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(x)) -> 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)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
min1(add2(n, nil)) -> n
min1(add2(n, add2(m, x))) -> if_min2(le2(n, m), add2(n, add2(m, x)))
if_min2(true, add2(n, add2(m, x))) -> min1(add2(n, x))
if_min2(false, add2(n, add2(m, x))) -> min1(add2(m, x))
rm2(n, nil) -> nil
rm2(n, add2(m, x)) -> if_rm3(eq2(n, m), n, add2(m, x))
if_rm3(true, n, add2(m, x)) -> rm2(n, x)
if_rm3(false, n, add2(m, x)) -> add2(m, rm2(n, x))
minsort2(nil, nil) -> nil
minsort2(add2(n, x), y) -> if_minsort3(eq2(n, min1(add2(n, x))), add2(n, x), y)
if_minsort3(true, add2(n, x), y) -> add2(n, minsort2(app2(rm2(n, x), y), nil))
if_minsort3(false, add2(n, x), y) -> minsort2(x, add2(n, y))

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(x)) -> 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)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
min1(add2(n, nil)) -> n
min1(add2(n, add2(m, x))) -> if_min2(le2(n, m), add2(n, add2(m, x)))
if_min2(true, add2(n, add2(m, x))) -> min1(add2(n, x))
if_min2(false, add2(n, add2(m, x))) -> min1(add2(m, x))
rm2(n, nil) -> nil
rm2(n, add2(m, x)) -> if_rm3(eq2(n, m), n, add2(m, x))
if_rm3(true, n, add2(m, x)) -> rm2(n, x)
if_rm3(false, n, add2(m, x)) -> add2(m, rm2(n, x))
minsort2(nil, nil) -> nil
minsort2(add2(n, x), y) -> if_minsort3(eq2(n, min1(add2(n, x))), add2(n, x), y)
if_minsort3(true, add2(n, x), y) -> add2(n, minsort2(app2(rm2(n, x), y), nil))
if_minsort3(false, add2(n, x), y) -> minsort2(x, add2(n, y))

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(x)) -> 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)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
min1(add2(n, nil)) -> n
min1(add2(n, add2(m, x))) -> if_min2(le2(n, m), add2(n, add2(m, x)))
if_min2(true, add2(n, add2(m, x))) -> min1(add2(n, x))
if_min2(false, add2(n, add2(m, x))) -> min1(add2(m, x))
rm2(n, nil) -> nil
rm2(n, add2(m, x)) -> if_rm3(eq2(n, m), n, add2(m, x))
if_rm3(true, n, add2(m, x)) -> rm2(n, x)
if_rm3(false, n, add2(m, x)) -> add2(m, rm2(n, x))
minsort2(nil, nil) -> nil
minsort2(add2(n, x), y) -> if_minsort3(eq2(n, min1(add2(n, x))), add2(n, x), y)
if_minsort3(true, add2(n, x), y) -> add2(n, minsort2(app2(rm2(n, x), y), nil))
if_minsort3(false, add2(n, x), y) -> minsort2(x, add2(n, y))

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))
app2(nil, x0)
app2(add2(x0, x1), x2)
min1(add2(x0, nil))
min1(add2(x0, add2(x1, x2)))
if_min2(true, add2(x0, add2(x1, x2)))
if_min2(false, add2(x0, add2(x1, x2)))
rm2(x0, nil)
rm2(x0, add2(x1, x2))
if_rm3(true, x0, add2(x1, x2))
if_rm3(false, x0, add2(x1, x2))
minsort2(nil, nil)
minsort2(add2(x0, x1), x2)
if_minsort3(true, add2(x0, x1), x2)
if_minsort3(false, add2(x0, x1), x2)


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

MINSORT2(add2(n, x), y) -> IF_MINSORT3(eq2(n, min1(add2(n, x))), add2(n, x), y)
MINSORT2(add2(n, x), y) -> MIN1(add2(n, x))
MIN1(add2(n, add2(m, x))) -> IF_MIN2(le2(n, m), add2(n, add2(m, x)))
IF_MINSORT3(true, add2(n, x), y) -> APP2(rm2(n, x), y)
IF_RM3(true, n, add2(m, x)) -> RM2(n, x)
IF_MIN2(false, add2(n, add2(m, x))) -> MIN1(add2(m, x))
IF_MINSORT3(false, add2(n, x), y) -> MINSORT2(x, add2(n, y))
IF_MIN2(true, add2(n, add2(m, x))) -> MIN1(add2(n, x))
RM2(n, add2(m, x)) -> EQ2(n, m)
EQ2(s1(x), s1(y)) -> EQ2(x, y)
IF_MINSORT3(true, add2(n, x), y) -> MINSORT2(app2(rm2(n, x), y), nil)
MIN1(add2(n, add2(m, x))) -> LE2(n, m)
RM2(n, add2(m, x)) -> IF_RM3(eq2(n, m), n, add2(m, x))
LE2(s1(x), s1(y)) -> LE2(x, y)
MINSORT2(add2(n, x), y) -> EQ2(n, min1(add2(n, x)))
IF_RM3(false, n, add2(m, x)) -> RM2(n, x)
IF_MINSORT3(true, add2(n, x), y) -> RM2(n, x)
APP2(add2(n, x), y) -> APP2(x, y)

The TRS R consists of the following rules:

eq2(0, 0) -> true
eq2(0, s1(x)) -> 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)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
min1(add2(n, nil)) -> n
min1(add2(n, add2(m, x))) -> if_min2(le2(n, m), add2(n, add2(m, x)))
if_min2(true, add2(n, add2(m, x))) -> min1(add2(n, x))
if_min2(false, add2(n, add2(m, x))) -> min1(add2(m, x))
rm2(n, nil) -> nil
rm2(n, add2(m, x)) -> if_rm3(eq2(n, m), n, add2(m, x))
if_rm3(true, n, add2(m, x)) -> rm2(n, x)
if_rm3(false, n, add2(m, x)) -> add2(m, rm2(n, x))
minsort2(nil, nil) -> nil
minsort2(add2(n, x), y) -> if_minsort3(eq2(n, min1(add2(n, x))), add2(n, x), y)
if_minsort3(true, add2(n, x), y) -> add2(n, minsort2(app2(rm2(n, x), y), nil))
if_minsort3(false, add2(n, x), y) -> minsort2(x, add2(n, y))

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))
app2(nil, x0)
app2(add2(x0, x1), x2)
min1(add2(x0, nil))
min1(add2(x0, add2(x1, x2)))
if_min2(true, add2(x0, add2(x1, x2)))
if_min2(false, add2(x0, add2(x1, x2)))
rm2(x0, nil)
rm2(x0, add2(x1, x2))
if_rm3(true, x0, add2(x1, x2))
if_rm3(false, x0, add2(x1, x2))
minsort2(nil, nil)
minsort2(add2(x0, x1), x2)
if_minsort3(true, add2(x0, x1), x2)
if_minsort3(false, add2(x0, x1), x2)

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:

MINSORT2(add2(n, x), y) -> IF_MINSORT3(eq2(n, min1(add2(n, x))), add2(n, x), y)
MINSORT2(add2(n, x), y) -> MIN1(add2(n, x))
MIN1(add2(n, add2(m, x))) -> IF_MIN2(le2(n, m), add2(n, add2(m, x)))
IF_MINSORT3(true, add2(n, x), y) -> APP2(rm2(n, x), y)
IF_RM3(true, n, add2(m, x)) -> RM2(n, x)
IF_MIN2(false, add2(n, add2(m, x))) -> MIN1(add2(m, x))
IF_MINSORT3(false, add2(n, x), y) -> MINSORT2(x, add2(n, y))
IF_MIN2(true, add2(n, add2(m, x))) -> MIN1(add2(n, x))
RM2(n, add2(m, x)) -> EQ2(n, m)
EQ2(s1(x), s1(y)) -> EQ2(x, y)
IF_MINSORT3(true, add2(n, x), y) -> MINSORT2(app2(rm2(n, x), y), nil)
MIN1(add2(n, add2(m, x))) -> LE2(n, m)
RM2(n, add2(m, x)) -> IF_RM3(eq2(n, m), n, add2(m, x))
LE2(s1(x), s1(y)) -> LE2(x, y)
MINSORT2(add2(n, x), y) -> EQ2(n, min1(add2(n, x)))
IF_RM3(false, n, add2(m, x)) -> RM2(n, x)
IF_MINSORT3(true, add2(n, x), y) -> RM2(n, x)
APP2(add2(n, x), y) -> APP2(x, y)

The TRS R consists of the following rules:

eq2(0, 0) -> true
eq2(0, s1(x)) -> 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)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
min1(add2(n, nil)) -> n
min1(add2(n, add2(m, x))) -> if_min2(le2(n, m), add2(n, add2(m, x)))
if_min2(true, add2(n, add2(m, x))) -> min1(add2(n, x))
if_min2(false, add2(n, add2(m, x))) -> min1(add2(m, x))
rm2(n, nil) -> nil
rm2(n, add2(m, x)) -> if_rm3(eq2(n, m), n, add2(m, x))
if_rm3(true, n, add2(m, x)) -> rm2(n, x)
if_rm3(false, n, add2(m, x)) -> add2(m, rm2(n, x))
minsort2(nil, nil) -> nil
minsort2(add2(n, x), y) -> if_minsort3(eq2(n, min1(add2(n, x))), add2(n, x), y)
if_minsort3(true, add2(n, x), y) -> add2(n, minsort2(app2(rm2(n, x), y), nil))
if_minsort3(false, add2(n, x), y) -> minsort2(x, add2(n, y))

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))
app2(nil, x0)
app2(add2(x0, x1), x2)
min1(add2(x0, nil))
min1(add2(x0, add2(x1, x2)))
if_min2(true, add2(x0, add2(x1, x2)))
if_min2(false, add2(x0, add2(x1, x2)))
rm2(x0, nil)
rm2(x0, add2(x1, x2))
if_rm3(true, x0, add2(x1, x2))
if_rm3(false, x0, add2(x1, x2))
minsort2(nil, nil)
minsort2(add2(x0, x1), x2)
if_minsort3(true, add2(x0, x1), x2)
if_minsort3(false, add2(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 6 SCCs with 6 less nodes.

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

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

APP2(add2(n, x), y) -> APP2(x, y)

The TRS R consists of the following rules:

eq2(0, 0) -> true
eq2(0, s1(x)) -> 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)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
min1(add2(n, nil)) -> n
min1(add2(n, add2(m, x))) -> if_min2(le2(n, m), add2(n, add2(m, x)))
if_min2(true, add2(n, add2(m, x))) -> min1(add2(n, x))
if_min2(false, add2(n, add2(m, x))) -> min1(add2(m, x))
rm2(n, nil) -> nil
rm2(n, add2(m, x)) -> if_rm3(eq2(n, m), n, add2(m, x))
if_rm3(true, n, add2(m, x)) -> rm2(n, x)
if_rm3(false, n, add2(m, x)) -> add2(m, rm2(n, x))
minsort2(nil, nil) -> nil
minsort2(add2(n, x), y) -> if_minsort3(eq2(n, min1(add2(n, x))), add2(n, x), y)
if_minsort3(true, add2(n, x), y) -> add2(n, minsort2(app2(rm2(n, x), y), nil))
if_minsort3(false, add2(n, x), y) -> minsort2(x, add2(n, y))

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))
app2(nil, x0)
app2(add2(x0, x1), x2)
min1(add2(x0, nil))
min1(add2(x0, add2(x1, x2)))
if_min2(true, add2(x0, add2(x1, x2)))
if_min2(false, add2(x0, add2(x1, x2)))
rm2(x0, nil)
rm2(x0, add2(x1, x2))
if_rm3(true, x0, add2(x1, x2))
if_rm3(false, x0, add2(x1, x2))
minsort2(nil, nil)
minsort2(add2(x0, x1), x2)
if_minsort3(true, add2(x0, x1), x2)
if_minsort3(false, add2(x0, x1), x2)

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.


APP2(add2(n, x), y) -> APP2(x, y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  APP1(x1)
add2(x1, x2)  =  add1(x2)

Lexicographic Path Order [19].
Precedence:
[APP1, add1]


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
              ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

eq2(0, 0) -> true
eq2(0, s1(x)) -> 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)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
min1(add2(n, nil)) -> n
min1(add2(n, add2(m, x))) -> if_min2(le2(n, m), add2(n, add2(m, x)))
if_min2(true, add2(n, add2(m, x))) -> min1(add2(n, x))
if_min2(false, add2(n, add2(m, x))) -> min1(add2(m, x))
rm2(n, nil) -> nil
rm2(n, add2(m, x)) -> if_rm3(eq2(n, m), n, add2(m, x))
if_rm3(true, n, add2(m, x)) -> rm2(n, x)
if_rm3(false, n, add2(m, x)) -> add2(m, rm2(n, x))
minsort2(nil, nil) -> nil
minsort2(add2(n, x), y) -> if_minsort3(eq2(n, min1(add2(n, x))), add2(n, x), y)
if_minsort3(true, add2(n, x), y) -> add2(n, minsort2(app2(rm2(n, x), y), nil))
if_minsort3(false, add2(n, x), y) -> minsort2(x, add2(n, y))

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))
app2(nil, x0)
app2(add2(x0, x1), x2)
min1(add2(x0, nil))
min1(add2(x0, add2(x1, x2)))
if_min2(true, add2(x0, add2(x1, x2)))
if_min2(false, add2(x0, add2(x1, x2)))
rm2(x0, nil)
rm2(x0, add2(x1, x2))
if_rm3(true, x0, add2(x1, x2))
if_rm3(false, x0, add2(x1, x2))
minsort2(nil, nil)
minsort2(add2(x0, x1), x2)
if_minsort3(true, add2(x0, x1), x2)
if_minsort3(false, add2(x0, x1), x2)

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
              ↳ 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(x)) -> 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)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
min1(add2(n, nil)) -> n
min1(add2(n, add2(m, x))) -> if_min2(le2(n, m), add2(n, add2(m, x)))
if_min2(true, add2(n, add2(m, x))) -> min1(add2(n, x))
if_min2(false, add2(n, add2(m, x))) -> min1(add2(m, x))
rm2(n, nil) -> nil
rm2(n, add2(m, x)) -> if_rm3(eq2(n, m), n, add2(m, x))
if_rm3(true, n, add2(m, x)) -> rm2(n, x)
if_rm3(false, n, add2(m, x)) -> add2(m, rm2(n, x))
minsort2(nil, nil) -> nil
minsort2(add2(n, x), y) -> if_minsort3(eq2(n, min1(add2(n, x))), add2(n, x), y)
if_minsort3(true, add2(n, x), y) -> add2(n, minsort2(app2(rm2(n, x), y), nil))
if_minsort3(false, add2(n, x), y) -> minsort2(x, add2(n, y))

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))
app2(nil, x0)
app2(add2(x0, x1), x2)
min1(add2(x0, nil))
min1(add2(x0, add2(x1, x2)))
if_min2(true, add2(x0, add2(x1, x2)))
if_min2(false, add2(x0, add2(x1, x2)))
rm2(x0, nil)
rm2(x0, add2(x1, x2))
if_rm3(true, x0, add2(x1, x2))
if_rm3(false, x0, add2(x1, x2))
minsort2(nil, nil)
minsort2(add2(x0, x1), x2)
if_minsort3(true, add2(x0, x1), x2)
if_minsort3(false, add2(x0, x1), x2)

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(x), s1(y)) -> LE2(x, y)
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
              ↳ 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(x)) -> 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)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
min1(add2(n, nil)) -> n
min1(add2(n, add2(m, x))) -> if_min2(le2(n, m), add2(n, add2(m, x)))
if_min2(true, add2(n, add2(m, x))) -> min1(add2(n, x))
if_min2(false, add2(n, add2(m, x))) -> min1(add2(m, x))
rm2(n, nil) -> nil
rm2(n, add2(m, x)) -> if_rm3(eq2(n, m), n, add2(m, x))
if_rm3(true, n, add2(m, x)) -> rm2(n, x)
if_rm3(false, n, add2(m, x)) -> add2(m, rm2(n, x))
minsort2(nil, nil) -> nil
minsort2(add2(n, x), y) -> if_minsort3(eq2(n, min1(add2(n, x))), add2(n, x), y)
if_minsort3(true, add2(n, x), y) -> add2(n, minsort2(app2(rm2(n, x), y), nil))
if_minsort3(false, add2(n, x), y) -> minsort2(x, add2(n, y))

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))
app2(nil, x0)
app2(add2(x0, x1), x2)
min1(add2(x0, nil))
min1(add2(x0, add2(x1, x2)))
if_min2(true, add2(x0, add2(x1, x2)))
if_min2(false, add2(x0, add2(x1, x2)))
rm2(x0, nil)
rm2(x0, add2(x1, x2))
if_rm3(true, x0, add2(x1, x2))
if_rm3(false, x0, add2(x1, x2))
minsort2(nil, nil)
minsort2(add2(x0, x1), x2)
if_minsort3(true, add2(x0, x1), x2)
if_minsort3(false, add2(x0, x1), x2)

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
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

MIN1(add2(n, add2(m, x))) -> IF_MIN2(le2(n, m), add2(n, add2(m, x)))
IF_MIN2(false, add2(n, add2(m, x))) -> MIN1(add2(m, x))
IF_MIN2(true, add2(n, add2(m, x))) -> MIN1(add2(n, x))

The TRS R consists of the following rules:

eq2(0, 0) -> true
eq2(0, s1(x)) -> 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)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
min1(add2(n, nil)) -> n
min1(add2(n, add2(m, x))) -> if_min2(le2(n, m), add2(n, add2(m, x)))
if_min2(true, add2(n, add2(m, x))) -> min1(add2(n, x))
if_min2(false, add2(n, add2(m, x))) -> min1(add2(m, x))
rm2(n, nil) -> nil
rm2(n, add2(m, x)) -> if_rm3(eq2(n, m), n, add2(m, x))
if_rm3(true, n, add2(m, x)) -> rm2(n, x)
if_rm3(false, n, add2(m, x)) -> add2(m, rm2(n, x))
minsort2(nil, nil) -> nil
minsort2(add2(n, x), y) -> if_minsort3(eq2(n, min1(add2(n, x))), add2(n, x), y)
if_minsort3(true, add2(n, x), y) -> add2(n, minsort2(app2(rm2(n, x), y), nil))
if_minsort3(false, add2(n, x), y) -> minsort2(x, add2(n, y))

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))
app2(nil, x0)
app2(add2(x0, x1), x2)
min1(add2(x0, nil))
min1(add2(x0, add2(x1, x2)))
if_min2(true, add2(x0, add2(x1, x2)))
if_min2(false, add2(x0, add2(x1, x2)))
rm2(x0, nil)
rm2(x0, add2(x1, x2))
if_rm3(true, x0, add2(x1, x2))
if_rm3(false, x0, add2(x1, x2))
minsort2(nil, nil)
minsort2(add2(x0, x1), x2)
if_minsort3(true, add2(x0, x1), x2)
if_minsort3(false, add2(x0, x1), x2)

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.


MIN1(add2(n, add2(m, x))) -> IF_MIN2(le2(n, m), add2(n, add2(m, x)))
IF_MIN2(false, add2(n, add2(m, x))) -> MIN1(add2(m, x))
IF_MIN2(true, add2(n, add2(m, x))) -> MIN1(add2(n, x))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MIN1(x1)  =  MIN1(x1)
add2(x1, x2)  =  add2(x1, x2)
IF_MIN2(x1, x2)  =  x2
le2(x1, x2)  =  le2(x1, x2)
false  =  false
true  =  true
0  =  0
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
add2 > [MIN1, le2, false]
0 > true > [MIN1, le2, false]
s1 > [MIN1, le2, false]


The following usable rules [14] were oriented:

le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

eq2(0, 0) -> true
eq2(0, s1(x)) -> 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)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
min1(add2(n, nil)) -> n
min1(add2(n, add2(m, x))) -> if_min2(le2(n, m), add2(n, add2(m, x)))
if_min2(true, add2(n, add2(m, x))) -> min1(add2(n, x))
if_min2(false, add2(n, add2(m, x))) -> min1(add2(m, x))
rm2(n, nil) -> nil
rm2(n, add2(m, x)) -> if_rm3(eq2(n, m), n, add2(m, x))
if_rm3(true, n, add2(m, x)) -> rm2(n, x)
if_rm3(false, n, add2(m, x)) -> add2(m, rm2(n, x))
minsort2(nil, nil) -> nil
minsort2(add2(n, x), y) -> if_minsort3(eq2(n, min1(add2(n, x))), add2(n, x), y)
if_minsort3(true, add2(n, x), y) -> add2(n, minsort2(app2(rm2(n, x), y), nil))
if_minsort3(false, add2(n, x), y) -> minsort2(x, add2(n, y))

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))
app2(nil, x0)
app2(add2(x0, x1), x2)
min1(add2(x0, nil))
min1(add2(x0, add2(x1, x2)))
if_min2(true, add2(x0, add2(x1, x2)))
if_min2(false, add2(x0, add2(x1, x2)))
rm2(x0, nil)
rm2(x0, add2(x1, x2))
if_rm3(true, x0, add2(x1, x2))
if_rm3(false, x0, add2(x1, x2))
minsort2(nil, nil)
minsort2(add2(x0, x1), x2)
if_minsort3(true, add2(x0, x1), x2)
if_minsort3(false, add2(x0, x1), x2)

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
              ↳ 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(x)) -> 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)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
min1(add2(n, nil)) -> n
min1(add2(n, add2(m, x))) -> if_min2(le2(n, m), add2(n, add2(m, x)))
if_min2(true, add2(n, add2(m, x))) -> min1(add2(n, x))
if_min2(false, add2(n, add2(m, x))) -> min1(add2(m, x))
rm2(n, nil) -> nil
rm2(n, add2(m, x)) -> if_rm3(eq2(n, m), n, add2(m, x))
if_rm3(true, n, add2(m, x)) -> rm2(n, x)
if_rm3(false, n, add2(m, x)) -> add2(m, rm2(n, x))
minsort2(nil, nil) -> nil
minsort2(add2(n, x), y) -> if_minsort3(eq2(n, min1(add2(n, x))), add2(n, x), y)
if_minsort3(true, add2(n, x), y) -> add2(n, minsort2(app2(rm2(n, x), y), nil))
if_minsort3(false, add2(n, x), y) -> minsort2(x, add2(n, y))

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))
app2(nil, x0)
app2(add2(x0, x1), x2)
min1(add2(x0, nil))
min1(add2(x0, add2(x1, x2)))
if_min2(true, add2(x0, add2(x1, x2)))
if_min2(false, add2(x0, add2(x1, x2)))
rm2(x0, nil)
rm2(x0, add2(x1, x2))
if_rm3(true, x0, add2(x1, x2))
if_rm3(false, x0, add2(x1, x2))
minsort2(nil, nil)
minsort2(add2(x0, x1), x2)
if_minsort3(true, add2(x0, x1), x2)
if_minsort3(false, add2(x0, x1), x2)

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(x), s1(y)) -> EQ2(x, y)
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
              ↳ 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(x)) -> 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)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
min1(add2(n, nil)) -> n
min1(add2(n, add2(m, x))) -> if_min2(le2(n, m), add2(n, add2(m, x)))
if_min2(true, add2(n, add2(m, x))) -> min1(add2(n, x))
if_min2(false, add2(n, add2(m, x))) -> min1(add2(m, x))
rm2(n, nil) -> nil
rm2(n, add2(m, x)) -> if_rm3(eq2(n, m), n, add2(m, x))
if_rm3(true, n, add2(m, x)) -> rm2(n, x)
if_rm3(false, n, add2(m, x)) -> add2(m, rm2(n, x))
minsort2(nil, nil) -> nil
minsort2(add2(n, x), y) -> if_minsort3(eq2(n, min1(add2(n, x))), add2(n, x), y)
if_minsort3(true, add2(n, x), y) -> add2(n, minsort2(app2(rm2(n, x), y), nil))
if_minsort3(false, add2(n, x), y) -> minsort2(x, add2(n, y))

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))
app2(nil, x0)
app2(add2(x0, x1), x2)
min1(add2(x0, nil))
min1(add2(x0, add2(x1, x2)))
if_min2(true, add2(x0, add2(x1, x2)))
if_min2(false, add2(x0, add2(x1, x2)))
rm2(x0, nil)
rm2(x0, add2(x1, x2))
if_rm3(true, x0, add2(x1, x2))
if_rm3(false, x0, add2(x1, x2))
minsort2(nil, nil)
minsort2(add2(x0, x1), x2)
if_minsort3(true, add2(x0, x1), x2)
if_minsort3(false, add2(x0, x1), x2)

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
QDP
                ↳ QDPOrderProof
              ↳ QDP

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

RM2(n, add2(m, x)) -> IF_RM3(eq2(n, m), n, add2(m, x))
IF_RM3(true, n, add2(m, x)) -> RM2(n, x)
IF_RM3(false, n, add2(m, x)) -> RM2(n, x)

The TRS R consists of the following rules:

eq2(0, 0) -> true
eq2(0, s1(x)) -> 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)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
min1(add2(n, nil)) -> n
min1(add2(n, add2(m, x))) -> if_min2(le2(n, m), add2(n, add2(m, x)))
if_min2(true, add2(n, add2(m, x))) -> min1(add2(n, x))
if_min2(false, add2(n, add2(m, x))) -> min1(add2(m, x))
rm2(n, nil) -> nil
rm2(n, add2(m, x)) -> if_rm3(eq2(n, m), n, add2(m, x))
if_rm3(true, n, add2(m, x)) -> rm2(n, x)
if_rm3(false, n, add2(m, x)) -> add2(m, rm2(n, x))
minsort2(nil, nil) -> nil
minsort2(add2(n, x), y) -> if_minsort3(eq2(n, min1(add2(n, x))), add2(n, x), y)
if_minsort3(true, add2(n, x), y) -> add2(n, minsort2(app2(rm2(n, x), y), nil))
if_minsort3(false, add2(n, x), y) -> minsort2(x, add2(n, y))

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))
app2(nil, x0)
app2(add2(x0, x1), x2)
min1(add2(x0, nil))
min1(add2(x0, add2(x1, x2)))
if_min2(true, add2(x0, add2(x1, x2)))
if_min2(false, add2(x0, add2(x1, x2)))
rm2(x0, nil)
rm2(x0, add2(x1, x2))
if_rm3(true, x0, add2(x1, x2))
if_rm3(false, x0, add2(x1, x2))
minsort2(nil, nil)
minsort2(add2(x0, x1), x2)
if_minsort3(true, add2(x0, x1), x2)
if_minsort3(false, add2(x0, x1), x2)

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_RM3(true, n, add2(m, x)) -> RM2(n, x)
IF_RM3(false, n, add2(m, x)) -> RM2(n, x)
The remaining pairs can at least by weakly be oriented.

RM2(n, add2(m, x)) -> IF_RM3(eq2(n, m), n, add2(m, x))
Used ordering: Combined order from the following AFS and order.
RM2(x1, x2)  =  x2
add2(x1, x2)  =  add1(x2)
IF_RM3(x1, x2, x3)  =  x3
eq2(x1, x2)  =  eq
true  =  true
false  =  false
0  =  0
s1(x1)  =  s

Lexicographic Path Order [19].
Precedence:
0 > [add1, eq, true] > [false, s]


The following usable rules [14] were oriented:

eq2(0, 0) -> true
eq2(0, s1(x)) -> false
eq2(s1(x), 0) -> false
eq2(s1(x), s1(y)) -> eq2(x, y)



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

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

RM2(n, add2(m, x)) -> IF_RM3(eq2(n, m), n, add2(m, x))

The TRS R consists of the following rules:

eq2(0, 0) -> true
eq2(0, s1(x)) -> 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)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
min1(add2(n, nil)) -> n
min1(add2(n, add2(m, x))) -> if_min2(le2(n, m), add2(n, add2(m, x)))
if_min2(true, add2(n, add2(m, x))) -> min1(add2(n, x))
if_min2(false, add2(n, add2(m, x))) -> min1(add2(m, x))
rm2(n, nil) -> nil
rm2(n, add2(m, x)) -> if_rm3(eq2(n, m), n, add2(m, x))
if_rm3(true, n, add2(m, x)) -> rm2(n, x)
if_rm3(false, n, add2(m, x)) -> add2(m, rm2(n, x))
minsort2(nil, nil) -> nil
minsort2(add2(n, x), y) -> if_minsort3(eq2(n, min1(add2(n, x))), add2(n, x), y)
if_minsort3(true, add2(n, x), y) -> add2(n, minsort2(app2(rm2(n, x), y), nil))
if_minsort3(false, add2(n, x), y) -> minsort2(x, add2(n, y))

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))
app2(nil, x0)
app2(add2(x0, x1), x2)
min1(add2(x0, nil))
min1(add2(x0, add2(x1, x2)))
if_min2(true, add2(x0, add2(x1, x2)))
if_min2(false, add2(x0, add2(x1, x2)))
rm2(x0, nil)
rm2(x0, add2(x1, x2))
if_rm3(true, x0, add2(x1, x2))
if_rm3(false, x0, add2(x1, x2))
minsort2(nil, nil)
minsort2(add2(x0, x1), x2)
if_minsort3(true, add2(x0, x1), x2)
if_minsort3(false, add2(x0, x1), x2)

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
QDP

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

MINSORT2(add2(n, x), y) -> IF_MINSORT3(eq2(n, min1(add2(n, x))), add2(n, x), y)
IF_MINSORT3(true, add2(n, x), y) -> MINSORT2(app2(rm2(n, x), y), nil)
IF_MINSORT3(false, add2(n, x), y) -> MINSORT2(x, add2(n, y))

The TRS R consists of the following rules:

eq2(0, 0) -> true
eq2(0, s1(x)) -> 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)
app2(nil, y) -> y
app2(add2(n, x), y) -> add2(n, app2(x, y))
min1(add2(n, nil)) -> n
min1(add2(n, add2(m, x))) -> if_min2(le2(n, m), add2(n, add2(m, x)))
if_min2(true, add2(n, add2(m, x))) -> min1(add2(n, x))
if_min2(false, add2(n, add2(m, x))) -> min1(add2(m, x))
rm2(n, nil) -> nil
rm2(n, add2(m, x)) -> if_rm3(eq2(n, m), n, add2(m, x))
if_rm3(true, n, add2(m, x)) -> rm2(n, x)
if_rm3(false, n, add2(m, x)) -> add2(m, rm2(n, x))
minsort2(nil, nil) -> nil
minsort2(add2(n, x), y) -> if_minsort3(eq2(n, min1(add2(n, x))), add2(n, x), y)
if_minsort3(true, add2(n, x), y) -> add2(n, minsort2(app2(rm2(n, x), y), nil))
if_minsort3(false, add2(n, x), y) -> minsort2(x, add2(n, y))

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))
app2(nil, x0)
app2(add2(x0, x1), x2)
min1(add2(x0, nil))
min1(add2(x0, add2(x1, x2)))
if_min2(true, add2(x0, add2(x1, x2)))
if_min2(false, add2(x0, add2(x1, x2)))
rm2(x0, nil)
rm2(x0, add2(x1, x2))
if_rm3(true, x0, add2(x1, x2))
if_rm3(false, x0, add2(x1, x2))
minsort2(nil, nil)
minsort2(add2(x0, x1), x2)
if_minsort3(true, add2(x0, x1), x2)
if_minsort3(false, add2(x0, x1), x2)

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