Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/RubioSLASHselsort.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(Y)) -> 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)
min₁(cons₂(0, nil)) -> 0
min₁(cons₂(s₁(N), nil)) -> s₁(N)
min₁(cons₂(N, cons₂(M, L))) -> ifmin₂(le₂(N, M), cons₂(N, cons₂(M, L)))
ifmin₂(true, cons₂(N, cons₂(M, L))) -> min₁(cons₂(N, L))
ifmin₂(false, cons₂(N, cons₂(M, L))) -> min₁(cons₂(M, L))
replace₃(N, M, nil) -> nil
replace₃(N, M, cons₂(K, L)) -> ifrepl₄(eq₂(N, K), N, M, cons₂(K, L))
ifrepl₄(true, N, M, cons₂(K, L)) -> cons₂(M, L)
ifrepl₄(false, N, M, cons₂(K, L)) -> cons₂(K, replace₃(N, M, L))
selsort₁(nil) -> nil
selsort₁(cons₂(N, L)) -> ifselsort₂(eq₂(N, min₁(cons₂(N, L))), cons₂(N, L))
ifselsort₂(true, cons₂(N, L)) -> cons₂(N, selsort₁(L))
ifselsort₂(false, cons₂(N, L)) -> cons₂(min₁(cons₂(N, L)), selsort₁(replace₃(min₁(cons₂(N, L)), N, L)))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

eq₂(0, 0) -> true
eq₂(0, s₁(Y)) -> 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)
min₁(cons₂(0, nil)) -> 0
min₁(cons₂(s₁(N), nil)) -> s₁(N)
min₁(cons₂(N, cons₂(M, L))) -> ifmin₂(le₂(N, M), cons₂(N, cons₂(M, L)))
ifmin₂(true, cons₂(N, cons₂(M, L))) -> min₁(cons₂(N, L))
ifmin₂(false, cons₂(N, cons₂(M, L))) -> min₁(cons₂(M, L))
replace₃(N, M, nil) -> nil
replace₃(N, M, cons₂(K, L)) -> ifrepl₄(eq₂(N, K), N, M, cons₂(K, L))
ifrepl₄(true, N, M, cons₂(K, L)) -> cons₂(M, L)
ifrepl₄(false, N, M, cons₂(K, L)) -> cons₂(K, replace₃(N, M, L))
selsort₁(nil) -> nil
selsort₁(cons₂(N, L)) -> ifselsort₂(eq₂(N, min₁(cons₂(N, L))), cons₂(N, L))
ifselsort₂(true, cons₂(N, L)) -> cons₂(N, selsort₁(L))
ifselsort₂(false, cons₂(N, L)) -> cons₂(min₁(cons₂(N, L)), selsort₁(replace₃(min₁(cons₂(N, L)), N, L)))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

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

eq₂(0, 0) -> true
eq₂(0, s₁(Y)) -> 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)
min₁(cons₂(0, nil)) -> 0
min₁(cons₂(s₁(N), nil)) -> s₁(N)
min₁(cons₂(N, cons₂(M, L))) -> ifmin₂(le₂(N, M), cons₂(N, cons₂(M, L)))
ifmin₂(true, cons₂(N, cons₂(M, L))) -> min₁(cons₂(N, L))
ifmin₂(false, cons₂(N, cons₂(M, L))) -> min₁(cons₂(M, L))
replace₃(N, M, nil) -> nil
replace₃(N, M, cons₂(K, L)) -> ifrepl₄(eq₂(N, K), N, M, cons₂(K, L))
ifrepl₄(true, N, M, cons₂(K, L)) -> cons₂(M, L)
ifrepl₄(false, N, M, cons₂(K, L)) -> cons₂(K, replace₃(N, M, L))
selsort₁(nil) -> nil
selsort₁(cons₂(N, L)) -> ifselsort₂(eq₂(N, min₁(cons₂(N, L))), cons₂(N, L))
ifselsort₂(true, cons₂(N, L)) -> cons₂(N, selsort₁(L))
ifselsort₂(false, cons₂(N, L)) -> cons₂(min₁(cons₂(N, L)), selsort₁(replace₃(min₁(cons₂(N, L)), N, L)))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
min₁(cons₂(0, nil))
min₁(cons₂(s₁(x0), nil))
min₁(cons₂(x0, cons₂(x1, x2)))
ifmin₂(true, cons₂(x0, cons₂(x1, x2)))
ifmin₂(false, cons₂(x0, cons₂(x1, x2)))
replace₃(x0, x1, nil)
replace₃(x0, x1, cons₂(x2, x3))
ifrepl₄(true, x0, x1, cons₂(x2, x3))
ifrepl₄(false, x0, x1, cons₂(x2, x3))
selsort₁(nil)
selsort₁(cons₂(x0, x1))
ifselsort₂(true, cons₂(x0, x1))
ifselsort₂(false, cons₂(x0, x1))

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

MIN₁(cons₂(N, cons₂(M, L))) -> LE₂(N, M)
IFMIN₂(true, cons₂(N, cons₂(M, L))) -> MIN₁(cons₂(N, L))
IFREPL₄(false, N, M, cons₂(K, L)) -> REPLACE₃(N, M, L)
REPLACE₃(N, M, cons₂(K, L)) -> EQ₂(N, K)
REPLACE₃(N, M, cons₂(K, L)) -> IFREPL₄(eq₂(N, K), N, M, cons₂(K, L))
IFSELSORT₂(false, cons₂(N, L)) -> SELSORT₁(replace₃(min₁(cons₂(N, L)), N, L))
IFMIN₂(false, cons₂(N, cons₂(M, L))) -> MIN₁(cons₂(M, L))
SELSORT₁(cons₂(N, L)) -> EQ₂(N, min₁(cons₂(N, L)))
EQ₂(s₁(X), s₁(Y)) -> EQ₂(X, Y)
SELSORT₁(cons₂(N, L)) -> MIN₁(cons₂(N, L))
IFSELSORT₂(false, cons₂(N, L)) -> REPLACE₃(min₁(cons₂(N, L)), N, L)
LE₂(s₁(X), s₁(Y)) -> LE₂(X, Y)
IFSELSORT₂(true, cons₂(N, L)) -> SELSORT₁(L)
MIN₁(cons₂(N, cons₂(M, L))) -> IFMIN₂(le₂(N, M), cons₂(N, cons₂(M, L)))
IFSELSORT₂(false, cons₂(N, L)) -> MIN₁(cons₂(N, L))
SELSORT₁(cons₂(N, L)) -> IFSELSORT₂(eq₂(N, min₁(cons₂(N, L))), cons₂(N, L))

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(Y)) -> 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)
min₁(cons₂(0, nil)) -> 0
min₁(cons₂(s₁(N), nil)) -> s₁(N)
min₁(cons₂(N, cons₂(M, L))) -> ifmin₂(le₂(N, M), cons₂(N, cons₂(M, L)))
ifmin₂(true, cons₂(N, cons₂(M, L))) -> min₁(cons₂(N, L))
ifmin₂(false, cons₂(N, cons₂(M, L))) -> min₁(cons₂(M, L))
replace₃(N, M, nil) -> nil
replace₃(N, M, cons₂(K, L)) -> ifrepl₄(eq₂(N, K), N, M, cons₂(K, L))
ifrepl₄(true, N, M, cons₂(K, L)) -> cons₂(M, L)
ifrepl₄(false, N, M, cons₂(K, L)) -> cons₂(K, replace₃(N, M, L))
selsort₁(nil) -> nil
selsort₁(cons₂(N, L)) -> ifselsort₂(eq₂(N, min₁(cons₂(N, L))), cons₂(N, L))
ifselsort₂(true, cons₂(N, L)) -> cons₂(N, selsort₁(L))
ifselsort₂(false, cons₂(N, L)) -> cons₂(min₁(cons₂(N, L)), selsort₁(replace₃(min₁(cons₂(N, L)), N, L)))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
min₁(cons₂(0, nil))
min₁(cons₂(s₁(x0), nil))
min₁(cons₂(x0, cons₂(x1, x2)))
ifmin₂(true, cons₂(x0, cons₂(x1, x2)))
ifmin₂(false, cons₂(x0, cons₂(x1, x2)))
replace₃(x0, x1, nil)
replace₃(x0, x1, cons₂(x2, x3))
ifrepl₄(true, x0, x1, cons₂(x2, x3))
ifrepl₄(false, x0, x1, cons₂(x2, x3))
selsort₁(nil)
selsort₁(cons₂(x0, x1))
ifselsort₂(true, cons₂(x0, x1))
ifselsort₂(false, cons₂(x0, x1))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

MIN₁(cons₂(N, cons₂(M, L))) -> LE₂(N, M)
IFMIN₂(true, cons₂(N, cons₂(M, L))) -> MIN₁(cons₂(N, L))
IFREPL₄(false, N, M, cons₂(K, L)) -> REPLACE₃(N, M, L)
REPLACE₃(N, M, cons₂(K, L)) -> EQ₂(N, K)
REPLACE₃(N, M, cons₂(K, L)) -> IFREPL₄(eq₂(N, K), N, M, cons₂(K, L))
IFSELSORT₂(false, cons₂(N, L)) -> SELSORT₁(replace₃(min₁(cons₂(N, L)), N, L))
IFMIN₂(false, cons₂(N, cons₂(M, L))) -> MIN₁(cons₂(M, L))
SELSORT₁(cons₂(N, L)) -> EQ₂(N, min₁(cons₂(N, L)))
EQ₂(s₁(X), s₁(Y)) -> EQ₂(X, Y)
SELSORT₁(cons₂(N, L)) -> MIN₁(cons₂(N, L))
IFSELSORT₂(false, cons₂(N, L)) -> REPLACE₃(min₁(cons₂(N, L)), N, L)
LE₂(s₁(X), s₁(Y)) -> LE₂(X, Y)
IFSELSORT₂(true, cons₂(N, L)) -> SELSORT₁(L)
MIN₁(cons₂(N, cons₂(M, L))) -> IFMIN₂(le₂(N, M), cons₂(N, cons₂(M, L)))
IFSELSORT₂(false, cons₂(N, L)) -> MIN₁(cons₂(N, L))
SELSORT₁(cons₂(N, L)) -> IFSELSORT₂(eq₂(N, min₁(cons₂(N, L))), cons₂(N, L))

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(Y)) -> 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)
min₁(cons₂(0, nil)) -> 0
min₁(cons₂(s₁(N), nil)) -> s₁(N)
min₁(cons₂(N, cons₂(M, L))) -> ifmin₂(le₂(N, M), cons₂(N, cons₂(M, L)))
ifmin₂(true, cons₂(N, cons₂(M, L))) -> min₁(cons₂(N, L))
ifmin₂(false, cons₂(N, cons₂(M, L))) -> min₁(cons₂(M, L))
replace₃(N, M, nil) -> nil
replace₃(N, M, cons₂(K, L)) -> ifrepl₄(eq₂(N, K), N, M, cons₂(K, L))
ifrepl₄(true, N, M, cons₂(K, L)) -> cons₂(M, L)
ifrepl₄(false, N, M, cons₂(K, L)) -> cons₂(K, replace₃(N, M, L))
selsort₁(nil) -> nil
selsort₁(cons₂(N, L)) -> ifselsort₂(eq₂(N, min₁(cons₂(N, L))), cons₂(N, L))
ifselsort₂(true, cons₂(N, L)) -> cons₂(N, selsort₁(L))
ifselsort₂(false, cons₂(N, L)) -> cons₂(min₁(cons₂(N, L)), selsort₁(replace₃(min₁(cons₂(N, L)), N, L)))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
min₁(cons₂(0, nil))
min₁(cons₂(s₁(x0), nil))
min₁(cons₂(x0, cons₂(x1, x2)))
ifmin₂(true, cons₂(x0, cons₂(x1, x2)))
ifmin₂(false, cons₂(x0, cons₂(x1, x2)))
replace₃(x0, x1, nil)
replace₃(x0, x1, cons₂(x2, x3))
ifrepl₄(true, x0, x1, cons₂(x2, x3))
ifrepl₄(false, x0, x1, cons₂(x2, x3))
selsort₁(nil)
selsort₁(cons₂(x0, x1))
ifselsort₂(true, cons₂(x0, x1))
ifselsort₂(false, cons₂(x0, x1))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ 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(Y)) -> 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)
min₁(cons₂(0, nil)) -> 0
min₁(cons₂(s₁(N), nil)) -> s₁(N)
min₁(cons₂(N, cons₂(M, L))) -> ifmin₂(le₂(N, M), cons₂(N, cons₂(M, L)))
ifmin₂(true, cons₂(N, cons₂(M, L))) -> min₁(cons₂(N, L))
ifmin₂(false, cons₂(N, cons₂(M, L))) -> min₁(cons₂(M, L))
replace₃(N, M, nil) -> nil
replace₃(N, M, cons₂(K, L)) -> ifrepl₄(eq₂(N, K), N, M, cons₂(K, L))
ifrepl₄(true, N, M, cons₂(K, L)) -> cons₂(M, L)
ifrepl₄(false, N, M, cons₂(K, L)) -> cons₂(K, replace₃(N, M, L))
selsort₁(nil) -> nil
selsort₁(cons₂(N, L)) -> ifselsort₂(eq₂(N, min₁(cons₂(N, L))), cons₂(N, L))
ifselsort₂(true, cons₂(N, L)) -> cons₂(N, selsort₁(L))
ifselsort₂(false, cons₂(N, L)) -> cons₂(min₁(cons₂(N, L)), selsort₁(replace₃(min₁(cons₂(N, L)), N, L)))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
min₁(cons₂(0, nil))
min₁(cons₂(s₁(x0), nil))
min₁(cons₂(x0, cons₂(x1, x2)))
ifmin₂(true, cons₂(x0, cons₂(x1, x2)))
ifmin₂(false, cons₂(x0, cons₂(x1, x2)))
replace₃(x0, x1, nil)
replace₃(x0, x1, cons₂(x2, x3))
ifrepl₄(true, x0, x1, cons₂(x2, x3))
ifrepl₄(false, x0, x1, cons₂(x2, x3))
selsort₁(nil)
selsort₁(cons₂(x0, x1))
ifselsort₂(true, cons₂(x0, x1))
ifselsort₂(false, cons₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

LE₂(s₁(X), s₁(Y)) -> LE₂(X, Y)

Used argument filtering: LE₂(x1, x2) = x2
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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(Y)) -> 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)
min₁(cons₂(0, nil)) -> 0
min₁(cons₂(s₁(N), nil)) -> s₁(N)
min₁(cons₂(N, cons₂(M, L))) -> ifmin₂(le₂(N, M), cons₂(N, cons₂(M, L)))
ifmin₂(true, cons₂(N, cons₂(M, L))) -> min₁(cons₂(N, L))
ifmin₂(false, cons₂(N, cons₂(M, L))) -> min₁(cons₂(M, L))
replace₃(N, M, nil) -> nil
replace₃(N, M, cons₂(K, L)) -> ifrepl₄(eq₂(N, K), N, M, cons₂(K, L))
ifrepl₄(true, N, M, cons₂(K, L)) -> cons₂(M, L)
ifrepl₄(false, N, M, cons₂(K, L)) -> cons₂(K, replace₃(N, M, L))
selsort₁(nil) -> nil
selsort₁(cons₂(N, L)) -> ifselsort₂(eq₂(N, min₁(cons₂(N, L))), cons₂(N, L))
ifselsort₂(true, cons₂(N, L)) -> cons₂(N, selsort₁(L))
ifselsort₂(false, cons₂(N, L)) -> cons₂(min₁(cons₂(N, L)), selsort₁(replace₃(min₁(cons₂(N, L)), N, L)))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
min₁(cons₂(0, nil))
min₁(cons₂(s₁(x0), nil))
min₁(cons₂(x0, cons₂(x1, x2)))
ifmin₂(true, cons₂(x0, cons₂(x1, x2)))
ifmin₂(false, cons₂(x0, cons₂(x1, x2)))
replace₃(x0, x1, nil)
replace₃(x0, x1, cons₂(x2, x3))
ifrepl₄(true, x0, x1, cons₂(x2, x3))
ifrepl₄(false, x0, x1, cons₂(x2, x3))
selsort₁(nil)
selsort₁(cons₂(x0, x1))
ifselsort₂(true, cons₂(x0, x1))
ifselsort₂(false, cons₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

IFMIN₂(false, cons₂(N, cons₂(M, L))) -> MIN₁(cons₂(M, L))
IFMIN₂(true, cons₂(N, cons₂(M, L))) -> MIN₁(cons₂(N, L))
MIN₁(cons₂(N, cons₂(M, L))) -> IFMIN₂(le₂(N, M), cons₂(N, cons₂(M, L)))

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(Y)) -> 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)
min₁(cons₂(0, nil)) -> 0
min₁(cons₂(s₁(N), nil)) -> s₁(N)
min₁(cons₂(N, cons₂(M, L))) -> ifmin₂(le₂(N, M), cons₂(N, cons₂(M, L)))
ifmin₂(true, cons₂(N, cons₂(M, L))) -> min₁(cons₂(N, L))
ifmin₂(false, cons₂(N, cons₂(M, L))) -> min₁(cons₂(M, L))
replace₃(N, M, nil) -> nil
replace₃(N, M, cons₂(K, L)) -> ifrepl₄(eq₂(N, K), N, M, cons₂(K, L))
ifrepl₄(true, N, M, cons₂(K, L)) -> cons₂(M, L)
ifrepl₄(false, N, M, cons₂(K, L)) -> cons₂(K, replace₃(N, M, L))
selsort₁(nil) -> nil
selsort₁(cons₂(N, L)) -> ifselsort₂(eq₂(N, min₁(cons₂(N, L))), cons₂(N, L))
ifselsort₂(true, cons₂(N, L)) -> cons₂(N, selsort₁(L))
ifselsort₂(false, cons₂(N, L)) -> cons₂(min₁(cons₂(N, L)), selsort₁(replace₃(min₁(cons₂(N, L)), N, L)))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
min₁(cons₂(0, nil))
min₁(cons₂(s₁(x0), nil))
min₁(cons₂(x0, cons₂(x1, x2)))
ifmin₂(true, cons₂(x0, cons₂(x1, x2)))
ifmin₂(false, cons₂(x0, cons₂(x1, x2)))
replace₃(x0, x1, nil)
replace₃(x0, x1, cons₂(x2, x3))
ifrepl₄(true, x0, x1, cons₂(x2, x3))
ifrepl₄(false, x0, x1, cons₂(x2, x3))
selsort₁(nil)
selsort₁(cons₂(x0, x1))
ifselsort₂(true, cons₂(x0, x1))
ifselsort₂(false, cons₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

IFMIN₂(false, cons₂(N, cons₂(M, L))) -> MIN₁(cons₂(M, L))
IFMIN₂(true, cons₂(N, cons₂(M, L))) -> MIN₁(cons₂(N, L))

Used argument filtering: IFMIN₂(x1, x2) = x2
cons₂(x1, x2) = cons₁(x2)
MIN₁(x1) = x1
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ DependencyGraphProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

MIN₁(cons₂(N, cons₂(M, L))) -> IFMIN₂(le₂(N, M), cons₂(N, cons₂(M, L)))

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(Y)) -> 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)
min₁(cons₂(0, nil)) -> 0
min₁(cons₂(s₁(N), nil)) -> s₁(N)
min₁(cons₂(N, cons₂(M, L))) -> ifmin₂(le₂(N, M), cons₂(N, cons₂(M, L)))
ifmin₂(true, cons₂(N, cons₂(M, L))) -> min₁(cons₂(N, L))
ifmin₂(false, cons₂(N, cons₂(M, L))) -> min₁(cons₂(M, L))
replace₃(N, M, nil) -> nil
replace₃(N, M, cons₂(K, L)) -> ifrepl₄(eq₂(N, K), N, M, cons₂(K, L))
ifrepl₄(true, N, M, cons₂(K, L)) -> cons₂(M, L)
ifrepl₄(false, N, M, cons₂(K, L)) -> cons₂(K, replace₃(N, M, L))
selsort₁(nil) -> nil
selsort₁(cons₂(N, L)) -> ifselsort₂(eq₂(N, min₁(cons₂(N, L))), cons₂(N, L))
ifselsort₂(true, cons₂(N, L)) -> cons₂(N, selsort₁(L))
ifselsort₂(false, cons₂(N, L)) -> cons₂(min₁(cons₂(N, L)), selsort₁(replace₃(min₁(cons₂(N, L)), N, L)))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
min₁(cons₂(0, nil))
min₁(cons₂(s₁(x0), nil))
min₁(cons₂(x0, cons₂(x1, x2)))
ifmin₂(true, cons₂(x0, cons₂(x1, x2)))
ifmin₂(false, cons₂(x0, cons₂(x1, x2)))
replace₃(x0, x1, nil)
replace₃(x0, x1, cons₂(x2, x3))
ifrepl₄(true, x0, x1, cons₂(x2, x3))
ifrepl₄(false, x0, x1, cons₂(x2, x3))
selsort₁(nil)
selsort₁(cons₂(x0, x1))
ifselsort₂(true, cons₂(x0, x1))
ifselsort₂(false, cons₂(x0, x1))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ 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(Y)) -> 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)
min₁(cons₂(0, nil)) -> 0
min₁(cons₂(s₁(N), nil)) -> s₁(N)
min₁(cons₂(N, cons₂(M, L))) -> ifmin₂(le₂(N, M), cons₂(N, cons₂(M, L)))
ifmin₂(true, cons₂(N, cons₂(M, L))) -> min₁(cons₂(N, L))
ifmin₂(false, cons₂(N, cons₂(M, L))) -> min₁(cons₂(M, L))
replace₃(N, M, nil) -> nil
replace₃(N, M, cons₂(K, L)) -> ifrepl₄(eq₂(N, K), N, M, cons₂(K, L))
ifrepl₄(true, N, M, cons₂(K, L)) -> cons₂(M, L)
ifrepl₄(false, N, M, cons₂(K, L)) -> cons₂(K, replace₃(N, M, L))
selsort₁(nil) -> nil
selsort₁(cons₂(N, L)) -> ifselsort₂(eq₂(N, min₁(cons₂(N, L))), cons₂(N, L))
ifselsort₂(true, cons₂(N, L)) -> cons₂(N, selsort₁(L))
ifselsort₂(false, cons₂(N, L)) -> cons₂(min₁(cons₂(N, L)), selsort₁(replace₃(min₁(cons₂(N, L)), N, L)))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
min₁(cons₂(0, nil))
min₁(cons₂(s₁(x0), nil))
min₁(cons₂(x0, cons₂(x1, x2)))
ifmin₂(true, cons₂(x0, cons₂(x1, x2)))
ifmin₂(false, cons₂(x0, cons₂(x1, x2)))
replace₃(x0, x1, nil)
replace₃(x0, x1, cons₂(x2, x3))
ifrepl₄(true, x0, x1, cons₂(x2, x3))
ifrepl₄(false, x0, x1, cons₂(x2, x3))
selsort₁(nil)
selsort₁(cons₂(x0, x1))
ifselsort₂(true, cons₂(x0, x1))
ifselsort₂(false, cons₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

EQ₂(s₁(X), s₁(Y)) -> EQ₂(X, Y)

Used argument filtering: EQ₂(x1, x2) = x2
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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(Y)) -> 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)
min₁(cons₂(0, nil)) -> 0
min₁(cons₂(s₁(N), nil)) -> s₁(N)
min₁(cons₂(N, cons₂(M, L))) -> ifmin₂(le₂(N, M), cons₂(N, cons₂(M, L)))
ifmin₂(true, cons₂(N, cons₂(M, L))) -> min₁(cons₂(N, L))
ifmin₂(false, cons₂(N, cons₂(M, L))) -> min₁(cons₂(M, L))
replace₃(N, M, nil) -> nil
replace₃(N, M, cons₂(K, L)) -> ifrepl₄(eq₂(N, K), N, M, cons₂(K, L))
ifrepl₄(true, N, M, cons₂(K, L)) -> cons₂(M, L)
ifrepl₄(false, N, M, cons₂(K, L)) -> cons₂(K, replace₃(N, M, L))
selsort₁(nil) -> nil
selsort₁(cons₂(N, L)) -> ifselsort₂(eq₂(N, min₁(cons₂(N, L))), cons₂(N, L))
ifselsort₂(true, cons₂(N, L)) -> cons₂(N, selsort₁(L))
ifselsort₂(false, cons₂(N, L)) -> cons₂(min₁(cons₂(N, L)), selsort₁(replace₃(min₁(cons₂(N, L)), N, L)))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
min₁(cons₂(0, nil))
min₁(cons₂(s₁(x0), nil))
min₁(cons₂(x0, cons₂(x1, x2)))
ifmin₂(true, cons₂(x0, cons₂(x1, x2)))
ifmin₂(false, cons₂(x0, cons₂(x1, x2)))
replace₃(x0, x1, nil)
replace₃(x0, x1, cons₂(x2, x3))
ifrepl₄(true, x0, x1, cons₂(x2, x3))
ifrepl₄(false, x0, x1, cons₂(x2, x3))
selsort₁(nil)
selsort₁(cons₂(x0, x1))
ifselsort₂(true, cons₂(x0, x1))
ifselsort₂(false, cons₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

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

IFREPL₄(false, N, M, cons₂(K, L)) -> REPLACE₃(N, M, L)
REPLACE₃(N, M, cons₂(K, L)) -> IFREPL₄(eq₂(N, K), N, M, cons₂(K, L))

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(Y)) -> 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)
min₁(cons₂(0, nil)) -> 0
min₁(cons₂(s₁(N), nil)) -> s₁(N)
min₁(cons₂(N, cons₂(M, L))) -> ifmin₂(le₂(N, M), cons₂(N, cons₂(M, L)))
ifmin₂(true, cons₂(N, cons₂(M, L))) -> min₁(cons₂(N, L))
ifmin₂(false, cons₂(N, cons₂(M, L))) -> min₁(cons₂(M, L))
replace₃(N, M, nil) -> nil
replace₃(N, M, cons₂(K, L)) -> ifrepl₄(eq₂(N, K), N, M, cons₂(K, L))
ifrepl₄(true, N, M, cons₂(K, L)) -> cons₂(M, L)
ifrepl₄(false, N, M, cons₂(K, L)) -> cons₂(K, replace₃(N, M, L))
selsort₁(nil) -> nil
selsort₁(cons₂(N, L)) -> ifselsort₂(eq₂(N, min₁(cons₂(N, L))), cons₂(N, L))
ifselsort₂(true, cons₂(N, L)) -> cons₂(N, selsort₁(L))
ifselsort₂(false, cons₂(N, L)) -> cons₂(min₁(cons₂(N, L)), selsort₁(replace₃(min₁(cons₂(N, L)), N, L)))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
min₁(cons₂(0, nil))
min₁(cons₂(s₁(x0), nil))
min₁(cons₂(x0, cons₂(x1, x2)))
ifmin₂(true, cons₂(x0, cons₂(x1, x2)))
ifmin₂(false, cons₂(x0, cons₂(x1, x2)))
replace₃(x0, x1, nil)
replace₃(x0, x1, cons₂(x2, x3))
ifrepl₄(true, x0, x1, cons₂(x2, x3))
ifrepl₄(false, x0, x1, cons₂(x2, x3))
selsort₁(nil)
selsort₁(cons₂(x0, x1))
ifselsort₂(true, cons₂(x0, x1))
ifselsort₂(false, cons₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

IFREPL₄(false, N, M, cons₂(K, L)) -> REPLACE₃(N, M, L)

Used argument filtering: IFREPL₄(x1, x2, x3, x4) = x4
cons₂(x1, x2) = cons₁(x2)
REPLACE₃(x1, x2, x3) = x3
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ DependencyGraphProof

              ↳ QDP

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

REPLACE₃(N, M, cons₂(K, L)) -> IFREPL₄(eq₂(N, K), N, M, cons₂(K, L))

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(Y)) -> 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)
min₁(cons₂(0, nil)) -> 0
min₁(cons₂(s₁(N), nil)) -> s₁(N)
min₁(cons₂(N, cons₂(M, L))) -> ifmin₂(le₂(N, M), cons₂(N, cons₂(M, L)))
ifmin₂(true, cons₂(N, cons₂(M, L))) -> min₁(cons₂(N, L))
ifmin₂(false, cons₂(N, cons₂(M, L))) -> min₁(cons₂(M, L))
replace₃(N, M, nil) -> nil
replace₃(N, M, cons₂(K, L)) -> ifrepl₄(eq₂(N, K), N, M, cons₂(K, L))
ifrepl₄(true, N, M, cons₂(K, L)) -> cons₂(M, L)
ifrepl₄(false, N, M, cons₂(K, L)) -> cons₂(K, replace₃(N, M, L))
selsort₁(nil) -> nil
selsort₁(cons₂(N, L)) -> ifselsort₂(eq₂(N, min₁(cons₂(N, L))), cons₂(N, L))
ifselsort₂(true, cons₂(N, L)) -> cons₂(N, selsort₁(L))
ifselsort₂(false, cons₂(N, L)) -> cons₂(min₁(cons₂(N, L)), selsort₁(replace₃(min₁(cons₂(N, L)), N, L)))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
min₁(cons₂(0, nil))
min₁(cons₂(s₁(x0), nil))
min₁(cons₂(x0, cons₂(x1, x2)))
ifmin₂(true, cons₂(x0, cons₂(x1, x2)))
ifmin₂(false, cons₂(x0, cons₂(x1, x2)))
replace₃(x0, x1, nil)
replace₃(x0, x1, cons₂(x2, x3))
ifrepl₄(true, x0, x1, cons₂(x2, x3))
ifrepl₄(false, x0, x1, cons₂(x2, x3))
selsort₁(nil)
selsort₁(cons₂(x0, x1))
ifselsort₂(true, cons₂(x0, x1))
ifselsort₂(false, cons₂(x0, x1))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

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

IFSELSORT₂(true, cons₂(N, L)) -> SELSORT₁(L)
IFSELSORT₂(false, cons₂(N, L)) -> SELSORT₁(replace₃(min₁(cons₂(N, L)), N, L))
SELSORT₁(cons₂(N, L)) -> IFSELSORT₂(eq₂(N, min₁(cons₂(N, L))), cons₂(N, L))

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(Y)) -> 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)
min₁(cons₂(0, nil)) -> 0
min₁(cons₂(s₁(N), nil)) -> s₁(N)
min₁(cons₂(N, cons₂(M, L))) -> ifmin₂(le₂(N, M), cons₂(N, cons₂(M, L)))
ifmin₂(true, cons₂(N, cons₂(M, L))) -> min₁(cons₂(N, L))
ifmin₂(false, cons₂(N, cons₂(M, L))) -> min₁(cons₂(M, L))
replace₃(N, M, nil) -> nil
replace₃(N, M, cons₂(K, L)) -> ifrepl₄(eq₂(N, K), N, M, cons₂(K, L))
ifrepl₄(true, N, M, cons₂(K, L)) -> cons₂(M, L)
ifrepl₄(false, N, M, cons₂(K, L)) -> cons₂(K, replace₃(N, M, L))
selsort₁(nil) -> nil
selsort₁(cons₂(N, L)) -> ifselsort₂(eq₂(N, min₁(cons₂(N, L))), cons₂(N, L))
ifselsort₂(true, cons₂(N, L)) -> cons₂(N, selsort₁(L))
ifselsort₂(false, cons₂(N, L)) -> cons₂(min₁(cons₂(N, L)), selsort₁(replace₃(min₁(cons₂(N, L)), N, L)))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
min₁(cons₂(0, nil))
min₁(cons₂(s₁(x0), nil))
min₁(cons₂(x0, cons₂(x1, x2)))
ifmin₂(true, cons₂(x0, cons₂(x1, x2)))
ifmin₂(false, cons₂(x0, cons₂(x1, x2)))
replace₃(x0, x1, nil)
replace₃(x0, x1, cons₂(x2, x3))
ifrepl₄(true, x0, x1, cons₂(x2, x3))
ifrepl₄(false, x0, x1, cons₂(x2, x3))
selsort₁(nil)
selsort₁(cons₂(x0, x1))
ifselsort₂(true, cons₂(x0, x1))
ifselsort₂(false, cons₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

IFSELSORT₂(true, cons₂(N, L)) -> SELSORT₁(L)
IFSELSORT₂(false, cons₂(N, L)) -> SELSORT₁(replace₃(min₁(cons₂(N, L)), N, L))

Used argument filtering: IFSELSORT₂(x1, x2) = x2
cons₂(x1, x2) = cons₁(x2)
SELSORT₁(x1) = x1
replace₃(x1, x2, x3) = x3
nil = nil
ifrepl₄(x1, x2, x3, x4) = x4
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ DependencyGraphProof

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

SELSORT₁(cons₂(N, L)) -> IFSELSORT₂(eq₂(N, min₁(cons₂(N, L))), cons₂(N, L))

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(Y)) -> 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)
min₁(cons₂(0, nil)) -> 0
min₁(cons₂(s₁(N), nil)) -> s₁(N)
min₁(cons₂(N, cons₂(M, L))) -> ifmin₂(le₂(N, M), cons₂(N, cons₂(M, L)))
ifmin₂(true, cons₂(N, cons₂(M, L))) -> min₁(cons₂(N, L))
ifmin₂(false, cons₂(N, cons₂(M, L))) -> min₁(cons₂(M, L))
replace₃(N, M, nil) -> nil
replace₃(N, M, cons₂(K, L)) -> ifrepl₄(eq₂(N, K), N, M, cons₂(K, L))
ifrepl₄(true, N, M, cons₂(K, L)) -> cons₂(M, L)
ifrepl₄(false, N, M, cons₂(K, L)) -> cons₂(K, replace₃(N, M, L))
selsort₁(nil) -> nil
selsort₁(cons₂(N, L)) -> ifselsort₂(eq₂(N, min₁(cons₂(N, L))), cons₂(N, L))
ifselsort₂(true, cons₂(N, L)) -> cons₂(N, selsort₁(L))
ifselsort₂(false, cons₂(N, L)) -> cons₂(min₁(cons₂(N, L)), selsort₁(replace₃(min₁(cons₂(N, L)), N, L)))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
min₁(cons₂(0, nil))
min₁(cons₂(s₁(x0), nil))
min₁(cons₂(x0, cons₂(x1, x2)))
ifmin₂(true, cons₂(x0, cons₂(x1, x2)))
ifmin₂(false, cons₂(x0, cons₂(x1, x2)))
replace₃(x0, x1, nil)
replace₃(x0, x1, cons₂(x2, x3))
ifrepl₄(true, x0, x1, cons₂(x2, x3))
ifrepl₄(false, x0, x1, cons₂(x2, x3))
selsort₁(nil)
selsort₁(cons₂(x0, x1))
ifselsort₂(true, cons₂(x0, x1))
ifselsort₂(false, cons₂(x0, x1))

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