Termination w.r.t. Q proof of /tmp/tmpwLbiKP/thiemann15.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))
inc(0) → 0
inc(s(x)) → s(inc(x))
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(x)) → x
bits(x) → bitIter(x, 0)
bitIter(x, y) → if(zero(x), x, inc(y))
if(true, x, y) → p(y)
if(false, x, y) → bitIter(half(x), y)

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))
inc(0) → 0
inc(s(x)) → s(inc(x))
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(x)) → x
bits(x) → bitIter(x, 0)
bitIter(x, y) → if(zero(x), x, inc(y))
if(true, x, y) → p(y)
if(false, x, y) → bitIter(half(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:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
inc(0) → 0
inc(s(x)) → s(inc(x))
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(x)) → x
bits(x) → bitIter(x, 0)
bitIter(x, y) → if(zero(x), x, inc(y))
if(true, x, y) → p(y)
if(false, x, y) → bitIter(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(0)
inc(s(x0))
zero(0)
zero(s(x0))
p(0)
p(s(x0))
bits(x0)
bitIter(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, y) → HALF(x)
BITITER(x, y) → INC(y)
IF(false, x, y) → BITITER(half(x), y)
IF(true, x, y) → P(y)
HALF(s(s(x))) → HALF(x)
BITITER(x, y) → IF(zero(x), x, inc(y))
INC(s(x)) → INC(x)
BITITER(x, y) → ZERO(x)
BITS(x) → BITITER(x, 0)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
inc(0) → 0
inc(s(x)) → s(inc(x))
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(x)) → x
bits(x) → bitIter(x, 0)
bitIter(x, y) → if(zero(x), x, inc(y))
if(true, x, y) → p(y)
if(false, x, y) → bitIter(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(0)
inc(s(x0))
zero(0)
zero(s(x0))
p(0)
p(s(x0))
bits(x0)
bitIter(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, y) → HALF(x)
BITITER(x, y) → INC(y)
IF(false, x, y) → BITITER(half(x), y)
IF(true, x, y) → P(y)
HALF(s(s(x))) → HALF(x)
BITITER(x, y) → IF(zero(x), x, inc(y))
INC(s(x)) → INC(x)
BITITER(x, y) → ZERO(x)
BITS(x) → BITITER(x, 0)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
inc(0) → 0
inc(s(x)) → s(inc(x))
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(x)) → x
bits(x) → bitIter(x, 0)
bitIter(x, y) → if(zero(x), x, inc(y))
if(true, x, y) → p(y)
if(false, x, y) → bitIter(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(0)
inc(s(x0))
zero(0)
zero(s(x0))
p(0)
p(s(x0))
bits(x0)
bitIter(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 5 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:

INC(s(x)) → INC(x)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
inc(0) → 0
inc(s(x)) → s(inc(x))
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(x)) → x
bits(x) → bitIter(x, 0)
bitIter(x, y) → if(zero(x), x, inc(y))
if(true, x, y) → p(y)
if(false, x, y) → bitIter(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(0)
inc(s(x0))
zero(0)
zero(s(x0))
p(0)
p(s(x0))
bits(x0)
bitIter(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:

INC(s(x)) → INC(x)

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

half(0)
half(s(0))
half(s(s(x0)))
inc(0)
inc(s(x0))
zero(0)
zero(s(x0))
p(0)
p(s(x0))
bits(x0)
bitIter(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)))
inc(0)
inc(s(x0))
zero(0)
zero(s(x0))
p(0)
p(s(x0))
bits(x0)
bitIter(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:

INC(s(x)) → INC(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:

INC(s(x)) → INC(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))
inc(0) → 0
inc(s(x)) → s(inc(x))
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(x)) → x
bits(x) → bitIter(x, 0)
bitIter(x, y) → if(zero(x), x, inc(y))
if(true, x, y) → p(y)
if(false, x, y) → bitIter(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(0)
inc(s(x0))
zero(0)
zero(s(x0))
p(0)
p(s(x0))
bits(x0)
bitIter(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)))
inc(0)
inc(s(x0))
zero(0)
zero(s(x0))
p(0)
p(s(x0))
bits(x0)
bitIter(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)))
inc(0)
inc(s(x0))
zero(0)
zero(s(x0))
p(0)
p(s(x0))
bits(x0)
bitIter(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:

IF(false, x, y) → BITITER(half(x), y)
BITITER(x, y) → IF(zero(x), x, inc(y))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
inc(0) → 0
inc(s(x)) → s(inc(x))
zero(0) → true
zero(s(x)) → false
p(0) → 0
p(s(x)) → x
bits(x) → bitIter(x, 0)
bitIter(x, y) → if(zero(x), x, inc(y))
if(true, x, y) → p(y)
if(false, x, y) → bitIter(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(0)
inc(s(x0))
zero(0)
zero(s(x0))
p(0)
p(s(x0))
bits(x0)
bitIter(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:

IF(false, x, y) → BITITER(half(x), y)
BITITER(x, y) → IF(zero(x), x, inc(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(0)
inc(s(x0))
zero(0)
zero(s(x0))
p(0)
p(s(x0))
bits(x0)
bitIter(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.

p(0)
p(s(x0))
bits(x0)
bitIter(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, y) → BITITER(half(x), y)
BITITER(x, y) → IF(zero(x), x, inc(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(0)
inc(s(x0))
zero(0)
zero(s(x0))

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

BITITER(s(x0), y1) → IF(false, s(x0), inc(y1))
BITITER(0, y1) → IF(true, 0, inc(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:

IF(false, x, y) → BITITER(half(x), y)
BITITER(0, y1) → IF(true, 0, inc(y1))
BITITER(s(x0), y1) → IF(false, s(x0), inc(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(0)
inc(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:

IF(false, x, y) → BITITER(half(x), y)
BITITER(s(x0), y1) → IF(false, s(x0), inc(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(0)
inc(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:

IF(false, x, y) → BITITER(half(x), y)
BITITER(s(x0), y1) → IF(false, s(x0), inc(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(0)
inc(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

                                        ↳ RuleRemovalProof

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

IF(false, x, y) → BITITER(half(x), y)
BITITER(s(x0), y1) → IF(false, s(x0), inc(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(0)
inc(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

half(s(0)) → 0

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 1
POL(BITITER(x₁, x₂)) = x₁ + x₂
POL(IF(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(false) = 0
POL(half(x₁)) = x₁
POL(inc(x₁)) = x₁
POL(s(x₁)) = 2·x₁

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ RuleRemovalProof

                                          ↳ QDP

                                            ↳ RuleRemovalProof

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

IF(false, x, y) → BITITER(half(x), y)
BITITER(s(x0), y1) → IF(false, s(x0), inc(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(0)
inc(s(x0))

half(s(s(x))) → s(half(x))

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(BITITER(x₁, x₂)) = x₁ + 2·x₂
POL(IF(x₁, x₂, x₃)) = 2·x₁ + x₂ + 2·x₃
POL(false) = 0
POL(half(x₁)) = x₁
POL(inc(x₁)) = x₁
POL(s(x₁)) = 2 + x₁

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ RuleRemovalProof

                                          ↳ QDP

                                            ↳ RuleRemovalProof

                                              ↳ QDP

                                                ↳ DependencyGraphProof

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

IF(false, x, y) → BITITER(half(x), y)
BITITER(s(x0), y1) → IF(false, s(x0), inc(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(0)
inc(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 2 less nodes.