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:

isEmpty1(cons2(x, xs)) -> false
isEmpty1(nil) -> true
isZero1(0) -> true
isZero1(s1(x)) -> false
head1(cons2(x, xs)) -> x
tail1(cons2(x, xs)) -> xs
tail1(nil) -> nil
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
p1(0) -> 0
inc1(s1(x)) -> s1(inc1(x))
inc1(0) -> s1(0)
sumList2(xs, y) -> if6(isEmpty1(xs), isZero1(head1(xs)), y, tail1(xs), cons2(p1(head1(xs)), tail1(xs)), inc1(y))
if6(true, b, y, xs, ys, x) -> y
if6(false, true, y, xs, ys, x) -> sumList2(xs, y)
if6(false, false, y, xs, ys, x) -> sumList2(ys, x)
sum1(xs) -> sumList2(xs, 0)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

isEmpty1(cons2(x, xs)) -> false
isEmpty1(nil) -> true
isZero1(0) -> true
isZero1(s1(x)) -> false
head1(cons2(x, xs)) -> x
tail1(cons2(x, xs)) -> xs
tail1(nil) -> nil
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
p1(0) -> 0
inc1(s1(x)) -> s1(inc1(x))
inc1(0) -> s1(0)
sumList2(xs, y) -> if6(isEmpty1(xs), isZero1(head1(xs)), y, tail1(xs), cons2(p1(head1(xs)), tail1(xs)), inc1(y))
if6(true, b, y, xs, ys, x) -> y
if6(false, true, y, xs, ys, x) -> sumList2(xs, y)
if6(false, false, y, xs, ys, x) -> sumList2(ys, x)
sum1(xs) -> sumList2(xs, 0)

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:

IF6(false, false, y, xs, ys, x) -> SUMLIST2(ys, x)
P1(s1(s1(x))) -> P1(s1(x))
SUMLIST2(xs, y) -> P1(head1(xs))
SUMLIST2(xs, y) -> HEAD1(xs)
SUMLIST2(xs, y) -> TAIL1(xs)
IF6(false, true, y, xs, ys, x) -> SUMLIST2(xs, y)
SUMLIST2(xs, y) -> INC1(y)
SUMLIST2(xs, y) -> ISZERO1(head1(xs))
SUMLIST2(xs, y) -> ISEMPTY1(xs)
SUM1(xs) -> SUMLIST2(xs, 0)
INC1(s1(x)) -> INC1(x)
SUMLIST2(xs, y) -> IF6(isEmpty1(xs), isZero1(head1(xs)), y, tail1(xs), cons2(p1(head1(xs)), tail1(xs)), inc1(y))

The TRS R consists of the following rules:

isEmpty1(cons2(x, xs)) -> false
isEmpty1(nil) -> true
isZero1(0) -> true
isZero1(s1(x)) -> false
head1(cons2(x, xs)) -> x
tail1(cons2(x, xs)) -> xs
tail1(nil) -> nil
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
p1(0) -> 0
inc1(s1(x)) -> s1(inc1(x))
inc1(0) -> s1(0)
sumList2(xs, y) -> if6(isEmpty1(xs), isZero1(head1(xs)), y, tail1(xs), cons2(p1(head1(xs)), tail1(xs)), inc1(y))
if6(true, b, y, xs, ys, x) -> y
if6(false, true, y, xs, ys, x) -> sumList2(xs, y)
if6(false, false, y, xs, ys, x) -> sumList2(ys, x)
sum1(xs) -> sumList2(xs, 0)

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:

IF6(false, false, y, xs, ys, x) -> SUMLIST2(ys, x)
P1(s1(s1(x))) -> P1(s1(x))
SUMLIST2(xs, y) -> P1(head1(xs))
SUMLIST2(xs, y) -> HEAD1(xs)
SUMLIST2(xs, y) -> TAIL1(xs)
IF6(false, true, y, xs, ys, x) -> SUMLIST2(xs, y)
SUMLIST2(xs, y) -> INC1(y)
SUMLIST2(xs, y) -> ISZERO1(head1(xs))
SUMLIST2(xs, y) -> ISEMPTY1(xs)
SUM1(xs) -> SUMLIST2(xs, 0)
INC1(s1(x)) -> INC1(x)
SUMLIST2(xs, y) -> IF6(isEmpty1(xs), isZero1(head1(xs)), y, tail1(xs), cons2(p1(head1(xs)), tail1(xs)), inc1(y))

The TRS R consists of the following rules:

isEmpty1(cons2(x, xs)) -> false
isEmpty1(nil) -> true
isZero1(0) -> true
isZero1(s1(x)) -> false
head1(cons2(x, xs)) -> x
tail1(cons2(x, xs)) -> xs
tail1(nil) -> nil
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
p1(0) -> 0
inc1(s1(x)) -> s1(inc1(x))
inc1(0) -> s1(0)
sumList2(xs, y) -> if6(isEmpty1(xs), isZero1(head1(xs)), y, tail1(xs), cons2(p1(head1(xs)), tail1(xs)), inc1(y))
if6(true, b, y, xs, ys, x) -> y
if6(false, true, y, xs, ys, x) -> sumList2(xs, y)
if6(false, false, y, xs, ys, x) -> sumList2(ys, x)
sum1(xs) -> sumList2(xs, 0)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 7 less nodes.

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

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

INC1(s1(x)) -> INC1(x)

The TRS R consists of the following rules:

isEmpty1(cons2(x, xs)) -> false
isEmpty1(nil) -> true
isZero1(0) -> true
isZero1(s1(x)) -> false
head1(cons2(x, xs)) -> x
tail1(cons2(x, xs)) -> xs
tail1(nil) -> nil
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
p1(0) -> 0
inc1(s1(x)) -> s1(inc1(x))
inc1(0) -> s1(0)
sumList2(xs, y) -> if6(isEmpty1(xs), isZero1(head1(xs)), y, tail1(xs), cons2(p1(head1(xs)), tail1(xs)), inc1(y))
if6(true, b, y, xs, ys, x) -> y
if6(false, true, y, xs, ys, x) -> sumList2(xs, y)
if6(false, false, y, xs, ys, x) -> sumList2(ys, x)
sum1(xs) -> sumList2(xs, 0)

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.


INC1(s1(x)) -> INC1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( INC1(x1) ) = max{0, x1 - 2}


POL( s1(x1) ) = x1 + 3



The following usable rules [14] were oriented: none



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

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

isEmpty1(cons2(x, xs)) -> false
isEmpty1(nil) -> true
isZero1(0) -> true
isZero1(s1(x)) -> false
head1(cons2(x, xs)) -> x
tail1(cons2(x, xs)) -> xs
tail1(nil) -> nil
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
p1(0) -> 0
inc1(s1(x)) -> s1(inc1(x))
inc1(0) -> s1(0)
sumList2(xs, y) -> if6(isEmpty1(xs), isZero1(head1(xs)), y, tail1(xs), cons2(p1(head1(xs)), tail1(xs)), inc1(y))
if6(true, b, y, xs, ys, x) -> y
if6(false, true, y, xs, ys, x) -> sumList2(xs, y)
if6(false, false, y, xs, ys, x) -> sumList2(ys, x)
sum1(xs) -> sumList2(xs, 0)

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

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

P1(s1(s1(x))) -> P1(s1(x))

The TRS R consists of the following rules:

isEmpty1(cons2(x, xs)) -> false
isEmpty1(nil) -> true
isZero1(0) -> true
isZero1(s1(x)) -> false
head1(cons2(x, xs)) -> x
tail1(cons2(x, xs)) -> xs
tail1(nil) -> nil
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
p1(0) -> 0
inc1(s1(x)) -> s1(inc1(x))
inc1(0) -> s1(0)
sumList2(xs, y) -> if6(isEmpty1(xs), isZero1(head1(xs)), y, tail1(xs), cons2(p1(head1(xs)), tail1(xs)), inc1(y))
if6(true, b, y, xs, ys, x) -> y
if6(false, true, y, xs, ys, x) -> sumList2(xs, y)
if6(false, false, y, xs, ys, x) -> sumList2(ys, x)
sum1(xs) -> sumList2(xs, 0)

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.


P1(s1(s1(x))) -> P1(s1(x))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( P1(x1) ) = max{0, x1 - 3}


POL( s1(x1) ) = x1 + 2



The following usable rules [14] were oriented: none



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

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

isEmpty1(cons2(x, xs)) -> false
isEmpty1(nil) -> true
isZero1(0) -> true
isZero1(s1(x)) -> false
head1(cons2(x, xs)) -> x
tail1(cons2(x, xs)) -> xs
tail1(nil) -> nil
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
p1(0) -> 0
inc1(s1(x)) -> s1(inc1(x))
inc1(0) -> s1(0)
sumList2(xs, y) -> if6(isEmpty1(xs), isZero1(head1(xs)), y, tail1(xs), cons2(p1(head1(xs)), tail1(xs)), inc1(y))
if6(true, b, y, xs, ys, x) -> y
if6(false, true, y, xs, ys, x) -> sumList2(xs, y)
if6(false, false, y, xs, ys, x) -> sumList2(ys, x)
sum1(xs) -> sumList2(xs, 0)

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

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

IF6(false, false, y, xs, ys, x) -> SUMLIST2(ys, x)
IF6(false, true, y, xs, ys, x) -> SUMLIST2(xs, y)
SUMLIST2(xs, y) -> IF6(isEmpty1(xs), isZero1(head1(xs)), y, tail1(xs), cons2(p1(head1(xs)), tail1(xs)), inc1(y))

The TRS R consists of the following rules:

isEmpty1(cons2(x, xs)) -> false
isEmpty1(nil) -> true
isZero1(0) -> true
isZero1(s1(x)) -> false
head1(cons2(x, xs)) -> x
tail1(cons2(x, xs)) -> xs
tail1(nil) -> nil
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
p1(0) -> 0
inc1(s1(x)) -> s1(inc1(x))
inc1(0) -> s1(0)
sumList2(xs, y) -> if6(isEmpty1(xs), isZero1(head1(xs)), y, tail1(xs), cons2(p1(head1(xs)), tail1(xs)), inc1(y))
if6(true, b, y, xs, ys, x) -> y
if6(false, true, y, xs, ys, x) -> sumList2(xs, y)
if6(false, false, y, xs, ys, x) -> sumList2(ys, x)
sum1(xs) -> sumList2(xs, 0)

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