Termination w.r.t. Q proof of /tmp/tmpFXhOch/#4.29.trs

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:

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 [19] 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 [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

TIMES(s(x), y) → IF_TIMES(even(s(x)), s(x), y)
HALF(s(s(x))) → HALF(x)
EVEN(s(s(x))) → EVEN(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) → TIMES(x, y)
IF_TIMES(true, s(x), y) → TIMES(half(s(x)), y)
IF_TIMES(true, s(x), y) → HALF(s(x))
IF_TIMES(false, s(x), y) → PLUS(y, times(x, y))
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.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 4 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ 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.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

PLUS(s(x), y) → PLUS(x, y)

R is empty.
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 the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

PLUS(s(x), y) → PLUS(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

PLUS(s(x), y) → PLUS(x, y)
The graph contains the following edges 1 > 1, 2 >= 2

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ 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)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

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

HALF(s(s(x))) → HALF(x)

R is empty.
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 the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

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

HALF(s(s(x))) → HALF(x)

From the DPs we obtained the following set of size-change graphs:

HALF(s(s(x))) → HALF(x)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ 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)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

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

EVEN(s(s(x))) → EVEN(x)

R is empty.
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 the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

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

EVEN(s(s(x))) → EVEN(x)

From the DPs we obtained the following set of size-change graphs:

EVEN(s(s(x))) → EVEN(x)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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(false, s(x), y) → TIMES(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)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

The TRS R consists of the following rules:

half(s(s(x))) → s(half(x))
half(0) → 0
even(s(0)) → false
even(s(s(x))) → even(x)
even(0) → true

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 the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

The TRS R consists of the following rules:

half(s(s(x))) → s(half(x))
half(0) → 0
even(s(0)) → false
even(s(s(x))) → even(x)
even(0) → true

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

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: Polynomial interpretation with max and min functions [25]:

POL(0) = 0
POL(IF_TIMES(x₁, x₂, x₃)) = x₂
POL(TIMES(x₁, x₂)) = x₁
POL(even(x₁)) = 0
POL(false) = 0
POL(half(x₁)) = x₁
POL(s(x₁)) = 1 + x₁
POL(true) = 0

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ 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)

The TRS R consists of the following rules:

half(s(s(x))) → s(half(x))
half(0) → 0
even(s(0)) → false
even(s(s(x))) → even(x)
even(0) → true

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Matrix interpretation [3]:
Non-tuple symbols:

M( half(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	1	0	/

x₁

M( even(x₁) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

M( true ) =

/	0	\
\	0	/

M( false ) =

/	0	\
\	0	/

M( s(x₁) ) =

/	0	\
\	1	/

/	1	1	\
\	1	1	/

x₁

M( 0 ) =

/	1	\
\	0	/

Tuple symbols:

M( TIMES(x₁, x₂) ) =

[

]

x₁

[

]

x₂

M( IF_TIMES(x₁, ..., x₃) ) =

[

]

x₁

[

]

x₂

[

]

x₃

Matrix type:
We used a basic matrix type which is not further parametrizeable.

As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

half(s(s(x))) → s(half(x))
half(0) → 0
even(s(0)) → false
even(s(s(x))) → even(x)
even(0) → true

The set Q consists of the following terms:

even(0)
even(s(0))
even(s(s(x0)))
half(0)
half(s(s(x0)))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.