Termination w.r.t. Q proof of /tmp/tmpaRF5UM/thiemann30.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:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(nil) → 0
min(add(n, x)) → minIter(add(n, x), add(n, x), 0)
minIter(nil, add(n, y), m) → minIter(add(n, y), add(n, y), s(m))
minIter(add(n, x), y, m) → if_min(le(n, m), x, y, m)
if_min(true, x, y, m) → m
if_min(false, x, y, m) → minIter(x, y, m)
head(add(n, x)) → n
tail(add(n, x)) → x
tail(nil) → nil
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(n, y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(nil) → 0
min(add(n, x)) → minIter(add(n, x), add(n, x), 0)
minIter(nil, add(n, y), m) → minIter(add(n, y), add(n, y), s(m))
minIter(add(n, x), y, m) → if_min(le(n, m), x, y, m)
if_min(true, x, y, m) → m
if_min(false, x, y, m) → minIter(x, y, m)
head(add(n, x)) → n
tail(add(n, x)) → x
tail(nil) → nil
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(n, y))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

IF_MINSORT(true, add(n, x), y) → RM(n, x)
MINITER(add(n, x), y, m) → IF_MIN(le(n, m), x, y, m)
RM(n, add(m, x)) → IF_RM(eq(n, m), n, add(m, x))
MIN(add(n, x)) → MINITER(add(n, x), add(n, x), 0)
IF_MIN(false, x, y, m) → MINITER(x, y, m)
IF_RM(false, n, add(m, x)) → RM(n, x)
MINSORT(add(n, x), y) → IF_MINSORT(eq(n, min(add(n, x))), add(n, x), y)
MINITER(add(n, x), y, m) → LE(n, m)
IF_MINSORT(false, add(n, x), y) → MINSORT(x, add(n, y))
IF_MINSORT(true, add(n, x), y) → APP(rm(n, x), y)
LE(s(x), s(y)) → LE(x, y)
IF_MINSORT(true, add(n, x), y) → MINSORT(app(rm(n, x), y), nil)
MINSORT(add(n, x), y) → EQ(n, min(add(n, x)))
APP(add(n, x), y) → APP(x, y)
MINSORT(add(n, x), y) → MIN(add(n, x))
MINITER(nil, add(n, y), m) → MINITER(add(n, y), add(n, y), s(m))
EQ(s(x), s(y)) → EQ(x, y)
IF_RM(true, n, add(m, x)) → RM(n, x)
RM(n, add(m, x)) → EQ(n, m)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(nil) → 0
min(add(n, x)) → minIter(add(n, x), add(n, x), 0)
minIter(nil, add(n, y), m) → minIter(add(n, y), add(n, y), s(m))
minIter(add(n, x), y, m) → if_min(le(n, m), x, y, m)
if_min(true, x, y, m) → m
if_min(false, x, y, m) → minIter(x, y, m)
head(add(n, x)) → n
tail(add(n, x)) → x
tail(nil) → nil
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(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:

IF_MINSORT(true, add(n, x), y) → RM(n, x)
MINITER(add(n, x), y, m) → IF_MIN(le(n, m), x, y, m)
RM(n, add(m, x)) → IF_RM(eq(n, m), n, add(m, x))
MIN(add(n, x)) → MINITER(add(n, x), add(n, x), 0)
IF_MIN(false, x, y, m) → MINITER(x, y, m)
IF_RM(false, n, add(m, x)) → RM(n, x)
MINSORT(add(n, x), y) → IF_MINSORT(eq(n, min(add(n, x))), add(n, x), y)
MINITER(add(n, x), y, m) → LE(n, m)
IF_MINSORT(false, add(n, x), y) → MINSORT(x, add(n, y))
IF_MINSORT(true, add(n, x), y) → APP(rm(n, x), y)
LE(s(x), s(y)) → LE(x, y)
IF_MINSORT(true, add(n, x), y) → MINSORT(app(rm(n, x), y), nil)
MINSORT(add(n, x), y) → EQ(n, min(add(n, x)))
APP(add(n, x), y) → APP(x, y)
MINSORT(add(n, x), y) → MIN(add(n, x))
MINITER(nil, add(n, y), m) → MINITER(add(n, y), add(n, y), s(m))
EQ(s(x), s(y)) → EQ(x, y)
IF_RM(true, n, add(m, x)) → RM(n, x)
RM(n, add(m, x)) → EQ(n, m)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(nil) → 0
min(add(n, x)) → minIter(add(n, x), add(n, x), 0)
minIter(nil, add(n, y), m) → minIter(add(n, y), add(n, y), s(m))
minIter(add(n, x), y, m) → if_min(le(n, m), x, y, m)
if_min(true, x, y, m) → m
if_min(false, x, y, m) → minIter(x, y, m)
head(add(n, x)) → n
tail(add(n, x)) → x
tail(nil) → nil
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(n, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 6 SCCs with 7 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:

APP(add(n, x), y) → APP(x, y)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(nil) → 0
min(add(n, x)) → minIter(add(n, x), add(n, x), 0)
minIter(nil, add(n, y), m) → minIter(add(n, y), add(n, y), s(m))
minIter(add(n, x), y, m) → if_min(le(n, m), x, y, m)
if_min(true, x, y, m) → m
if_min(false, x, y, m) → minIter(x, y, m)
head(add(n, x)) → n
tail(add(n, x)) → x
tail(nil) → nil
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(n, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

APP(add(n, x), y) → APP(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(APP(x₁, x₂)) = (2)x_1
POL(add(x₁, x₂)) = 1/4 + (7/2)x_2

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] 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:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(nil) → 0
min(add(n, x)) → minIter(add(n, x), add(n, x), 0)
minIter(nil, add(n, y), m) → minIter(add(n, y), add(n, y), s(m))
minIter(add(n, x), y, m) → if_min(le(n, m), x, y, m)
if_min(true, x, y, m) → m
if_min(false, x, y, m) → minIter(x, y, m)
head(add(n, x)) → n
tail(add(n, x)) → x
tail(nil) → nil
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(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:

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(nil) → 0
min(add(n, x)) → minIter(add(n, x), add(n, x), 0)
minIter(nil, add(n, y), m) → minIter(add(n, y), add(n, y), s(m))
minIter(add(n, x), y, m) → if_min(le(n, m), x, y, m)
if_min(true, x, y, m) → m
if_min(false, x, y, m) → minIter(x, y, m)
head(add(n, x)) → n
tail(add(n, x)) → x
tail(nil) → nil
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(n, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

LE(s(x), s(y)) → LE(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(s(x₁)) = 1/2 + (13/4)x_1
POL(LE(x₁, x₂)) = (15/4)x_2

The value of delta used in the strict ordering is 15/8.
The following usable rules [17] 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:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(nil) → 0
min(add(n, x)) → minIter(add(n, x), add(n, x), 0)
minIter(nil, add(n, y), m) → minIter(add(n, y), add(n, y), s(m))
minIter(add(n, x), y, m) → if_min(le(n, m), x, y, m)
if_min(true, x, y, m) → m
if_min(false, x, y, m) → minIter(x, y, m)
head(add(n, x)) → n
tail(add(n, x)) → x
tail(nil) → nil
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(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

          ↳ QDP

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

MINITER(add(n, x), y, m) → IF_MIN(le(n, m), x, y, m)
MINITER(nil, add(n, y), m) → MINITER(add(n, y), add(n, y), s(m))
IF_MIN(false, x, y, m) → MINITER(x, y, m)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(nil) → 0
min(add(n, x)) → minIter(add(n, x), add(n, x), 0)
minIter(nil, add(n, y), m) → minIter(add(n, y), add(n, y), s(m))
minIter(add(n, x), y, m) → if_min(le(n, m), x, y, m)
if_min(true, x, y, m) → m
if_min(false, x, y, m) → minIter(x, y, m)
head(add(n, x)) → n
tail(add(n, x)) → x
tail(nil) → nil
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(n, y))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

EQ(s(x), s(y)) → EQ(x, y)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(nil) → 0
min(add(n, x)) → minIter(add(n, x), add(n, x), 0)
minIter(nil, add(n, y), m) → minIter(add(n, y), add(n, y), s(m))
minIter(add(n, x), y, m) → if_min(le(n, m), x, y, m)
if_min(true, x, y, m) → m
if_min(false, x, y, m) → minIter(x, y, m)
head(add(n, x)) → n
tail(add(n, x)) → x
tail(nil) → nil
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(n, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

EQ(s(x), s(y)) → EQ(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(EQ(x₁, x₂)) = x_1 + (13/4)x_2
POL(s(x₁)) = 5/4 + (15/4)x_1

The value of delta used in the strict ordering is 85/16.
The following usable rules [17] 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:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(nil) → 0
min(add(n, x)) → minIter(add(n, x), add(n, x), 0)
minIter(nil, add(n, y), m) → minIter(add(n, y), add(n, y), s(m))
minIter(add(n, x), y, m) → if_min(le(n, m), x, y, m)
if_min(true, x, y, m) → m
if_min(false, x, y, m) → minIter(x, y, m)
head(add(n, x)) → n
tail(add(n, x)) → x
tail(nil) → nil
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(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:

RM(n, add(m, x)) → IF_RM(eq(n, m), n, add(m, x))
IF_RM(true, n, add(m, x)) → RM(n, x)
IF_RM(false, n, add(m, x)) → RM(n, x)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(nil) → 0
min(add(n, x)) → minIter(add(n, x), add(n, x), 0)
minIter(nil, add(n, y), m) → minIter(add(n, y), add(n, y), s(m))
minIter(add(n, x), y, m) → if_min(le(n, m), x, y, m)
if_min(true, x, y, m) → m
if_min(false, x, y, m) → minIter(x, y, m)
head(add(n, x)) → n
tail(add(n, x)) → x
tail(nil) → nil
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(n, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

RM(n, add(m, x)) → IF_RM(eq(n, m), n, add(m, x))
IF_RM(true, n, add(m, x)) → RM(n, x)
IF_RM(false, n, add(m, x)) → RM(n, x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(RM(x₁, x₂)) = 15/4 + (11/4)x_2
POL(add(x₁, x₂)) = 1/4 + (15/4)x_1 + (4)x_2
POL(eq(x₁, x₂)) = 2 + (4)x_1 + (9/4)x_2
POL(IF_RM(x₁, x₂, x₃)) = 7/2 + (2)x_3
POL(true) = 15/4
POL(false) = 11/4
POL(s(x₁)) = 7/4 + (2)x_1
POL(0) = 3

The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented: none

↳ 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:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(nil) → 0
min(add(n, x)) → minIter(add(n, x), add(n, x), 0)
minIter(nil, add(n, y), m) → minIter(add(n, y), add(n, y), s(m))
minIter(add(n, x), y, m) → if_min(le(n, m), x, y, m)
if_min(true, x, y, m) → m
if_min(false, x, y, m) → minIter(x, y, m)
head(add(n, x)) → n
tail(add(n, x)) → x
tail(nil) → nil
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(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

          ↳ QDP

            ↳ QDPOrderProof

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

IF_MINSORT(true, add(n, x), y) → MINSORT(app(rm(n, x), y), nil)
MINSORT(add(n, x), y) → IF_MINSORT(eq(n, min(add(n, x))), add(n, x), y)
IF_MINSORT(false, add(n, x), y) → MINSORT(x, add(n, y))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(nil) → 0
min(add(n, x)) → minIter(add(n, x), add(n, x), 0)
minIter(nil, add(n, y), m) → minIter(add(n, y), add(n, y), s(m))
minIter(add(n, x), y, m) → if_min(le(n, m), x, y, m)
if_min(true, x, y, m) → m
if_min(false, x, y, m) → minIter(x, y, m)
head(add(n, x)) → n
tail(add(n, x)) → x
tail(nil) → nil
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(n, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

IF_MINSORT(true, add(n, x), y) → MINSORT(app(rm(n, x), y), nil)

The remaining pairs can at least be oriented weakly.

MINSORT(add(n, x), y) → IF_MINSORT(eq(n, min(add(n, x))), add(n, x), y)
IF_MINSORT(false, add(n, x), y) → MINSORT(x, add(n, y))

Used ordering: Polynomial interpretation [25,35]:

POL(eq(x₁, x₂)) = 0
POL(MINSORT(x₁, x₂)) = 1/4 + (4)x_1 + (4)x_2
POL(le(x₁, x₂)) = 11/4 + (4)x_1 + (4)x_2
POL(true) = 0
POL(rm(x₁, x₂)) = x_2
POL(if_rm(x₁, x₂, x₃)) = x_3
POL(0) = 1/4
POL(IF_MINSORT(x₁, x₂, x₃)) = 1/4 + (4)x_2 + (4)x_3
POL(add(x₁, x₂)) = 4 + x_2
POL(false) = 1
POL(if_min(x₁, x₂, x₃, x₄)) = 3 + (1/4)x_1 + (7/2)x_2 + (2)x_3
POL(s(x₁)) = 0
POL(app(x₁, x₂)) = x_1 + x_2
POL(min(x₁)) = 0
POL(minIter(x₁, x₂, x₃)) = 1/4 + (7/4)x_1 + (4)x_2
POL(nil) = 0

The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented:

if_rm(false, n, add(m, x)) → add(m, rm(n, x))
rm(n, nil) → nil
if_rm(true, n, add(m, x)) → rm(n, x)
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, 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:

MINSORT(add(n, x), y) → IF_MINSORT(eq(n, min(add(n, x))), add(n, x), y)
IF_MINSORT(false, add(n, x), y) → MINSORT(x, add(n, y))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(nil) → 0
min(add(n, x)) → minIter(add(n, x), add(n, x), 0)
minIter(nil, add(n, y), m) → minIter(add(n, y), add(n, y), s(m))
minIter(add(n, x), y, m) → if_min(le(n, m), x, y, m)
if_min(true, x, y, m) → m
if_min(false, x, y, m) → minIter(x, y, m)
head(add(n, x)) → n
tail(add(n, x)) → x
tail(nil) → nil
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(n, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

MINSORT(add(n, x), y) → IF_MINSORT(eq(n, min(add(n, x))), add(n, x), y)
IF_MINSORT(false, add(n, x), y) → MINSORT(x, add(n, y))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(eq(x₁, x₂)) = (1/4)x_1
POL(MINSORT(x₁, x₂)) = 1 + (1/2)x_1
POL(le(x₁, x₂)) = 15/4 + (4)x_2
POL(true) = 1/4
POL(0) = 1
POL(IF_MINSORT(x₁, x₂, x₃)) = 3/4 + (4)x_1 + (1/4)x_2
POL(add(x₁, x₂)) = 1/4 + (4)x_1 + (4)x_2
POL(false) = 1/4
POL(if_min(x₁, x₂, x₃, x₄)) = 4 + (13/4)x_2 + (5/4)x_3 + (2)x_4
POL(s(x₁)) = 1 + (4)x_1
POL(min(x₁)) = 5/2 + (4)x_1
POL(minIter(x₁, x₂, x₃)) = 3 + (1/2)x_1 + (1/4)x_2 + (4)x_3
POL(nil) = 0

The value of delta used in the strict ordering is 5/16.
The following usable rules [17] were oriented:

eq(0, s(x)) → false
eq(0, 0) → true
eq(s(x), s(y)) → eq(x, y)
eq(s(x), 0) → false

↳ 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:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
min(nil) → 0
min(add(n, x)) → minIter(add(n, x), add(n, x), 0)
minIter(nil, add(n, y), m) → minIter(add(n, y), add(n, y), s(m))
minIter(add(n, x), y, m) → if_min(le(n, m), x, y, m)
if_min(true, x, y, m) → m
if_min(false, x, y, m) → minIter(x, y, m)
head(add(n, x)) → n
tail(add(n, x)) → x
tail(nil) → nil
null(nil) → true
null(add(n, x)) → false
rm(n, nil) → nil
rm(n, add(m, x)) → if_rm(eq(n, m), n, add(m, x))
if_rm(true, n, add(m, x)) → rm(n, x)
if_rm(false, n, add(m, x)) → add(m, rm(n, x))
minsort(nil, nil) → nil
minsort(add(n, x), y) → if_minsort(eq(n, min(add(n, x))), add(n, x), y)
if_minsort(true, add(n, x), y) → add(n, minsort(app(rm(n, x), y), nil))
if_minsort(false, add(n, x), y) → minsort(x, add(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.