Termination w.r.t. Q proof of /tmp/tmpSTcOdb/thiemann10.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:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), l))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), l))

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:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), l))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(x0, x1)
if(true, x0, x1)
if(false, 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:

IF(false, x, l) → HALF(x)
HALF(s(s(x))) → HALF(x)
CONVITER(x, l) → ZERO(x)
LASTBIT(s(s(x))) → LASTBIT(x)
IF(false, x, l) → LASTBIT(x)
CONV(x) → CONVITER(x, cons(0, nil))
IF(false, x, l) → CONVITER(half(x), cons(lastbit(x), l))
CONVITER(x, l) → IF(zero(x), x, l)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), l))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(x0, x1)
if(true, x0, x1)
if(false, 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:

IF(false, x, l) → HALF(x)
HALF(s(s(x))) → HALF(x)
CONVITER(x, l) → ZERO(x)
LASTBIT(s(s(x))) → LASTBIT(x)
IF(false, x, l) → LASTBIT(x)
CONV(x) → CONVITER(x, cons(0, nil))
IF(false, x, l) → CONVITER(half(x), cons(lastbit(x), l))
CONVITER(x, l) → IF(zero(x), x, l)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), l))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 4 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

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

LASTBIT(s(s(x))) → LASTBIT(x)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), l))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(x0, x1)
if(true, x0, x1)
if(false, 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

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

LASTBIT(s(s(x))) → LASTBIT(x)

R is empty.
The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(x0, x1)
if(true, x0, x1)
if(false, 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.

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

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

LASTBIT(s(s(x))) → LASTBIT(x)

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:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ 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:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), l))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ 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:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(x0, x1)
if(true, x0, x1)
if(false, 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.

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ 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

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

CONVITER(x, l) → IF(zero(x), x, l)
IF(false, x, l) → CONVITER(half(x), cons(lastbit(x), l))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), l))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

CONVITER(x, l) → IF(zero(x), x, l)
IF(false, x, l) → CONVITER(half(x), cons(lastbit(x), l))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(x0, x1)
if(true, x0, x1)
if(false, 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.

conv(x0)
conviter(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

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

IF(false, x, l) → CONVITER(half(x), cons(lastbit(x), l))
CONVITER(x, l) → IF(zero(x), x, l)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule CONVITER(x, l) → IF(zero(x), x, l) at position [0] we obtained the following new rules:

CONVITER(0, y1) → IF(true, 0, y1)
CONVITER(s(x0), y1) → IF(false, s(x0), y1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

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

CONVITER(0, y1) → IF(true, 0, y1)
CONVITER(s(x0), y1) → IF(false, s(x0), y1)
IF(false, x, l) → CONVITER(half(x), cons(lastbit(x), l))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

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

CONVITER(s(x0), y1) → IF(false, s(x0), y1)
IF(false, x, l) → CONVITER(half(x), cons(lastbit(x), l))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

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

CONVITER(s(x0), y1) → IF(false, s(x0), y1)
IF(false, x, l) → CONVITER(half(x), cons(lastbit(x), l))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))

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.

zero(0)
zero(s(x0))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ QDPOrderProof

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

CONVITER(s(x0), y1) → IF(false, s(x0), y1)
IF(false, x, l) → CONVITER(half(x), cons(lastbit(x), l))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(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.

CONVITER(s(x0), y1) → IF(false, s(x0), y1)

The remaining pairs can at least be oriented weakly.

IF(false, x, l) → CONVITER(half(x), cons(lastbit(x), l))

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

M( cons(x₁, x₂) ) =

/	0	\
\	0	/

/	1	1	\
\	1	1	/

x₁

/	0	0	\
\	0	0	/

x₂

M( half(x₁) ) =

/	0	\
\	0	/

/	0	1	\
\	1	0	/

x₁

M( false ) =

/	0	\
\	0	/

M( s(x₁) ) =

/	0	\
\	1	/

/	1	1	\
\	1	1	/

x₁

M( lastbit(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	1	1	/

x₁

M( 0 ) =

/	0	\
\	0	/

Tuple symbols:

M( CONVITER(x₁, x₂) ) =

[

]

x₁

[

]

x₂

M( IF(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(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ QDPOrderProof

                                          ↳ QDP

                                            ↳ DependencyGraphProof

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

IF(false, x, l) → CONVITER(half(x), cons(lastbit(x), l))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(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.