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(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
  ↳ Non-Overlap Check

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.

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(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)))

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)))
if_min2(true, cons2(x0, cons2(x1, x2)))
if_min2(false, cons2(x0, cons2(x1, x2)))
replace3(x0, x1, nil)
replace3(x0, x1, cons2(x2, x3))
if_replace4(true, x0, x1, cons2(x2, x3))
if_replace4(false, x0, x1, cons2(x2, x3))
sort1(nil)
sort1(cons2(x0, x1))


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)))

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)))
if_min2(true, cons2(x0, cons2(x1, x2)))
if_min2(false, cons2(x0, cons2(x1, x2)))
replace3(x0, x1, nil)
replace3(x0, x1, cons2(x2, x3))
if_replace4(true, x0, x1, cons2(x2, x3))
if_replace4(false, x0, x1, cons2(x2, x3))
sort1(nil)
sort1(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:

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)))

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)))
if_min2(true, cons2(x0, cons2(x1, x2)))
if_min2(false, cons2(x0, cons2(x1, x2)))
replace3(x0, x1, nil)
replace3(x0, x1, cons2(x2, x3))
if_replace4(true, x0, x1, cons2(x2, x3))
if_replace4(false, x0, x1, cons2(x2, x3))
sort1(nil)
sort1(cons2(x0, x1))

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
  ↳ Non-Overlap Check
    ↳ 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)))

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)))
if_min2(true, cons2(x0, cons2(x1, x2)))
if_min2(false, cons2(x0, cons2(x1, x2)))
replace3(x0, x1, nil)
replace3(x0, x1, cons2(x2, x3))
if_replace4(true, x0, x1, cons2(x2, x3))
if_replace4(false, x0, x1, cons2(x2, x3))
sort1(nil)
sort1(cons2(x0, x1))

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


The following pairs can be strictly oriented and are deleted.


LE2(s1(n), s1(m)) -> LE2(n, m)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
LE2(x1, x2)  =  LE1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)))

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)))
if_min2(true, cons2(x0, cons2(x1, x2)))
if_min2(false, cons2(x0, cons2(x1, x2)))
replace3(x0, x1, nil)
replace3(x0, x1, cons2(x2, x3))
if_replace4(true, x0, x1, cons2(x2, x3))
if_replace4(false, x0, x1, cons2(x2, x3))
sort1(nil)
sort1(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
                ↳ QDPOrderProof
              ↳ 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)))

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)))
if_min2(true, cons2(x0, cons2(x1, x2)))
if_min2(false, cons2(x0, cons2(x1, x2)))
replace3(x0, x1, nil)
replace3(x0, x1, cons2(x2, x3))
if_replace4(true, x0, x1, cons2(x2, x3))
if_replace4(false, x0, x1, cons2(x2, x3))
sort1(nil)
sort1(cons2(x0, x1))

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


The following pairs can be strictly oriented and are deleted.


IF_MIN2(true, cons2(n, cons2(m, x))) -> MIN1(cons2(n, x))
IF_MIN2(false, cons2(n, cons2(m, x))) -> MIN1(cons2(m, x))
The remaining pairs can at least by weakly be oriented.

MIN1(cons2(n, cons2(m, x))) -> IF_MIN2(le2(n, m), cons2(n, cons2(m, x)))
Used ordering: Combined order from the following AFS and order.
IF_MIN2(x1, x2)  =  x2
true  =  true
cons2(x1, x2)  =  cons1(x2)
MIN1(x1)  =  x1
le2(x1, x2)  =  le
false  =  false
0  =  0
s1(x1)  =  s

Lexicographic Path Order [19].
Precedence:
[cons1, le] > [true, false, 0, s]


The following usable rules [14] were oriented:

le2(0, m) -> true
le2(s1(n), 0) -> false
le2(s1(n), s1(m)) -> le2(n, m)



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ DependencyGraphProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

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

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)))
if_min2(true, cons2(x0, cons2(x1, x2)))
if_min2(false, cons2(x0, cons2(x1, x2)))
replace3(x0, x1, nil)
replace3(x0, x1, cons2(x2, x3))
if_replace4(true, x0, x1, cons2(x2, x3))
if_replace4(false, x0, x1, cons2(x2, x3))
sort1(nil)
sort1(cons2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
QDP
                ↳ QDPOrderProof
              ↳ 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)))

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)))
if_min2(true, cons2(x0, cons2(x1, x2)))
if_min2(false, cons2(x0, cons2(x1, x2)))
replace3(x0, x1, nil)
replace3(x0, x1, cons2(x2, x3))
if_replace4(true, x0, x1, cons2(x2, x3))
if_replace4(false, x0, x1, cons2(x2, x3))
sort1(nil)
sort1(cons2(x0, x1))

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


The following pairs can be strictly oriented and are deleted.


EQ2(s1(n), s1(m)) -> EQ2(n, m)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
EQ2(x1, x2)  =  EQ1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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:

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)))

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)))
if_min2(true, cons2(x0, cons2(x1, x2)))
if_min2(false, cons2(x0, cons2(x1, x2)))
replace3(x0, x1, nil)
replace3(x0, x1, cons2(x2, x3))
if_replace4(true, x0, x1, cons2(x2, x3))
if_replace4(false, x0, x1, cons2(x2, x3))
sort1(nil)
sort1(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
                ↳ QDPOrderProof
              ↳ 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)))

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)))
if_min2(true, cons2(x0, cons2(x1, x2)))
if_min2(false, cons2(x0, cons2(x1, x2)))
replace3(x0, x1, nil)
replace3(x0, x1, cons2(x2, x3))
if_replace4(true, x0, x1, cons2(x2, x3))
if_replace4(false, x0, x1, cons2(x2, x3))
sort1(nil)
sort1(cons2(x0, x1))

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


The following pairs can be strictly oriented and are deleted.


REPLACE3(n, m, cons2(k, x)) -> IF_REPLACE4(eq2(n, k), n, m, cons2(k, x))
The remaining pairs can at least by weakly be oriented.

IF_REPLACE4(false, n, m, cons2(k, x)) -> REPLACE3(n, m, x)
Used ordering: Combined order from the following AFS and order.
REPLACE3(x1, x2, x3)  =  REPLACE1(x3)
cons2(x1, x2)  =  cons1(x2)
IF_REPLACE4(x1, x2, x3, x4)  =  x4
eq2(x1, x2)  =  eq1(x2)
false  =  false
0  =  0
true  =  true
s1(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
[REPLACE1, cons1] > eq1 > false
[REPLACE1, cons1] > eq1 > true
0 > false
0 > true


The following usable rules [14] were oriented:

eq2(0, 0) -> true
eq2(0, s1(m)) -> false
eq2(s1(n), 0) -> false
eq2(s1(n), s1(m)) -> eq2(n, m)



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ DependencyGraphProof
              ↳ QDP

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

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)))

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)))
if_min2(true, cons2(x0, cons2(x1, x2)))
if_min2(false, cons2(x0, cons2(x1, x2)))
replace3(x0, x1, nil)
replace3(x0, x1, cons2(x2, x3))
if_replace4(true, x0, x1, cons2(x2, x3))
if_replace4(false, x0, x1, cons2(x2, x3))
sort1(nil)
sort1(cons2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
QDP
                ↳ QDPOrderProof

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)))

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)))
if_min2(true, cons2(x0, cons2(x1, x2)))
if_min2(false, cons2(x0, cons2(x1, x2)))
replace3(x0, x1, nil)
replace3(x0, x1, cons2(x2, x3))
if_replace4(true, x0, x1, cons2(x2, x3))
if_replace4(false, x0, x1, cons2(x2, x3))
sort1(nil)
sort1(cons2(x0, x1))

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


The following pairs can be strictly oriented and are deleted.


SORT1(cons2(n, x)) -> SORT1(replace3(min1(cons2(n, x)), n, x))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
SORT1(x1)  =  SORT1(x1)
cons2(x1, x2)  =  cons1(x2)
replace3(x1, x2, x3)  =  x3
min1(x1)  =  min
0  =  0
nil  =  nil
s1(x1)  =  s
if_min2(x1, x2)  =  x1
false  =  false
le2(x1, x2)  =  le
true  =  true
if_replace4(x1, x2, x3, x4)  =  x4
eq2(x1, x2)  =  eq

Lexicographic Path Order [19].
Precedence:
cons1 > SORT1 > [min, s, false, le, true] > 0
nil > 0
eq > [min, s, false, le, true] > 0


The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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:

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)))

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)))
if_min2(true, cons2(x0, cons2(x1, x2)))
if_min2(false, cons2(x0, cons2(x1, x2)))
replace3(x0, x1, nil)
replace3(x0, x1, cons2(x2, x3))
if_replace4(true, x0, x1, cons2(x2, x3))
if_replace4(false, x0, x1, cons2(x2, x3))
sort1(nil)
sort1(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.