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:

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
exp2(x, 0) -> s1(0)
exp2(x, s1(y)) -> times2(x, exp2(x, y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
tower2(x, y) -> towerIter3(x, y, s1(0))
towerIter3(0, y, z) -> z
towerIter3(s1(x), y, z) -> towerIter3(p1(s1(x)), y, exp2(y, z))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
exp2(x, 0) -> s1(0)
exp2(x, s1(y)) -> times2(x, exp2(x, y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
tower2(x, y) -> towerIter3(x, y, s1(0))
towerIter3(0, y, z) -> z
towerIter3(s1(x), y, z) -> towerIter3(p1(s1(x)), y, exp2(y, z))

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:

TOWERITER3(s1(x), y, z) -> P1(s1(x))
TOWERITER3(s1(x), y, z) -> TOWERITER3(p1(s1(x)), y, exp2(y, z))
PLUS2(s1(x), y) -> PLUS2(p1(s1(x)), y)
PLUS2(s1(x), y) -> P1(s1(x))
P1(s1(s1(x))) -> P1(s1(x))
TOWER2(x, y) -> TOWERITER3(x, y, s1(0))
TIMES2(s1(x), y) -> TIMES2(p1(s1(x)), y)
TIMES2(s1(x), y) -> P1(s1(x))
TOWERITER3(s1(x), y, z) -> EXP2(y, z)
EXP2(x, s1(y)) -> EXP2(x, y)
EXP2(x, s1(y)) -> TIMES2(x, exp2(x, y))
TIMES2(s1(x), y) -> PLUS2(y, times2(p1(s1(x)), y))

The TRS R consists of the following rules:

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
exp2(x, 0) -> s1(0)
exp2(x, s1(y)) -> times2(x, exp2(x, y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
tower2(x, y) -> towerIter3(x, y, s1(0))
towerIter3(0, y, z) -> z
towerIter3(s1(x), y, z) -> towerIter3(p1(s1(x)), y, exp2(y, z))

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:

TOWERITER3(s1(x), y, z) -> P1(s1(x))
TOWERITER3(s1(x), y, z) -> TOWERITER3(p1(s1(x)), y, exp2(y, z))
PLUS2(s1(x), y) -> PLUS2(p1(s1(x)), y)
PLUS2(s1(x), y) -> P1(s1(x))
P1(s1(s1(x))) -> P1(s1(x))
TOWER2(x, y) -> TOWERITER3(x, y, s1(0))
TIMES2(s1(x), y) -> TIMES2(p1(s1(x)), y)
TIMES2(s1(x), y) -> P1(s1(x))
TOWERITER3(s1(x), y, z) -> EXP2(y, z)
EXP2(x, s1(y)) -> EXP2(x, y)
EXP2(x, s1(y)) -> TIMES2(x, exp2(x, y))
TIMES2(s1(x), y) -> PLUS2(y, times2(p1(s1(x)), y))

The TRS R consists of the following rules:

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
exp2(x, 0) -> s1(0)
exp2(x, s1(y)) -> times2(x, exp2(x, y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
tower2(x, y) -> towerIter3(x, y, s1(0))
towerIter3(0, y, z) -> z
towerIter3(s1(x), y, z) -> towerIter3(p1(s1(x)), y, exp2(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 5 SCCs with 7 less nodes.

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

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
exp2(x, 0) -> s1(0)
exp2(x, s1(y)) -> times2(x, exp2(x, y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
tower2(x, y) -> towerIter3(x, y, s1(0))
towerIter3(0, y, z) -> z
towerIter3(s1(x), y, z) -> towerIter3(p1(s1(x)), y, exp2(y, z))

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 - 2}


POL( s1(x1) ) = x1 + 2



The following usable rules [14] were oriented: none



↳ 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:

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
exp2(x, 0) -> s1(0)
exp2(x, s1(y)) -> times2(x, exp2(x, y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
tower2(x, y) -> towerIter3(x, y, s1(0))
towerIter3(0, y, z) -> z
towerIter3(s1(x), y, z) -> towerIter3(p1(s1(x)), y, exp2(y, z))

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

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

PLUS2(s1(x), y) -> PLUS2(p1(s1(x)), y)

The TRS R consists of the following rules:

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
exp2(x, 0) -> s1(0)
exp2(x, s1(y)) -> times2(x, exp2(x, y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
tower2(x, y) -> towerIter3(x, y, s1(0))
towerIter3(0, y, z) -> z
towerIter3(s1(x), y, z) -> towerIter3(p1(s1(x)), y, exp2(y, z))

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(p1(s1(x)), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( PLUS2(x1, x2) ) = max{0, x1 - 1}


POL( s1(x1) ) = x1 + 2


POL( p1(x1) ) = max{0, x1 - 2}


POL( 0 ) = 0



The following usable rules [14] were oriented:

p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0



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

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

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
exp2(x, 0) -> s1(0)
exp2(x, s1(y)) -> times2(x, exp2(x, y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
tower2(x, y) -> towerIter3(x, y, s1(0))
towerIter3(0, y, z) -> z
towerIter3(s1(x), y, z) -> towerIter3(p1(s1(x)), y, exp2(y, z))

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

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

TIMES2(s1(x), y) -> TIMES2(p1(s1(x)), y)

The TRS R consists of the following rules:

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
exp2(x, 0) -> s1(0)
exp2(x, s1(y)) -> times2(x, exp2(x, y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
tower2(x, y) -> towerIter3(x, y, s1(0))
towerIter3(0, y, z) -> z
towerIter3(s1(x), y, z) -> towerIter3(p1(s1(x)), y, exp2(y, z))

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.


TIMES2(s1(x), y) -> TIMES2(p1(s1(x)), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( TIMES2(x1, x2) ) = max{0, x1 - 1}


POL( s1(x1) ) = x1 + 2


POL( p1(x1) ) = max{0, x1 - 2}


POL( 0 ) = 0



The following usable rules [14] were oriented:

p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0



↳ 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:

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
exp2(x, 0) -> s1(0)
exp2(x, s1(y)) -> times2(x, exp2(x, y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
tower2(x, y) -> towerIter3(x, y, s1(0))
towerIter3(0, y, z) -> z
towerIter3(s1(x), y, z) -> towerIter3(p1(s1(x)), y, exp2(y, z))

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

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

EXP2(x, s1(y)) -> EXP2(x, y)

The TRS R consists of the following rules:

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
exp2(x, 0) -> s1(0)
exp2(x, s1(y)) -> times2(x, exp2(x, y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
tower2(x, y) -> towerIter3(x, y, s1(0))
towerIter3(0, y, z) -> z
towerIter3(s1(x), y, z) -> towerIter3(p1(s1(x)), y, exp2(y, z))

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.


EXP2(x, s1(y)) -> EXP2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( EXP2(x1, x2) ) = max{0, x2 - 1}


POL( s1(x1) ) = x1 + 2



The following usable rules [14] were oriented: none



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

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

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
exp2(x, 0) -> s1(0)
exp2(x, s1(y)) -> times2(x, exp2(x, y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
tower2(x, y) -> towerIter3(x, y, s1(0))
towerIter3(0, y, z) -> z
towerIter3(s1(x), y, z) -> towerIter3(p1(s1(x)), y, exp2(y, z))

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

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

TOWERITER3(s1(x), y, z) -> TOWERITER3(p1(s1(x)), y, exp2(y, z))

The TRS R consists of the following rules:

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
exp2(x, 0) -> s1(0)
exp2(x, s1(y)) -> times2(x, exp2(x, y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
tower2(x, y) -> towerIter3(x, y, s1(0))
towerIter3(0, y, z) -> z
towerIter3(s1(x), y, z) -> towerIter3(p1(s1(x)), y, exp2(y, z))

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.


TOWERITER3(s1(x), y, z) -> TOWERITER3(p1(s1(x)), y, exp2(y, z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( TOWERITER3(x1, ..., x3) ) = max{0, x1 - 1}


POL( s1(x1) ) = x1 + 2


POL( p1(x1) ) = max{0, x1 - 2}


POL( 0 ) = 0



The following usable rules [14] were oriented:

p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0



↳ 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:

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
exp2(x, 0) -> s1(0)
exp2(x, s1(y)) -> times2(x, exp2(x, y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
tower2(x, y) -> towerIter3(x, y, s1(0))
towerIter3(0, y, z) -> z
towerIter3(s1(x), y, z) -> towerIter3(p1(s1(x)), y, exp2(y, z))

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.