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:

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
le2(x1, x2)  =  le
0  =  0
true  =  true
s1(x1)  =  s
false  =  false
Used ordering: Quasi Precedence: [le, true] > false


↳ 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
eq2(x1, x2)  =  eq
0  =  0
true  =  true
s1(x1)  =  s
false  =  false
Used ordering: Quasi Precedence: [eq, false] > true


↳ 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

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.