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(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
  ↳ 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))

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:

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))

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:

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))

Q is empty.
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
  ↳ 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))

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.


APP2(add2(n, x), y) -> APP2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( APP2(x1, x2) ) = x1 + 3


POL( add2(x1, x2) ) = 2x2 + 1



The following usable rules [14] were oriented: none



↳ 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))

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
            ↳ 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))

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(x), s1(y)) -> LE2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( LE2(x1, x2) ) = 3x2 + 3


POL( s1(x1) ) = x1 + 3



The following usable rules [14] were oriented: none



↳ 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))

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
            ↳ 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))

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.


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 be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MIN1(x1) ) = max{0, 3x1 - 1}


POL( true ) = 0


POL( false ) = 1


POL( add2(x1, x2) ) = 2x2 + 3


POL( 0 ) = 0


POL( s1(x1) ) = max{0, -3}


POL( le2(x1, x2) ) = max{0, -1}


POL( IF_MIN2(x1, x2) ) = 2x2 + 1



The following usable rules [14] were oriented: none



↳ 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))

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
            ↳ 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))

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.


EQ2(s1(x), s1(y)) -> EQ2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s1(x1) ) = x1 + 3


POL( EQ2(x1, x2) ) = 3x2 + 3



The following usable rules [14] were oriented: none



↳ 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))

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
            ↳ 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))

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.


RM2(n, add2(m, x)) -> IF_RM3(eq2(n, m), n, add2(m, x))
The remaining pairs can at least be oriented weakly.

IF_RM3(true, n, add2(m, x)) -> RM2(n, x)
IF_RM3(false, n, add2(m, x)) -> RM2(n, x)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( true ) = max{0, -3}


POL( false ) = 0


POL( add2(x1, x2) ) = x2 + 1


POL( eq2(x1, x2) ) = max{0, x1 - 3}


POL( s1(x1) ) = 0


POL( 0 ) = 0


POL( IF_RM3(x1, ..., x3) ) = x2 + 2x3


POL( RM2(x1, x2) ) = x1 + 2x2 + 2



The following usable rules [14] were oriented: none



↳ 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:

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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 2 less nodes.

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

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))

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.


IF_MINSORT3(true, add2(n, x), y) -> MINSORT2(app2(rm2(n, x), y), nil)
The remaining pairs can at least be oriented weakly.

MINSORT2(add2(n, x), y) -> IF_MINSORT3(eq2(n, min1(add2(n, x))), add2(n, x), y)
IF_MINSORT3(false, add2(n, x), y) -> MINSORT2(x, add2(n, y))
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( rm2(x1, x2) ) = x2


POL( eq2(x1, x2) ) = max{0, -3}


POL( 0 ) = 0


POL( nil ) = max{0, -3}


POL( if_min2(x1, x2) ) = 2x2 + 1


POL( MINSORT2(x1, x2) ) = 2x1 + 2x2 + 2


POL( true ) = 0


POL( IF_MINSORT3(x1, ..., x3) ) = 2x2 + 2x3 + 2


POL( false ) = 2


POL( app2(x1, x2) ) = x1 + x2


POL( if_rm3(x1, ..., x3) ) = x3


POL( add2(x1, x2) ) = 3x1 + x2 + 1


POL( min1(x1) ) = x1 + 1


POL( s1(x1) ) = max{0, x1 - 2}


POL( le2(x1, x2) ) = 3



The following usable rules [14] were oriented:

if_rm3(true, n, add2(m, x)) -> rm2(n, x)
rm2(n, add2(m, x)) -> if_rm3(eq2(n, m), n, add2(m, x))
app2(add2(n, x), y) -> add2(n, app2(x, y))
rm2(n, nil) -> nil
if_rm3(false, n, add2(m, x)) -> add2(m, rm2(n, x))
app2(nil, y) -> y



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

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(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))

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.


MINSORT2(add2(n, x), y) -> IF_MINSORT3(eq2(n, min1(add2(n, x))), add2(n, x), y)
IF_MINSORT3(false, add2(n, x), y) -> MINSORT2(x, add2(n, y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( true ) = max{0, -3}


POL( IF_MINSORT3(x1, ..., x3) ) = max{0, 2x2 - 3}


POL( false ) = max{0, -3}


POL( add2(x1, x2) ) = 3x1 + x2 + 3


POL( eq2(x1, x2) ) = max{0, -3}


POL( s1(x1) ) = x1 + 3


POL( 0 ) = 1


POL( min1(x1) ) = 2x1


POL( le2(x1, x2) ) = 0


POL( nil ) = 1


POL( if_min2(x1, x2) ) = max{0, -3}


POL( MINSORT2(x1, x2) ) = max{0, 2x1 - 2}



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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(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.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.