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:

double1(x) -> permute3(x, x, a)
permute3(x, y, a) -> permute3(isZero1(x), x, b)
permute3(false, x, b) -> permute3(ack2(x, x), p1(x), c)
permute3(true, x, b) -> 0
permute3(y, x, c) -> s1(s1(permute3(x, y, a)))
p1(0) -> 0
p1(s1(x)) -> x
ack2(0, x) -> plus2(x, s1(0))
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, ack2(s1(x), y))
plus2(0, y) -> y
plus2(s1(x), y) -> plus2(x, s1(y))
plus2(x, s1(s1(y))) -> s1(plus2(s1(x), y))
plus2(x, s1(0)) -> s1(x)
plus2(x, 0) -> x
isZero1(0) -> true
isZero1(s1(x)) -> false

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

double1(x) -> permute3(x, x, a)
permute3(x, y, a) -> permute3(isZero1(x), x, b)
permute3(false, x, b) -> permute3(ack2(x, x), p1(x), c)
permute3(true, x, b) -> 0
permute3(y, x, c) -> s1(s1(permute3(x, y, a)))
p1(0) -> 0
p1(s1(x)) -> x
ack2(0, x) -> plus2(x, s1(0))
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, ack2(s1(x), y))
plus2(0, y) -> y
plus2(s1(x), y) -> plus2(x, s1(y))
plus2(x, s1(s1(y))) -> s1(plus2(s1(x), y))
plus2(x, s1(0)) -> s1(x)
plus2(x, 0) -> x
isZero1(0) -> true
isZero1(s1(x)) -> false

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:

PERMUTE3(x, y, a) -> ISZERO1(x)
PLUS2(x, s1(s1(y))) -> PLUS2(s1(x), y)
PLUS2(s1(x), y) -> PLUS2(x, s1(y))
PERMUTE3(false, x, b) -> PERMUTE3(ack2(x, x), p1(x), c)
ACK2(s1(x), 0) -> ACK2(x, s1(0))
ACK2(0, x) -> PLUS2(x, s1(0))
PERMUTE3(x, y, a) -> PERMUTE3(isZero1(x), x, b)
ACK2(s1(x), s1(y)) -> ACK2(s1(x), y)
DOUBLE1(x) -> PERMUTE3(x, x, a)
ACK2(s1(x), s1(y)) -> ACK2(x, ack2(s1(x), y))
PERMUTE3(false, x, b) -> ACK2(x, x)
PERMUTE3(y, x, c) -> PERMUTE3(x, y, a)
PERMUTE3(false, x, b) -> P1(x)

The TRS R consists of the following rules:

double1(x) -> permute3(x, x, a)
permute3(x, y, a) -> permute3(isZero1(x), x, b)
permute3(false, x, b) -> permute3(ack2(x, x), p1(x), c)
permute3(true, x, b) -> 0
permute3(y, x, c) -> s1(s1(permute3(x, y, a)))
p1(0) -> 0
p1(s1(x)) -> x
ack2(0, x) -> plus2(x, s1(0))
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, ack2(s1(x), y))
plus2(0, y) -> y
plus2(s1(x), y) -> plus2(x, s1(y))
plus2(x, s1(s1(y))) -> s1(plus2(s1(x), y))
plus2(x, s1(0)) -> s1(x)
plus2(x, 0) -> x
isZero1(0) -> true
isZero1(s1(x)) -> false

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:

PERMUTE3(x, y, a) -> ISZERO1(x)
PLUS2(x, s1(s1(y))) -> PLUS2(s1(x), y)
PLUS2(s1(x), y) -> PLUS2(x, s1(y))
PERMUTE3(false, x, b) -> PERMUTE3(ack2(x, x), p1(x), c)
ACK2(s1(x), 0) -> ACK2(x, s1(0))
ACK2(0, x) -> PLUS2(x, s1(0))
PERMUTE3(x, y, a) -> PERMUTE3(isZero1(x), x, b)
ACK2(s1(x), s1(y)) -> ACK2(s1(x), y)
DOUBLE1(x) -> PERMUTE3(x, x, a)
ACK2(s1(x), s1(y)) -> ACK2(x, ack2(s1(x), y))
PERMUTE3(false, x, b) -> ACK2(x, x)
PERMUTE3(y, x, c) -> PERMUTE3(x, y, a)
PERMUTE3(false, x, b) -> P1(x)

The TRS R consists of the following rules:

double1(x) -> permute3(x, x, a)
permute3(x, y, a) -> permute3(isZero1(x), x, b)
permute3(false, x, b) -> permute3(ack2(x, x), p1(x), c)
permute3(true, x, b) -> 0
permute3(y, x, c) -> s1(s1(permute3(x, y, a)))
p1(0) -> 0
p1(s1(x)) -> x
ack2(0, x) -> plus2(x, s1(0))
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, ack2(s1(x), y))
plus2(0, y) -> y
plus2(s1(x), y) -> plus2(x, s1(y))
plus2(x, s1(s1(y))) -> s1(plus2(s1(x), y))
plus2(x, s1(0)) -> s1(x)
plus2(x, 0) -> x
isZero1(0) -> true
isZero1(s1(x)) -> false

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 5 less nodes.

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

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

PLUS2(x, s1(s1(y))) -> PLUS2(s1(x), y)
PLUS2(s1(x), y) -> PLUS2(x, s1(y))

The TRS R consists of the following rules:

double1(x) -> permute3(x, x, a)
permute3(x, y, a) -> permute3(isZero1(x), x, b)
permute3(false, x, b) -> permute3(ack2(x, x), p1(x), c)
permute3(true, x, b) -> 0
permute3(y, x, c) -> s1(s1(permute3(x, y, a)))
p1(0) -> 0
p1(s1(x)) -> x
ack2(0, x) -> plus2(x, s1(0))
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, ack2(s1(x), y))
plus2(0, y) -> y
plus2(s1(x), y) -> plus2(x, s1(y))
plus2(x, s1(s1(y))) -> s1(plus2(s1(x), y))
plus2(x, s1(0)) -> s1(x)
plus2(x, 0) -> x
isZero1(0) -> true
isZero1(s1(x)) -> false

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.


PLUS2(s1(x), y) -> PLUS2(x, s1(y))
The remaining pairs can at least be oriented weakly.

PLUS2(x, s1(s1(y))) -> PLUS2(s1(x), y)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s1(x1) ) = x1 + 2


POL( PLUS2(x1, x2) ) = max{0, 2x1 + x2 - 3}



The following usable rules [14] were oriented: none



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

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

PLUS2(x, s1(s1(y))) -> PLUS2(s1(x), y)

The TRS R consists of the following rules:

double1(x) -> permute3(x, x, a)
permute3(x, y, a) -> permute3(isZero1(x), x, b)
permute3(false, x, b) -> permute3(ack2(x, x), p1(x), c)
permute3(true, x, b) -> 0
permute3(y, x, c) -> s1(s1(permute3(x, y, a)))
p1(0) -> 0
p1(s1(x)) -> x
ack2(0, x) -> plus2(x, s1(0))
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, ack2(s1(x), y))
plus2(0, y) -> y
plus2(s1(x), y) -> plus2(x, s1(y))
plus2(x, s1(s1(y))) -> s1(plus2(s1(x), y))
plus2(x, s1(0)) -> s1(x)
plus2(x, 0) -> x
isZero1(0) -> true
isZero1(s1(x)) -> false

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.


PLUS2(x, s1(s1(y))) -> PLUS2(s1(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( PLUS2(x1, x2) ) = x2 + 3



The following usable rules [14] were oriented: none



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

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

double1(x) -> permute3(x, x, a)
permute3(x, y, a) -> permute3(isZero1(x), x, b)
permute3(false, x, b) -> permute3(ack2(x, x), p1(x), c)
permute3(true, x, b) -> 0
permute3(y, x, c) -> s1(s1(permute3(x, y, a)))
p1(0) -> 0
p1(s1(x)) -> x
ack2(0, x) -> plus2(x, s1(0))
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, ack2(s1(x), y))
plus2(0, y) -> y
plus2(s1(x), y) -> plus2(x, s1(y))
plus2(x, s1(s1(y))) -> s1(plus2(s1(x), y))
plus2(x, s1(0)) -> s1(x)
plus2(x, 0) -> x
isZero1(0) -> true
isZero1(s1(x)) -> false

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:

ACK2(s1(x), s1(y)) -> ACK2(s1(x), y)
ACK2(s1(x), s1(y)) -> ACK2(x, ack2(s1(x), y))
ACK2(s1(x), 0) -> ACK2(x, s1(0))

The TRS R consists of the following rules:

double1(x) -> permute3(x, x, a)
permute3(x, y, a) -> permute3(isZero1(x), x, b)
permute3(false, x, b) -> permute3(ack2(x, x), p1(x), c)
permute3(true, x, b) -> 0
permute3(y, x, c) -> s1(s1(permute3(x, y, a)))
p1(0) -> 0
p1(s1(x)) -> x
ack2(0, x) -> plus2(x, s1(0))
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, ack2(s1(x), y))
plus2(0, y) -> y
plus2(s1(x), y) -> plus2(x, s1(y))
plus2(x, s1(s1(y))) -> s1(plus2(s1(x), y))
plus2(x, s1(0)) -> s1(x)
plus2(x, 0) -> x
isZero1(0) -> true
isZero1(s1(x)) -> false

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.


ACK2(s1(x), s1(y)) -> ACK2(x, ack2(s1(x), y))
ACK2(s1(x), 0) -> ACK2(x, s1(0))
The remaining pairs can at least be oriented weakly.

ACK2(s1(x), s1(y)) -> ACK2(s1(x), y)
Used ordering: Polynomial Order [17,21] with Interpretation:

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


POL( s1(x1) ) = 2x1 + 3


POL( ack2(x1, x2) ) = 3


POL( plus2(x1, x2) ) = 1


POL( ACK2(x1, x2) ) = 3x1 + 2



The following usable rules [14] were oriented: none



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

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

ACK2(s1(x), s1(y)) -> ACK2(s1(x), y)

The TRS R consists of the following rules:

double1(x) -> permute3(x, x, a)
permute3(x, y, a) -> permute3(isZero1(x), x, b)
permute3(false, x, b) -> permute3(ack2(x, x), p1(x), c)
permute3(true, x, b) -> 0
permute3(y, x, c) -> s1(s1(permute3(x, y, a)))
p1(0) -> 0
p1(s1(x)) -> x
ack2(0, x) -> plus2(x, s1(0))
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, ack2(s1(x), y))
plus2(0, y) -> y
plus2(s1(x), y) -> plus2(x, s1(y))
plus2(x, s1(s1(y))) -> s1(plus2(s1(x), y))
plus2(x, s1(0)) -> s1(x)
plus2(x, 0) -> x
isZero1(0) -> true
isZero1(s1(x)) -> false

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.


ACK2(s1(x), s1(y)) -> ACK2(s1(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( ACK2(x1, x2) ) = max{0, 3x1 + x2 - 1}



The following usable rules [14] were oriented: none



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

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

double1(x) -> permute3(x, x, a)
permute3(x, y, a) -> permute3(isZero1(x), x, b)
permute3(false, x, b) -> permute3(ack2(x, x), p1(x), c)
permute3(true, x, b) -> 0
permute3(y, x, c) -> s1(s1(permute3(x, y, a)))
p1(0) -> 0
p1(s1(x)) -> x
ack2(0, x) -> plus2(x, s1(0))
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, ack2(s1(x), y))
plus2(0, y) -> y
plus2(s1(x), y) -> plus2(x, s1(y))
plus2(x, s1(s1(y))) -> s1(plus2(s1(x), y))
plus2(x, s1(0)) -> s1(x)
plus2(x, 0) -> x
isZero1(0) -> true
isZero1(s1(x)) -> false

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:

PERMUTE3(y, x, c) -> PERMUTE3(x, y, a)
PERMUTE3(false, x, b) -> PERMUTE3(ack2(x, x), p1(x), c)
PERMUTE3(x, y, a) -> PERMUTE3(isZero1(x), x, b)

The TRS R consists of the following rules:

double1(x) -> permute3(x, x, a)
permute3(x, y, a) -> permute3(isZero1(x), x, b)
permute3(false, x, b) -> permute3(ack2(x, x), p1(x), c)
permute3(true, x, b) -> 0
permute3(y, x, c) -> s1(s1(permute3(x, y, a)))
p1(0) -> 0
p1(s1(x)) -> x
ack2(0, x) -> plus2(x, s1(0))
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, ack2(s1(x), y))
plus2(0, y) -> y
plus2(s1(x), y) -> plus2(x, s1(y))
plus2(x, s1(s1(y))) -> s1(plus2(s1(x), y))
plus2(x, s1(0)) -> s1(x)
plus2(x, 0) -> x
isZero1(0) -> true
isZero1(s1(x)) -> false

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