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
  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
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))

The set Q consists of the following terms:

even1(0)
even1(s1(0))
even1(s1(s1(x0)))
half1(0)
half1(s1(s1(x0)))
plus2(0, x0)
plus2(s1(x0), x1)
times2(0, x0)
times2(s1(x0), x1)
if_times3(true, s1(x0), x1)
if_times3(false, s1(x0), x1)


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))

The set Q consists of the following terms:

even1(0)
even1(s1(0))
even1(s1(s1(x0)))
half1(0)
half1(s1(s1(x0)))
plus2(0, x0)
plus2(s1(x0), x1)
times2(0, x0)
times2(s1(x0), x1)
if_times3(true, s1(x0), x1)
if_times3(false, s1(x0), x1)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

even1(0)
even1(s1(0))
even1(s1(s1(x0)))
half1(0)
half1(s1(s1(x0)))
plus2(0, x0)
plus2(s1(x0), x1)
times2(0, x0)
times2(s1(x0), x1)
if_times3(true, s1(x0), x1)
if_times3(false, s1(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 4 SCCs with 4 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
QDP
                ↳ QDPAfsSolverProof
              ↳ 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))

The set Q consists of the following terms:

even1(0)
even1(s1(0))
even1(s1(s1(x0)))
half1(0)
half1(s1(s1(x0)))
plus2(0, x0)
plus2(s1(x0), x1)
times2(0, x0)
times2(s1(x0), x1)
if_times3(true, s1(x0), x1)
if_times3(false, s1(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PLUS2(s1(x), y) -> PLUS2(x, y)
Used argument filtering: PLUS2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
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))

The set Q consists of the following terms:

even1(0)
even1(s1(0))
even1(s1(s1(x0)))
half1(0)
half1(s1(s1(x0)))
plus2(0, x0)
plus2(s1(x0), x1)
times2(0, x0)
times2(s1(x0), x1)
if_times3(true, s1(x0), x1)
if_times3(false, s1(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
QDP
                ↳ QDPAfsSolverProof
              ↳ 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))

The set Q consists of the following terms:

even1(0)
even1(s1(0))
even1(s1(s1(x0)))
half1(0)
half1(s1(s1(x0)))
plus2(0, x0)
plus2(s1(x0), x1)
times2(0, x0)
times2(s1(x0), x1)
if_times3(true, s1(x0), x1)
if_times3(false, s1(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

HALF1(s1(s1(x))) -> HALF1(x)
Used argument filtering: HALF1(x1)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
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))

The set Q consists of the following terms:

even1(0)
even1(s1(0))
even1(s1(s1(x0)))
half1(0)
half1(s1(s1(x0)))
plus2(0, x0)
plus2(s1(x0), x1)
times2(0, x0)
times2(s1(x0), x1)
if_times3(true, s1(x0), x1)
if_times3(false, s1(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
QDP
                ↳ QDPAfsSolverProof
              ↳ 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))

The set Q consists of the following terms:

even1(0)
even1(s1(0))
even1(s1(s1(x0)))
half1(0)
half1(s1(s1(x0)))
plus2(0, x0)
plus2(s1(x0), x1)
times2(0, x0)
times2(s1(x0), x1)
if_times3(true, s1(x0), x1)
if_times3(false, s1(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

EVEN1(s1(s1(x))) -> EVEN1(x)
Used argument filtering: EVEN1(x1)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
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))

The set Q consists of the following terms:

even1(0)
even1(s1(0))
even1(s1(s1(x0)))
half1(0)
half1(s1(s1(x0)))
plus2(0, x0)
plus2(s1(x0), x1)
times2(0, x0)
times2(s1(x0), x1)
if_times3(true, s1(x0), x1)
if_times3(false, s1(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
QDP
                ↳ QDPAfsSolverProof

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))

The set Q consists of the following terms:

even1(0)
even1(s1(0))
even1(s1(s1(x0)))
half1(0)
half1(s1(s1(x0)))
plus2(0, x0)
plus2(s1(x0), x1)
times2(0, x0)
times2(s1(x0), x1)
if_times3(true, s1(x0), x1)
if_times3(false, s1(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

IF_TIMES3(false, s1(x), y) -> TIMES2(x, y)
Used argument filtering: TIMES2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
IF_TIMES3(x1, x2, x3)  =  x2
half1(x1)  =  x1
even1(x1)  =  even
0  =  0
false  =  false
true  =  true
Used ordering: Quasi Precedence: [even, false, true]


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
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))

The set Q consists of the following terms:

even1(0)
even1(s1(0))
even1(s1(s1(x0)))
half1(0)
half1(s1(s1(x0)))
plus2(0, x0)
plus2(s1(x0), x1)
times2(0, x0)
times2(s1(x0), x1)
if_times3(true, s1(x0), x1)
if_times3(false, s1(x0), x1)

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