) -> if_rm₃(eq₂(n, m), n, add₂(m, x))
if_rm₃(true, n, add₂(m, x)) -> rm₂(n, x)
if_rm₃(false, n, add₂(m, x)) -> add₂(m, rm₂(n, x))
minsort₂(nil, nil) -> nil
minsort₂(add₂(n, x), y) -> if_minsort₃(eq₂(n, min₁(add₂(n, x))), add₂(n, x), y)
if_minsort₃(true, add₂(n, x), y) -> add₂(n, minsort₂(app₂(rm₂(n, x), y), nil))
if_minsort₃(false, add₂(n, x), y) -> minsort₂(x, add₂(n, y))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
min₁(add₂(x0, nil))
min₁(add₂(x0, add₂(x1, x2)))
if_min₂(true, add₂(x0, add₂(x1, x2)))
if_min₂(false, add₂(x0, add₂(x1, x2)))
rm₂(x0, nil)
rm₂(x0, add₂(x1, x2))
if_rm₃(true, x0, add₂(x1, x2))
if_rm₃(false, x0, add₂(x1, x2))
minsort₂(nil, nil)
minsort₂(add₂(x0, x1), x2)
if_minsort₃(true, add₂(x0, x1), x2)
if_minsort₃(false, add₂(x0, x1), x2)

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.

EQ₂(s₁(x), s₁(y)) -> EQ₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
EQ₂(x1, x2) = x2
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

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

eq₂(0, 0) -> true
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
min₁(add₂(n, nil)) -> n
min₁(add₂(n, add₂(m, x))) -> if_min₂(le₂(n, m), add₂(n, add₂(m, x)))
if_min₂(true, add₂(n, add₂(m, x))) -> min₁(add₂(n, x))
if_min₂(false, add₂(n, add₂(m, x))) -> min₁(add₂(m, x))
rm₂(n, nil) -> nil
rm₂(n, add₂(m, x)) -> if_rm₃(eq₂(n, m), n, add₂(m, x))
if_rm₃(true, n, add₂(m, x)) -> rm₂(n, x)
if_rm₃(false, n, add₂(m, x)) -> add₂(m, rm₂(n, x))
minsort₂(nil, nil) -> nil
minsort₂(add₂(n, x), y) -> if_minsort₃(eq₂(n, min₁(add₂(n, x))), add₂(n, x), y)
if_minsort₃(true, add₂(n, x), y) -> add₂(n, minsort₂(app₂(rm₂(n, x), y), nil))
if_minsort₃(false, add₂(n, x), y) -> minsort₂(x, add₂(n, y))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
min₁(add₂(x0, nil))
min₁(add₂(x0, add₂(x1, x2)))
if_min₂(true, add₂(x0, add₂(x1, x2)))
if_min₂(false, add₂(x0, add₂(x1, x2)))
rm₂(x0, nil)
rm₂(x0, add₂(x1, x2))
if_rm₃(true, x0, add₂(x1, x2))
if_rm₃(false, x0, add₂(x1, x2))
minsort₂(nil, nil)
minsort₂(add₂(x0, x1), x2)
if_minsort₃(true, add₂(x0, x1), x2)
if_minsort₃(false, add₂(x0, x1), x2)

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

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

RM₂(n, add₂(m, x)) -> IF_RM₃(eq₂(n, m), n, add₂(m, x))
IF_RM₃(true, n, add₂(m, x)) -> RM₂(n, x)
IF_RM₃(false, n, add₂(m, x)) -> RM₂(n, x)

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
min₁(add₂(n, nil)) -> n
min₁(add₂(n, add₂(m, x))) -> if_min₂(le₂(n, m), add₂(n, add₂(m, x)))
if_min₂(true, add₂(n, add₂(m, x))) -> min₁(add₂(n, x))
if_min₂(false, add₂(n, add₂(m, x))) -> min₁(add₂(m, x))
rm₂(n, nil) -> nil
rm₂(n, add₂(m, x)) -> if_rm₃(eq₂(n, m), n, add₂(m, x))
if_rm₃(true, n, add₂(m, x)) -> rm₂(n, x)
if_rm₃(false, n, add₂(m, x)) -> add₂(m, rm₂(n, x))
minsort₂(nil, nil) -> nil
minsort₂(add₂(n, x), y) -> if_minsort₃(eq₂(n, min₁(add₂(n, x))), add₂(n, x), y)
if_minsort₃(true, add₂(n, x), y) -> add₂(n, minsort₂(app₂(rm₂(n, x), y), nil))
if_minsort₃(false, add₂(n, x), y) -> minsort₂(x, add₂(n, y))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
min₁(add₂(x0, nil))
min₁(add₂(x0, add₂(x1, x2)))
if_min₂(true, add₂(x0, add₂(x1, x2)))
if_min₂(false, add₂(x0, add₂(x1, x2)))
rm₂(x0, nil)
rm₂(x0, add₂(x1, x2))
if_rm₃(true, x0, add₂(x1, x2))
if_rm₃(false, x0, add₂(x1, x2))
minsort₂(nil, nil)
minsort₂(add₂(x0, x1), x2)
if_minsort₃(true, add₂(x0, x1), x2)
if_minsort₃(false, add₂(x0, x1), x2)

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.

RM₂(n, add₂(m, x)) -> IF_RM₃(eq₂(n, m), n, add₂(m, x))
IF_RM₃(true, n, add₂(m, x)) -> RM₂(n, x)
IF_RM₃(false, n, add₂(m, x)) -> RM₂(n, x)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
RM₂(x1, x2) = RM₁(x2)
add₂(x1, x2) = add₁(x2)
IF_RM₃(x1, x2, x3) = x3
eq₂(x1, x2) = x2
true = true
false = false
0 = 0
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

add₁ > RM₁
0 > true
0 > false > RM₁
s₁ > false > RM₁

The following usable rules [14] were oriented:

eq₂(0, 0) -> true
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

eq₂(0, 0) -> true
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
min₁(add₂(n, nil)) -> n
min₁(add₂(n, add₂(m, x))) -> if_min₂(le₂(n, m), add₂(n, add₂(m, x)))
if_min₂(true, add₂(n, add₂(m, x))) -> min₁(add₂(n, x))
if_min₂(false, add₂(n, add₂(m, x))) -> min₁(add₂(m, x))
rm₂(n, nil) -> nil
rm₂(n, add₂(m, x)) -> if_rm₃(eq₂(n, m), n, add₂(m, x))
if_rm₃(true, n, add₂(m, x)) -> rm₂(n, x)
if_rm₃(false, n, add₂(m, x)) -> add₂(m, rm₂(n, x))
minsort₂(nil, nil) -> nil
minsort₂(add₂(n, x), y) -> if_minsort₃(eq₂(n, min₁(add₂(n, x))), add₂(n, x), y)
if_minsort₃(true, add₂(n, x), y) -> add₂(n, minsort₂(app₂(rm₂(n, x), y), nil))
if_minsort₃(false, add₂(n, x), y) -> minsort₂(x, add₂(n, y))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
min₁(add₂(x0, nil))
min₁(add₂(x0, add₂(x1, x2)))
if_min₂(true, add₂(x0, add₂(x1, x2)))
if_min₂(false, add₂(x0, add₂(x1, x2)))
rm₂(x0, nil)
rm₂(x0, add₂(x1, x2))
if_rm₃(true, x0, add₂(x1, x2))
if_rm₃(false, x0, add₂(x1, x2))
minsort₂(nil, nil)
minsort₂(add₂(x0, x1), x2)
if_minsort₃(true, add₂(x0, x1), x2)
if_minsort₃(false, add₂(x0, x1), x2)

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

              ↳ QDP

              ↳ QDP

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

MINSORT₂(add₂(n, x), y) -> IF_MINSORT₃(eq₂(n, min₁(add₂(n, x))), add₂(n, x), y)
IF_MINSORT₃(true, add₂(n, x), y) -> MINSORT₂(app₂(rm₂(n, x), y), nil)
IF_MINSORT₃(false, add₂(n, x), y) -> MINSORT₂(x, add₂(n, y))

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
min₁(add₂(n, nil)) -> n
min₁(add₂(n, add₂(m, x))) -> if_min₂(le₂(n, m), add₂(n, add₂(m, x)))
if_min₂(true, add₂(n, add₂(m, x))) -> min₁(add₂(n, x))
if_min₂(false, add₂(n, add₂(m, x))) -> min₁(add₂(m, x))
rm₂(n, nil) -> nil
rm₂(n, add₂(m, x)) -> if_rm₃(eq₂(n, m), n, add₂(m, x))
if_rm₃(true, n, add₂(m, x)) -> rm₂(n, x)
if_rm₃(false, n, add₂(m, x)) -> add₂(m, rm₂(n, x))
minsort₂(nil, nil) -> nil
minsort₂(add₂(n, x), y) -> if_minsort₃(eq₂(n, min₁(add₂(n, x))), add₂(n, x), y)
if_minsort₃(true, add₂(n, x), y) -> add₂(n, minsort₂(app₂(rm₂(n, x), y), nil))
if_minsort₃(false, add₂(n, x), y) -> minsort₂(x, add₂(n, y))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
min₁(add₂(x0, nil))
min₁(add₂(x0, add₂(x1, x2)))
if_min₂(true, add₂(x0, add₂(x1, x2)))
if_min₂(false, add₂(x0, add₂(x1, x2)))
rm₂(x0, nil)
rm₂(x0, add₂(x1, x2))
if_rm₃(true, x0, add₂(x1, x2))
if_rm₃(false, x0, add₂(x1, x2))
minsort₂(nil, nil)
minsort₂(add₂(x0, x1), x2)
if_minsort₃(true, add₂(x0, x1), x2)
if_minsort₃(false, add₂(x0, x1), x2)

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