Termination w.r.t. Q proof of /tmp/tmpRjWEew/selsort.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:

eq(0, 0) → true
eq(0, s(Y)) → 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)
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
selsort(nil) → nil
selsort(cons(N, L)) → ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) → cons(N, selsort(L))
ifselsort(false, cons(N, L)) → cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

eq(0, 0) → true
eq(0, s(Y)) → 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)
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
selsort(nil) → nil
selsort(cons(N, L)) → ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) → cons(N, selsort(L))
ifselsort(false, cons(N, L)) → cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

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:

IFSELSORT(false, cons(N, L)) → REPLACE(min(cons(N, L)), N, L)
SELSORT(cons(N, L)) → EQ(N, min(cons(N, L)))
IFSELSORT(false, cons(N, L)) → SELSORT(replace(min(cons(N, L)), N, L))
IFMIN(false, cons(N, cons(M, L))) → MIN(cons(M, L))
LE(s(X), s(Y)) → LE(X, Y)
IFSELSORT(true, cons(N, L)) → SELSORT(L)
SELSORT(cons(N, L)) → IFSELSORT(eq(N, min(cons(N, L))), cons(N, L))
IFMIN(true, cons(N, cons(M, L))) → MIN(cons(N, L))
IFSELSORT(false, cons(N, L)) → MIN(cons(N, L))
EQ(s(X), s(Y)) → EQ(X, Y)
IFREPL(false, N, M, cons(K, L)) → REPLACE(N, M, L)
MIN(cons(N, cons(M, L))) → IFMIN(le(N, M), cons(N, cons(M, L)))
REPLACE(N, M, cons(K, L)) → IFREPL(eq(N, K), N, M, cons(K, L))
SELSORT(cons(N, L)) → MIN(cons(N, L))
MIN(cons(N, cons(M, L))) → LE(N, M)
REPLACE(N, M, cons(K, L)) → EQ(N, K)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(Y)) → 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)
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
selsort(nil) → nil
selsort(cons(N, L)) → ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) → cons(N, selsort(L))
ifselsort(false, cons(N, L)) → cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

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:

IFSELSORT(false, cons(N, L)) → REPLACE(min(cons(N, L)), N, L)
SELSORT(cons(N, L)) → EQ(N, min(cons(N, L)))
IFSELSORT(false, cons(N, L)) → SELSORT(replace(min(cons(N, L)), N, L))
IFMIN(false, cons(N, cons(M, L))) → MIN(cons(M, L))
LE(s(X), s(Y)) → LE(X, Y)
IFSELSORT(true, cons(N, L)) → SELSORT(L)
SELSORT(cons(N, L)) → IFSELSORT(eq(N, min(cons(N, L))), cons(N, L))
IFMIN(true, cons(N, cons(M, L))) → MIN(cons(N, L))
IFSELSORT(false, cons(N, L)) → MIN(cons(N, L))
EQ(s(X), s(Y)) → EQ(X, Y)
IFREPL(false, N, M, cons(K, L)) → REPLACE(N, M, L)
MIN(cons(N, cons(M, L))) → IFMIN(le(N, M), cons(N, cons(M, L)))
REPLACE(N, M, cons(K, L)) → IFREPL(eq(N, K), N, M, cons(K, L))
SELSORT(cons(N, L)) → MIN(cons(N, L))
MIN(cons(N, cons(M, L))) → LE(N, M)
REPLACE(N, M, cons(K, L)) → EQ(N, K)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(Y)) → 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)
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
selsort(nil) → nil
selsort(cons(N, L)) → ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) → cons(N, selsort(L))
ifselsort(false, cons(N, L)) → cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 5 SCCs with 6 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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(Y)) → 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)
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
selsort(nil) → nil
selsort(cons(N, L)) → ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) → cons(N, selsort(L))
ifselsort(false, cons(N, L)) → cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

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)) = 4 + (2)x_1
POL(LE(x₁, x₂)) = (3)x_2

The value of delta used in the strict ordering is 12.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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(Y)) → 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)
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
selsort(nil) → nil
selsort(cons(N, L)) → ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) → cons(N, selsort(L))
ifselsort(false, cons(N, L)) → cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

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

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

IFMIN(true, cons(N, cons(M, L))) → MIN(cons(N, L))
MIN(cons(N, cons(M, L))) → IFMIN(le(N, M), cons(N, cons(M, L)))
IFMIN(false, cons(N, cons(M, L))) → MIN(cons(M, L))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(Y)) → 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)
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
selsort(nil) → nil
selsort(cons(N, L)) → ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) → cons(N, selsort(L))
ifselsort(false, cons(N, L)) → cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

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.

IFMIN(true, cons(N, cons(M, L))) → MIN(cons(N, L))
IFMIN(false, cons(N, cons(M, L))) → MIN(cons(M, L))

The remaining pairs can at least be oriented weakly.

MIN(cons(N, cons(M, L))) → IFMIN(le(N, M), cons(N, cons(M, L)))

Used ordering: Polynomial interpretation [25,35]:

POL(cons(x₁, x₂)) = 1 + x_2
POL(le(x₁, x₂)) = 0
POL(true) = 0
POL(MIN(x₁)) = x_1
POL(false) = 0
POL(s(x₁)) = 2 + (2)x_1
POL(0) = 3
POL(IFMIN(x₁, x₂)) = x_2

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

MIN(cons(N, cons(M, L))) → IFMIN(le(N, M), cons(N, cons(M, L)))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(Y)) → 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)
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
selsort(nil) → nil
selsort(cons(N, L)) → ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) → cons(N, selsort(L))
ifselsort(false, cons(N, L)) → cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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(Y)) → 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)
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
selsort(nil) → nil
selsort(cons(N, L)) → ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) → cons(N, selsort(L))
ifselsort(false, cons(N, L)) → cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

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)) = (3)x_2
POL(s(x₁)) = 1 + (4)x_1

The value of delta used in the strict ordering is 3.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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(Y)) → 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)
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
selsort(nil) → nil
selsort(cons(N, L)) → ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) → cons(N, selsort(L))
ifselsort(false, cons(N, L)) → cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

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

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

IFREPL(false, N, M, cons(K, L)) → REPLACE(N, M, L)
REPLACE(N, M, cons(K, L)) → IFREPL(eq(N, K), N, M, cons(K, L))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(Y)) → 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)
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
selsort(nil) → nil
selsort(cons(N, L)) → ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) → cons(N, selsort(L))
ifselsort(false, cons(N, L)) → cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

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.

IFREPL(false, N, M, cons(K, L)) → REPLACE(N, M, L)

The remaining pairs can at least be oriented weakly.

REPLACE(N, M, cons(K, L)) → IFREPL(eq(N, K), N, M, cons(K, L))

Used ordering: Polynomial interpretation [25,35]:

POL(IFREPL(x₁, x₂, x₃, x₄)) = x_1 + (2)x_4
POL(cons(x₁, x₂)) = (2)x_1 + (3)x_2
POL(eq(x₁, x₂)) = x_2
POL(REPLACE(x₁, x₂, x₃)) = (3)x_3
POL(true) = 3
POL(false) = 1
POL(s(x₁)) = 1 + (3)x_1
POL(0) = 4

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

eq(0, 0) → true
eq(s(X), 0) → false
eq(0, s(Y)) → false
eq(s(X), s(Y)) → eq(X, Y)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ QDP

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

REPLACE(N, M, cons(K, L)) → IFREPL(eq(N, K), N, M, cons(K, L))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(Y)) → 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)
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
selsort(nil) → nil
selsort(cons(N, L)) → ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) → cons(N, selsort(L))
ifselsort(false, cons(N, L)) → cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

SELSORT(cons(N, L)) → IFSELSORT(eq(N, min(cons(N, L))), cons(N, L))
IFSELSORT(false, cons(N, L)) → SELSORT(replace(min(cons(N, L)), N, L))
IFSELSORT(true, cons(N, L)) → SELSORT(L)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(Y)) → 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)
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
selsort(nil) → nil
selsort(cons(N, L)) → ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) → cons(N, selsort(L))
ifselsort(false, cons(N, L)) → cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

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.

SELSORT(cons(N, L)) → IFSELSORT(eq(N, min(cons(N, L))), cons(N, L))
IFSELSORT(false, cons(N, L)) → SELSORT(replace(min(cons(N, L)), N, L))
IFSELSORT(true, cons(N, L)) → SELSORT(L)

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

POL(eq(x₁, x₂)) = 0
POL(le(x₁, x₂)) = 0
POL(true) = 0
POL(0) = 1
POL(IFSELSORT(x₁, x₂)) = 2 + (4)x_2
POL(ifmin(x₁, x₂)) = 1
POL(cons(x₁, x₂)) = 4 + x_2
POL(ifrepl(x₁, x₂, x₃, x₄)) = 2 + x_2 + x_4
POL(false) = 0
POL(s(x₁)) = 0
POL(min(x₁)) = 1
POL(nil) = 2
POL(replace(x₁, x₂, x₃)) = 2 + x_1 + x_3
POL(SELSORT(x₁)) = 4 + (4)x_1

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

ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
min(cons(s(N), nil)) → s(N)
min(cons(0, nil)) → 0
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
replace(N, M, nil) → nil
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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(Y)) → 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)
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
selsort(nil) → nil
selsort(cons(N, L)) → ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) → cons(N, selsort(L))
ifselsort(false, cons(N, L)) → cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), N, L)))

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.