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(m)) -> false
eq2(s1(n), 0) -> false
eq2(s1(n), s1(m)) -> eq2(n, m)
le2(0, m) -> true
le2(s1(n), 0) -> false
le2(s1(n), s1(m)) -> le2(n, m)
min1(cons2(0, nil)) -> 0
min1(cons2(s1(n), nil)) -> s1(n)
min1(cons2(n, cons2(m, x))) -> if_min2(le2(n, m), cons2(n, cons2(m, x)))
if_min2(true, cons2(n, cons2(m, x))) -> min1(cons2(n, x))
if_min2(false, cons2(n, cons2(m, x))) -> min1(cons2(m, x))
replace3(n, m, nil) -> nil
replace3(n, m, cons2(k, x)) -> if_replace4(eq2(n, k), n, m, cons2(k, x))
if_replace4(true, n, m, cons2(k, x)) -> cons2(m, x)
if_replace4(false, n, m, cons2(k, x)) -> cons2(k, replace3(n, m, x))
sort1(nil) -> nil
sort1(cons2(n, x)) -> cons2(min1(cons2(n, x)), sort1(replace3(min1(cons2(n, x)), n, x)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

eq2(0, 0) -> true
eq2(0, s1(m)) -> false
eq2(s1(n), 0) -> false
eq2(s1(n), s1(m)) -> eq2(n, m)
le2(0, m) -> true
le2(s1(n), 0) -> false
le2(s1(n), s1(m)) -> le2(n, m)
min1(cons2(0, nil)) -> 0
min1(cons2(s1(n), nil)) -> s1(n)
min1(cons2(n, cons2(m, x))) -> if_min2(le2(n, m), cons2(n, cons2(m, x)))
if_min2(true, cons2(n, cons2(m, x))) -> min1(cons2(n, x))
if_min2(false, cons2(n, cons2(m, x))) -> min1(cons2(m, x))
replace3(n, m, nil) -> nil
replace3(n, m, cons2(k, x)) -> if_replace4(eq2(n, k), n, m, cons2(k, x))
if_replace4(true, n, m, cons2(k, x)) -> cons2(m, x)
if_replace4(false, n, m, cons2(k, x)) -> cons2(k, replace3(n, m, x))
sort1(nil) -> nil
sort1(cons2(n, x)) -> cons2(min1(cons2(n, x)), sort1(replace3(min1(cons2(n, x)), n, x)))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

REPLACE3(n, m, cons2(k, x)) -> IF_REPLACE4(eq2(n, k), n, m, cons2(k, x))
EQ2(s1(n), s1(m)) -> EQ2(n, m)
SORT1(cons2(n, x)) -> REPLACE3(min1(cons2(n, x)), n, x)
MIN1(cons2(n, cons2(m, x))) -> LE2(n, m)
LE2(s1(n), s1(m)) -> LE2(n, m)
SORT1(cons2(n, x)) -> MIN1(cons2(n, x))
REPLACE3(n, m, cons2(k, x)) -> EQ2(n, k)
SORT1(cons2(n, x)) -> SORT1(replace3(min1(cons2(n, x)), n, x))
IF_MIN2(true, cons2(n, cons2(m, x))) -> MIN1(cons2(n, x))
MIN1(cons2(n, cons2(m, x))) -> IF_MIN2(le2(n, m), cons2(n, cons2(m, x)))
IF_MIN2(false, cons2(n, cons2(m, x))) -> MIN1(cons2(m, x))
IF_REPLACE4(false, n, m, cons2(k, x)) -> REPLACE3(n, m, x)

The TRS R consists of the following rules:

eq2(0, 0) -> true
eq2(0, s1(m)) -> false
eq2(s1(n), 0) -> false
eq2(s1(n), s1(m)) -> eq2(n, m)
le2(0, m) -> true
le2(s1(n), 0) -> false
le2(s1(n), s1(m)) -> le2(n, m)
min1(cons2(0, nil)) -> 0
min1(cons2(s1(n), nil)) -> s1(n)
min1(cons2(n, cons2(m, x))) -> if_min2(le2(n, m), cons2(n, cons2(m, x)))
if_min2(true, cons2(n, cons2(m, x))) -> min1(cons2(n, x))
if_min2(false, cons2(n, cons2(m, x))) -> min1(cons2(m, x))
replace3(n, m, nil) -> nil
replace3(n, m, cons2(k, x)) -> if_replace4(eq2(n, k), n, m, cons2(k, x))
if_replace4(true, n, m, cons2(k, x)) -> cons2(m, x)
if_replace4(false, n, m, cons2(k, x)) -> cons2(k, replace3(n, m, x))
sort1(nil) -> nil
sort1(cons2(n, x)) -> cons2(min1(cons2(n, x)), sort1(replace3(min1(cons2(n, x)), n, x)))

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:

REPLACE3(n, m, cons2(k, x)) -> IF_REPLACE4(eq2(n, k), n, m, cons2(k, x))
EQ2(s1(n), s1(m)) -> EQ2(n, m)
SORT1(cons2(n, x)) -> REPLACE3(min1(cons2(n, x)), n, x)
MIN1(cons2(n, cons2(m, x))) -> LE2(n, m)
LE2(s1(n), s1(m)) -> LE2(n, m)
SORT1(cons2(n, x)) -> MIN1(cons2(n, x))
REPLACE3(n, m, cons2(k, x)) -> EQ2(n, k)
SORT1(cons2(n, x)) -> SORT1(replace3(min1(cons2(n, x)), n, x))
IF_MIN2(true, cons2(n, cons2(m, x))) -> MIN1(cons2(n, x))
MIN1(cons2(n, cons2(m, x))) -> IF_MIN2(le2(n, m), cons2(n, cons2(m, x)))
IF_MIN2(false, cons2(n, cons2(m, x))) -> MIN1(cons2(m, x))
IF_REPLACE4(false, n, m, cons2(k, x)) -> REPLACE3(n, m, x)

The TRS R consists of the following rules:

eq2(0, 0) -> true
eq2(0, s1(m)) -> false
eq2(s1(n), 0) -> false
eq2(s1(n), s1(m)) -> eq2(n, m)
le2(0, m) -> true
le2(s1(n), 0) -> false
le2(s1(n), s1(m)) -> le2(n, m)
min1(cons2(0, nil)) -> 0
min1(cons2(s1(n), nil)) -> s1(n)
min1(cons2(n, cons2(m, x))) -> if_min2(le2(n, m), cons2(n, cons2(m, x)))
if_min2(true, cons2(n, cons2(m, x))) -> min1(cons2(n, x))
if_min2(false, cons2(n, cons2(m, x))) -> min1(cons2(m, x))
replace3(n, m, nil) -> nil
replace3(n, m, cons2(k, x)) -> if_replace4(eq2(n, k), n, m, cons2(k, x))
if_replace4(true, n, m, cons2(k, x)) -> cons2(m, x)
if_replace4(false, n, m, cons2(k, x)) -> cons2(k, replace3(n, m, x))
sort1(nil) -> nil
sort1(cons2(n, x)) -> cons2(min1(cons2(n, x)), sort1(replace3(min1(cons2(n, x)), n, x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 5 SCCs with 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

LE2(s1(n), s1(m)) -> LE2(n, m)

The TRS R consists of the following rules:

eq2(0, 0) -> true
eq2(0, s1(m)) -> false
eq2(s1(n), 0) -> false
eq2(s1(n), s1(m)) -> eq2(n, m)
le2(0, m) -> true
le2(s1(n), 0) -> false
le2(s1(n), s1(m)) -> le2(n, m)
min1(cons2(0, nil)) -> 0
min1(cons2(s1(n), nil)) -> s1(n)
min1(cons2(n, cons2(m, x))) -> if_min2(le2(n, m), cons2(n, cons2(m, x)))
if_min2(true, cons2(n, cons2(m, x))) -> min1(cons2(n, x))
if_min2(false, cons2(n, cons2(m, x))) -> min1(cons2(m, x))
replace3(n, m, nil) -> nil
replace3(n, m, cons2(k, x)) -> if_replace4(eq2(n, k), n, m, cons2(k, x))
if_replace4(true, n, m, cons2(k, x)) -> cons2(m, x)
if_replace4(false, n, m, cons2(k, x)) -> cons2(k, replace3(n, m, x))
sort1(nil) -> nil
sort1(cons2(n, x)) -> cons2(min1(cons2(n, x)), sort1(replace3(min1(cons2(n, x)), n, x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


LE2(s1(n), s1(m)) -> LE2(n, m)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(LE2(x1, x2)) = x1·x2   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] 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:

eq2(0, 0) -> true
eq2(0, s1(m)) -> false
eq2(s1(n), 0) -> false
eq2(s1(n), s1(m)) -> eq2(n, m)
le2(0, m) -> true
le2(s1(n), 0) -> false
le2(s1(n), s1(m)) -> le2(n, m)
min1(cons2(0, nil)) -> 0
min1(cons2(s1(n), nil)) -> s1(n)
min1(cons2(n, cons2(m, x))) -> if_min2(le2(n, m), cons2(n, cons2(m, x)))
if_min2(true, cons2(n, cons2(m, x))) -> min1(cons2(n, x))
if_min2(false, cons2(n, cons2(m, x))) -> min1(cons2(m, x))
replace3(n, m, nil) -> nil
replace3(n, m, cons2(k, x)) -> if_replace4(eq2(n, k), n, m, cons2(k, x))
if_replace4(true, n, m, cons2(k, x)) -> cons2(m, x)
if_replace4(false, n, m, cons2(k, x)) -> cons2(k, replace3(n, m, x))
sort1(nil) -> nil
sort1(cons2(n, x)) -> cons2(min1(cons2(n, x)), sort1(replace3(min1(cons2(n, x)), n, x)))

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

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

IF_MIN2(true, cons2(n, cons2(m, x))) -> MIN1(cons2(n, x))
MIN1(cons2(n, cons2(m, x))) -> IF_MIN2(le2(n, m), cons2(n, cons2(m, x)))
IF_MIN2(false, cons2(n, cons2(m, x))) -> MIN1(cons2(m, x))

The TRS R consists of the following rules:

eq2(0, 0) -> true
eq2(0, s1(m)) -> false
eq2(s1(n), 0) -> false
eq2(s1(n), s1(m)) -> eq2(n, m)
le2(0, m) -> true
le2(s1(n), 0) -> false
le2(s1(n), s1(m)) -> le2(n, m)
min1(cons2(0, nil)) -> 0
min1(cons2(s1(n), nil)) -> s1(n)
min1(cons2(n, cons2(m, x))) -> if_min2(le2(n, m), cons2(n, cons2(m, x)))
if_min2(true, cons2(n, cons2(m, x))) -> min1(cons2(n, x))
if_min2(false, cons2(n, cons2(m, x))) -> min1(cons2(m, x))
replace3(n, m, nil) -> nil
replace3(n, m, cons2(k, x)) -> if_replace4(eq2(n, k), n, m, cons2(k, x))
if_replace4(true, n, m, cons2(k, x)) -> cons2(m, x)
if_replace4(false, n, m, cons2(k, x)) -> cons2(k, replace3(n, m, x))
sort1(nil) -> nil
sort1(cons2(n, x)) -> cons2(min1(cons2(n, x)), sort1(replace3(min1(cons2(n, x)), n, x)))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

EQ2(s1(n), s1(m)) -> EQ2(n, m)

The TRS R consists of the following rules:

eq2(0, 0) -> true
eq2(0, s1(m)) -> false
eq2(s1(n), 0) -> false
eq2(s1(n), s1(m)) -> eq2(n, m)
le2(0, m) -> true
le2(s1(n), 0) -> false
le2(s1(n), s1(m)) -> le2(n, m)
min1(cons2(0, nil)) -> 0
min1(cons2(s1(n), nil)) -> s1(n)
min1(cons2(n, cons2(m, x))) -> if_min2(le2(n, m), cons2(n, cons2(m, x)))
if_min2(true, cons2(n, cons2(m, x))) -> min1(cons2(n, x))
if_min2(false, cons2(n, cons2(m, x))) -> min1(cons2(m, x))
replace3(n, m, nil) -> nil
replace3(n, m, cons2(k, x)) -> if_replace4(eq2(n, k), n, m, cons2(k, x))
if_replace4(true, n, m, cons2(k, x)) -> cons2(m, x)
if_replace4(false, n, m, cons2(k, x)) -> cons2(k, replace3(n, m, x))
sort1(nil) -> nil
sort1(cons2(n, x)) -> cons2(min1(cons2(n, x)), sort1(replace3(min1(cons2(n, x)), n, x)))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

REPLACE3(n, m, cons2(k, x)) -> IF_REPLACE4(eq2(n, k), n, m, cons2(k, x))
IF_REPLACE4(false, n, m, cons2(k, x)) -> REPLACE3(n, m, x)

The TRS R consists of the following rules:

eq2(0, 0) -> true
eq2(0, s1(m)) -> false
eq2(s1(n), 0) -> false
eq2(s1(n), s1(m)) -> eq2(n, m)
le2(0, m) -> true
le2(s1(n), 0) -> false
le2(s1(n), s1(m)) -> le2(n, m)
min1(cons2(0, nil)) -> 0
min1(cons2(s1(n), nil)) -> s1(n)
min1(cons2(n, cons2(m, x))) -> if_min2(le2(n, m), cons2(n, cons2(m, x)))
if_min2(true, cons2(n, cons2(m, x))) -> min1(cons2(n, x))
if_min2(false, cons2(n, cons2(m, x))) -> min1(cons2(m, x))
replace3(n, m, nil) -> nil
replace3(n, m, cons2(k, x)) -> if_replace4(eq2(n, k), n, m, cons2(k, x))
if_replace4(true, n, m, cons2(k, x)) -> cons2(m, x)
if_replace4(false, n, m, cons2(k, x)) -> cons2(k, replace3(n, m, x))
sort1(nil) -> nil
sort1(cons2(n, x)) -> cons2(min1(cons2(n, x)), sort1(replace3(min1(cons2(n, x)), n, x)))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

SORT1(cons2(n, x)) -> SORT1(replace3(min1(cons2(n, x)), n, x))

The TRS R consists of the following rules:

eq2(0, 0) -> true
eq2(0, s1(m)) -> false
eq2(s1(n), 0) -> false
eq2(s1(n), s1(m)) -> eq2(n, m)
le2(0, m) -> true
le2(s1(n), 0) -> false
le2(s1(n), s1(m)) -> le2(n, m)
min1(cons2(0, nil)) -> 0
min1(cons2(s1(n), nil)) -> s1(n)
min1(cons2(n, cons2(m, x))) -> if_min2(le2(n, m), cons2(n, cons2(m, x)))
if_min2(true, cons2(n, cons2(m, x))) -> min1(cons2(n, x))
if_min2(false, cons2(n, cons2(m, x))) -> min1(cons2(m, x))
replace3(n, m, nil) -> nil
replace3(n, m, cons2(k, x)) -> if_replace4(eq2(n, k), n, m, cons2(k, x))
if_replace4(true, n, m, cons2(k, x)) -> cons2(m, x)
if_replace4(false, n, m, cons2(k, x)) -> cons2(k, replace3(n, m, x))
sort1(nil) -> nil
sort1(cons2(n, x)) -> cons2(min1(cons2(n, x)), sort1(replace3(min1(cons2(n, x)), n, x)))

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