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:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

Q is empty.


QTRS
  ↳ Overlay + Local Confluence

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

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
QTRS
      ↳ DependencyPairsProof

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

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

EVEN(s(s(x))) → EVEN(x)
TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)
PLUS(s(x), y) → PLUS(x, y)
IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)
IF_TIMES(true, s(x), y) → HALF(s(x))
IF_TIMES(true, s(x), y) → PLUS(times(half(s(x)), y), times(half(s(x)), y))
IF_TIMES(false, s(x), y) → PLUS(y, times(x, y))
TIMES(s(x), y) → EVEN(s(x))
IF_TIMES(false, s(x), y) → TIMES(x, y)
HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ EdgeDeletionProof

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

EVEN(s(s(x))) → EVEN(x)
TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)
PLUS(s(x), y) → PLUS(x, y)
IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)
IF_TIMES(true, s(x), y) → HALF(s(x))
IF_TIMES(true, s(x), y) → PLUS(times(half(s(x)), y), times(half(s(x)), y))
IF_TIMES(false, s(x), y) → PLUS(y, times(x, y))
TIMES(s(x), y) → EVEN(s(x))
IF_TIMES(false, s(x), y) → TIMES(x, y)
HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
QDP
              ↳ DependencyGraphProof

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

EVEN(s(s(x))) → EVEN(x)
PLUS(s(x), y) → PLUS(x, y)
TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)
IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)
IF_TIMES(true, s(x), y) → HALF(s(x))
IF_TIMES(true, s(x), y) → PLUS(times(half(s(x)), y), times(half(s(x)), y))
TIMES(s(x), y) → EVEN(s(x))
IF_TIMES(false, s(x), y) → PLUS(y, times(x, y))
HALF(s(s(x))) → HALF(x)
IF_TIMES(false, s(x), y) → TIMES(x, y)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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

PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

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.


PLUS(s(x), y) → PLUS(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2)  =  x1
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP

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

HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

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.


HALF(s(s(x))) → HALF(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
HALF(x1)  =  x1
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP

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

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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

EVEN(s(s(x))) → EVEN(x)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

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.


EVEN(s(s(x))) → EVEN(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
EVEN(x1)  =  x1
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof
                  ↳ QDP

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

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
QDP
                    ↳ QDPOrderProof

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

TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)
IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)
IF_TIMES(false, s(x), y) → TIMES(x, y)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

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_TIMES(false, s(x), y) → TIMES(x, y)
The remaining pairs can at least be oriented weakly.

TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)
IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)
Used ordering: Combined order from the following AFS and order.
TIMES(x1, x2)  =  x1
s(x1)  =  s(x1)
IF_TIMES(x1, x2, x3)  =  x2
half(x1)  =  x1
0  =  0

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented:

half(s(s(x))) → s(half(x))
half(0) → 0



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
QDP

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

TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)
IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
half(0) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → if_times(even(s(x)), s(x), y)
if_times(true, s(x), y) → plus(times(half(s(x)), y), times(half(s(x)), y))
if_times(false, s(x), y) → plus(y, times(x, y))

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
if_times(true, s(x0), x1)
if_times(false, s(x0), x1)

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