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:

even1(0) -> true
even1(s1(0)) -> false
even1(s1(s1(x))) -> even1(x)
half1(0) -> 0
half1(s1(s1(x))) -> s1(half1(x))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(x), y) -> if_times3(even1(s1(x)), s1(x), y)
if_times3(true, s1(x), y) -> plus2(times2(half1(s1(x)), y), times2(half1(s1(x)), y))
if_times3(false, s1(x), y) -> plus2(y, times2(x, y))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

even1(0) -> true
even1(s1(0)) -> false
even1(s1(s1(x))) -> even1(x)
half1(0) -> 0
half1(s1(s1(x))) -> s1(half1(x))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(x), y) -> if_times3(even1(s1(x)), s1(x), y)
if_times3(true, s1(x), y) -> plus2(times2(half1(s1(x)), y), times2(half1(s1(x)), y))
if_times3(false, s1(x), y) -> plus2(y, times2(x, 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:

EVEN1(s1(s1(x))) -> EVEN1(x)
TIMES2(s1(x), y) -> IF_TIMES3(even1(s1(x)), s1(x), y)
PLUS2(s1(x), y) -> PLUS2(x, y)
IF_TIMES3(true, s1(x), y) -> TIMES2(half1(s1(x)), y)
IF_TIMES3(true, s1(x), y) -> HALF1(s1(x))
IF_TIMES3(true, s1(x), y) -> PLUS2(times2(half1(s1(x)), y), times2(half1(s1(x)), y))
IF_TIMES3(false, s1(x), y) -> PLUS2(y, times2(x, y))
TIMES2(s1(x), y) -> EVEN1(s1(x))
IF_TIMES3(false, s1(x), y) -> TIMES2(x, y)
HALF1(s1(s1(x))) -> HALF1(x)

The TRS R consists of the following rules:

even1(0) -> true
even1(s1(0)) -> false
even1(s1(s1(x))) -> even1(x)
half1(0) -> 0
half1(s1(s1(x))) -> s1(half1(x))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(x), y) -> if_times3(even1(s1(x)), s1(x), y)
if_times3(true, s1(x), y) -> plus2(times2(half1(s1(x)), y), times2(half1(s1(x)), y))
if_times3(false, s1(x), y) -> plus2(y, times2(x, 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:

EVEN1(s1(s1(x))) -> EVEN1(x)
TIMES2(s1(x), y) -> IF_TIMES3(even1(s1(x)), s1(x), y)
PLUS2(s1(x), y) -> PLUS2(x, y)
IF_TIMES3(true, s1(x), y) -> TIMES2(half1(s1(x)), y)
IF_TIMES3(true, s1(x), y) -> HALF1(s1(x))
IF_TIMES3(true, s1(x), y) -> PLUS2(times2(half1(s1(x)), y), times2(half1(s1(x)), y))
IF_TIMES3(false, s1(x), y) -> PLUS2(y, times2(x, y))
TIMES2(s1(x), y) -> EVEN1(s1(x))
IF_TIMES3(false, s1(x), y) -> TIMES2(x, y)
HALF1(s1(s1(x))) -> HALF1(x)

The TRS R consists of the following rules:

even1(0) -> true
even1(s1(0)) -> false
even1(s1(s1(x))) -> even1(x)
half1(0) -> 0
half1(s1(s1(x))) -> s1(half1(x))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(x), y) -> if_times3(even1(s1(x)), s1(x), y)
if_times3(true, s1(x), y) -> plus2(times2(half1(s1(x)), y), times2(half1(s1(x)), y))
if_times3(false, s1(x), y) -> plus2(y, times2(x, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 4 less nodes.

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

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

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

The TRS R consists of the following rules:

even1(0) -> true
even1(s1(0)) -> false
even1(s1(s1(x))) -> even1(x)
half1(0) -> 0
half1(s1(s1(x))) -> s1(half1(x))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(x), y) -> if_times3(even1(s1(x)), s1(x), y)
if_times3(true, s1(x), y) -> plus2(times2(half1(s1(x)), y), times2(half1(s1(x)), y))
if_times3(false, s1(x), y) -> plus2(y, times2(x, 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.


PLUS2(s1(x), y) -> PLUS2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(PLUS2(x1, x2)) = 2·x1   
POL(s1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



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

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

even1(0) -> true
even1(s1(0)) -> false
even1(s1(s1(x))) -> even1(x)
half1(0) -> 0
half1(s1(s1(x))) -> s1(half1(x))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(x), y) -> if_times3(even1(s1(x)), s1(x), y)
if_times3(true, s1(x), y) -> plus2(times2(half1(s1(x)), y), times2(half1(s1(x)), y))
if_times3(false, s1(x), y) -> plus2(y, times2(x, 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

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

HALF1(s1(s1(x))) -> HALF1(x)

The TRS R consists of the following rules:

even1(0) -> true
even1(s1(0)) -> false
even1(s1(s1(x))) -> even1(x)
half1(0) -> 0
half1(s1(s1(x))) -> s1(half1(x))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(x), y) -> if_times3(even1(s1(x)), s1(x), y)
if_times3(true, s1(x), y) -> plus2(times2(half1(s1(x)), y), times2(half1(s1(x)), y))
if_times3(false, s1(x), y) -> plus2(y, times2(x, 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.


HALF1(s1(s1(x))) -> HALF1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(HALF1(x1)) = 2·x12   
POL(s1(x1)) = 2 + 2·x12   

The following usable rules [14] were oriented: none



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

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

even1(0) -> true
even1(s1(0)) -> false
even1(s1(s1(x))) -> even1(x)
half1(0) -> 0
half1(s1(s1(x))) -> s1(half1(x))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(x), y) -> if_times3(even1(s1(x)), s1(x), y)
if_times3(true, s1(x), y) -> plus2(times2(half1(s1(x)), y), times2(half1(s1(x)), y))
if_times3(false, s1(x), y) -> plus2(y, times2(x, 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

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

EVEN1(s1(s1(x))) -> EVEN1(x)

The TRS R consists of the following rules:

even1(0) -> true
even1(s1(0)) -> false
even1(s1(s1(x))) -> even1(x)
half1(0) -> 0
half1(s1(s1(x))) -> s1(half1(x))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(x), y) -> if_times3(even1(s1(x)), s1(x), y)
if_times3(true, s1(x), y) -> plus2(times2(half1(s1(x)), y), times2(half1(s1(x)), y))
if_times3(false, s1(x), y) -> plus2(y, times2(x, 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.


EVEN1(s1(s1(x))) -> EVEN1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(EVEN1(x1)) = 2·x12   
POL(s1(x1)) = 2 + 2·x12   

The following usable rules [14] were oriented: none



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

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

even1(0) -> true
even1(s1(0)) -> false
even1(s1(s1(x))) -> even1(x)
half1(0) -> 0
half1(s1(s1(x))) -> s1(half1(x))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(x), y) -> if_times3(even1(s1(x)), s1(x), y)
if_times3(true, s1(x), y) -> plus2(times2(half1(s1(x)), y), times2(half1(s1(x)), y))
if_times3(false, s1(x), y) -> plus2(y, times2(x, 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

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

TIMES2(s1(x), y) -> IF_TIMES3(even1(s1(x)), s1(x), y)
IF_TIMES3(true, s1(x), y) -> TIMES2(half1(s1(x)), y)
IF_TIMES3(false, s1(x), y) -> TIMES2(x, y)

The TRS R consists of the following rules:

even1(0) -> true
even1(s1(0)) -> false
even1(s1(s1(x))) -> even1(x)
half1(0) -> 0
half1(s1(s1(x))) -> s1(half1(x))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(x), y) -> if_times3(even1(s1(x)), s1(x), y)
if_times3(true, s1(x), y) -> plus2(times2(half1(s1(x)), y), times2(half1(s1(x)), y))
if_times3(false, s1(x), y) -> plus2(y, times2(x, 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_TIMES3(false, s1(x), y) -> TIMES2(x, y)
The remaining pairs can at least be oriented weakly.

TIMES2(s1(x), y) -> IF_TIMES3(even1(s1(x)), s1(x), y)
IF_TIMES3(true, s1(x), y) -> TIMES2(half1(s1(x)), y)
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(IF_TIMES3(x1, x2, x3)) = x2   
POL(TIMES2(x1, x2)) = x1   
POL(even1(x1)) = 0   
POL(false) = 0   
POL(half1(x1)) = x1   
POL(s1(x1)) = 1 + 2·x1   
POL(true) = 0   

The following usable rules [14] were oriented:

half1(0) -> 0
half1(s1(s1(x))) -> s1(half1(x))



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

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

TIMES2(s1(x), y) -> IF_TIMES3(even1(s1(x)), s1(x), y)
IF_TIMES3(true, s1(x), y) -> TIMES2(half1(s1(x)), y)

The TRS R consists of the following rules:

even1(0) -> true
even1(s1(0)) -> false
even1(s1(s1(x))) -> even1(x)
half1(0) -> 0
half1(s1(s1(x))) -> s1(half1(x))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(x), y) -> if_times3(even1(s1(x)), s1(x), y)
if_times3(true, s1(x), y) -> plus2(times2(half1(s1(x)), y), times2(half1(s1(x)), y))
if_times3(false, s1(x), y) -> plus2(y, times2(x, y))

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