Termination w.r.t. Q proof of TRS/AG01#3.10.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:

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

  ↳ 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))

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:

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))

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:

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))

Q is empty.
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

  ↳ 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))

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.

APP₂(add₂(n, x), y) -> APP₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(APP₂(x₁, x₂)) = 2·x₁
POL(add₂(x₁, x₂)) = 1 + 2·x₂

The following usable rules [14] were oriented: none

↳ 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))

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.

↳ 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))

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.

LE₂(s₁(x), s₁(y)) -> LE₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(LE₂(x₁, x₂)) = 2·x₁·x₂
POL(s₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented: none

↳ 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))

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.

↳ 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))

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.

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 be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(IF_MIN₂(x₁, x₂)) = x₂
POL(MIN₁(x₁)) = x₁
POL(add₂(x₁, x₂)) = 2 + x₂
POL(false) = 0
POL(le₂(x₁, x₂)) = 0
POL(s₁(x₁)) = 0
POL(true) = 0

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ 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)))

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.
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

  ↳ 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))

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.

EQ₂(s₁(x), s₁(y)) -> EQ₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(EQ₂(x₁, x₂)) = 2·x₁·x₂
POL(s₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented: none

↳ 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))

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.

↳ 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))

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.

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 be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(IF_RM₃(x₁, x₂, x₃)) = 2·x₃
POL(RM₂(x₁, x₂)) = 2·x₂
POL(add₂(x₁, x₂)) = 2 + 2·x₂
POL(eq₂(x₁, x₂)) = 0
POL(false) = 0
POL(s₁(x₁)) = 0
POL(true) = 0

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ 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))

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.
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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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))

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.

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 remaining pairs can at least be oriented weakly.

MINSORT₂(add₂(n, x), y) -> IF_MINSORT₃(eq₂(n, min₁(add₂(n, x))), add₂(n, x), y)

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(IF_MINSORT₃(x₁, x₂, x₃)) = x₂ + x₂·x₃
POL(MINSORT₂(x₁, x₂)) = x₁ + x₁·x₂
POL(add₂(x₁, x₂)) = 1 + 2·x₂
POL(app₂(x₁, x₂)) = 2·x₁ + x₁·x₂ + x₂
POL(eq₂(x₁, x₂)) = 0
POL(false) = 0
POL(if_min₂(x₁, x₂)) = 0
POL(if_rm₃(x₁, x₂, x₃)) = x₃
POL(le₂(x₁, x₂)) = 0
POL(min₁(x₁)) = 0
POL(nil) = 0
POL(rm₂(x₁, x₂)) = x₂
POL(s₁(x₁)) = 0
POL(true) = 0

The following usable rules [14] were oriented:

app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
if_rm₃(true, n, add₂(m, x)) -> rm₂(n, x)
rm₂(n, add₂(m, x)) -> if_rm₃(eq₂(n, m), n, add₂(m, x))
rm₂(n, nil) -> nil
if_rm₃(false, n, add₂(m, x)) -> add₂(m, rm₂(n, x))
app₂(nil, y) -> y

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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)

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.
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.