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 Order [17,21] with Interpretation:

POL( APP₂(x₁, x₂) ) = x₁ + 3

POL( add₂(x₁, x₂) ) = 2x₂ + 1

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 Order [17,21] with Interpretation:

POL( LE₂(x₁, x₂) ) = 3x₂ + 3

POL( s₁(x₁) ) = x₁ + 3

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.

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 be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MIN₁(x₁) ) = max{0, 3x₁ - 1}

POL( true ) = 0

POL( false ) = 1

POL( add₂(x₁, x₂) ) = 2x₂ + 3

POL( 0 ) = 0

POL( s₁(x₁) ) = max{0, -3}

POL( le₂(x₁, x₂) ) = max{0, -1}

POL( IF_MIN₂(x₁, x₂) ) = 2x₂ + 1

The following usable rules [14] were oriented: none

↳ 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))

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

            ↳ 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 Order [17,21] with Interpretation:

POL( s₁(x₁) ) = x₁ + 3

POL( EQ₂(x₁, x₂) ) = 3x₂ + 3

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.

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( true ) = max{0, -3}

POL( false ) = 0

POL( add₂(x₁, x₂) ) = x₂ + 1

POL( eq₂(x₁, x₂) ) = max{0, x₁ - 3}

POL( s₁(x₁) ) = 0

POL( 0 ) = 0

POL( IF_RM₃(x₁, ..., x₃) ) = x₂ + 2x₃

POL( RM₂(x₁, x₂) ) = x₁ + 2x₂ + 2

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:

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.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 2 less nodes.

↳ 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)

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)
IF_MINSORT₃(false, add₂(n, x), y) -> MINSORT₂(x, add₂(n, y))

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( rm₂(x₁, x₂) ) = x₂

POL( eq₂(x₁, x₂) ) = max{0, -3}

POL( 0 ) = 0

POL( nil ) = max{0, -3}

POL( if_min₂(x₁, x₂) ) = 2x₂ + 1

POL( MINSORT₂(x₁, x₂) ) = 2x₁ + 2x₂ + 2

POL( true ) = 0

POL( IF_MINSORT₃(x₁, ..., x₃) ) = 2x₂ + 2x₃ + 2

POL( false ) = 2

POL( app₂(x₁, x₂) ) = x₁ + x₂

POL( if_rm₃(x₁, ..., x₃) ) = x₃

POL( add₂(x₁, x₂) ) = 3x₁ + x₂ + 1

POL( min₁(x₁) ) = x₁ + 1

POL( s₁(x₁) ) = max{0, x₁ - 2}

POL( le₂(x₁, x₂) ) = 3

The following usable rules [14] were oriented:

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

                ↳ 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(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.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( true ) = max{0, -3}

POL( IF_MINSORT₃(x₁, ..., x₃) ) = max{0, 2x₂ - 3}

POL( false ) = max{0, -3}

POL( add₂(x₁, x₂) ) = 3x₁ + x₂ + 3

POL( eq₂(x₁, x₂) ) = max{0, -3}

POL( s₁(x₁) ) = x₁ + 3

POL( 0 ) = 1

POL( min₁(x₁) ) = 2x₁

POL( le₂(x₁, x₂) ) = 0

POL( nil ) = 1

POL( if_min₂(x₁, x₂) ) = max{0, -3}

POL( MINSORT₂(x₁, x₂) ) = max{0, 2x₁ - 2}

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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.