Termination w.r.t. Q proof of /tmp/tmpFp7z4g/thiemann12.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))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(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))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(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))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(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:

LOG2(x, y) → IF(le(x, s(0)), x, inc(y))
IF(false, x, y) → HALF(x)
LOG2(x, y) → LE(x, s(0))
HALF(s(s(x))) → HALF(x)
LOG(x) → LOG2(x, 0)
LOG2(x, y) → INC(y)
IF(false, x, y) → LOG2(half(x), y)
INC(s(x)) → INC(x)
LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(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:

LOG2(x, y) → IF(le(x, s(0)), x, inc(y))
IF(false, x, y) → HALF(x)
LOG2(x, y) → LE(x, s(0))
HALF(s(s(x))) → HALF(x)
LOG(x) → LOG2(x, 0)
LOG2(x, y) → INC(y)
IF(false, x, y) → LOG2(half(x), y)
INC(s(x)) → INC(x)
LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(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 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:

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))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(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

              ↳ 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)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(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)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(false, 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:

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

              ↳ QDP

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

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(false, 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:

LE(s(x), s(y)) → LE(x, y)

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

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(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)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(false, 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:

LE(s(x), s(y)) → LE(x, y)

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

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

↳ 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:

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))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(false, 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:

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)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(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)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(false, 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:

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

              ↳ QDP

                ↳ UsableRulesProof

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

LOG2(x, y) → IF(le(x, s(0)), x, inc(y))
IF(false, x, y) → LOG2(half(x), y)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(false, 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:

LOG2(x, y) → IF(le(x, s(0)), x, inc(y))
IF(false, x, y) → LOG2(half(x), y)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
le(s(x), 0) → false

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(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.

log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(false, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

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

LOG2(x, y) → IF(le(x, s(0)), x, inc(y))
IF(false, x, y) → LOG2(half(x), y)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
le(s(x), 0) → false

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule LOG2(x, y) → IF(le(x, s(0)), x, inc(y)) at position [0] we obtained the following new rules:

LOG2(0, y1) → IF(true, 0, inc(y1))
LOG2(s(x0), y1) → IF(le(x0, 0), s(x0), inc(y1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

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

LOG2(0, y1) → IF(true, 0, inc(y1))
LOG2(s(x0), y1) → IF(le(x0, 0), s(x0), inc(y1))
IF(false, x, y) → LOG2(half(x), y)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
le(s(x), 0) → false

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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 1 SCC with 1 less node.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

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

LOG2(s(x0), y1) → IF(le(x0, 0), s(x0), inc(y1))
IF(false, x, y) → LOG2(half(x), y)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
le(s(x), 0) → false

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ Narrowing

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

LOG2(s(x0), y1) → IF(le(x0, 0), s(x0), inc(y1))
IF(false, x, y) → LOG2(half(x), y)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → 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)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule LOG2(s(x0), y1) → IF(le(x0, 0), s(x0), inc(y1)) at position [0] we obtained the following new rules:

LOG2(s(0), y1) → IF(true, s(0), inc(y1))
LOG2(s(s(x0)), y1) → IF(false, s(s(x0)), inc(y1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

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

LOG2(s(0), y1) → IF(true, s(0), inc(y1))
LOG2(s(s(x0)), y1) → IF(false, s(s(x0)), inc(y1))
IF(false, x, y) → LOG2(half(x), y)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → 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)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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 1 SCC with 1 less node.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ UsableRulesProof

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

LOG2(s(s(x0)), y1) → IF(false, s(s(x0)), inc(y1))
IF(false, x, y) → LOG2(half(x), y)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → 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)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ UsableRulesProof

                                              ↳ QDP

                                                ↳ QReductionProof

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

LOG2(s(s(x0)), y1) → IF(false, s(s(x0)), inc(y1))
IF(false, x, y) → LOG2(half(x), 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))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(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.

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ UsableRulesProof

                                              ↳ QDP

                                                ↳ QReductionProof

                                                  ↳ QDP

                                                    ↳ Narrowing

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

LOG2(s(s(x0)), y1) → IF(false, s(s(x0)), inc(y1))
IF(false, x, y) → LOG2(half(x), 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))

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 narrowing [15] the rule IF(false, x, y) → LOG2(half(x), y) at position [0] we obtained the following new rules:

IF(false, s(0), y1) → LOG2(0, y1)
IF(false, 0, y1) → LOG2(0, y1)
IF(false, s(s(x0)), y1) → LOG2(s(half(x0)), y1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ UsableRulesProof

                                              ↳ QDP

                                                ↳ QReductionProof

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

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

IF(false, 0, y1) → LOG2(0, y1)
LOG2(s(s(x0)), y1) → IF(false, s(s(x0)), inc(y1))
IF(false, s(0), y1) → LOG2(0, y1)
IF(false, s(s(x0)), y1) → LOG2(s(half(x0)), y1)

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))

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 1 SCC with 2 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ UsableRulesProof

                                              ↳ QDP

                                                ↳ QReductionProof

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ RuleRemovalProof

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

LOG2(s(s(x0)), y1) → IF(false, s(s(x0)), inc(y1))
IF(false, s(s(x0)), y1) → LOG2(s(half(x0)), y1)

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))

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(0) → 0
half(s(0)) → 0

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 1
POL(IF(x₁, x₂, x₃)) = 2·x₁ + 2·x₂ + x₃
POL(LOG2(x₁, x₂)) = 2·x₁ + x₂
POL(false) = 0
POL(half(x₁)) = 2·x₁
POL(inc(x₁)) = x₁
POL(s(x₁)) = 2·x₁

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ UsableRulesProof

                                              ↳ QDP

                                                ↳ QReductionProof

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ RuleRemovalProof

                                                              ↳ QDP

                                                                ↳ RuleRemovalProof

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

LOG2(s(s(x0)), y1) → IF(false, s(s(x0)), inc(y1))
IF(false, s(s(x0)), y1) → LOG2(s(half(x0)), y1)

The TRS R consists of the following rules:

half(s(s(x))) → s(half(x))
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))

half(s(s(x))) → s(half(x))

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(IF(x₁, x₂, x₃)) = x₁ + 2·x₂ + x₃
POL(LOG2(x₁, x₂)) = 2·x₁ + x₂
POL(false) = 0
POL(half(x₁)) = 2 + x₁
POL(inc(x₁)) = x₁
POL(s(x₁)) = 2 + 2·x₁

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ UsableRulesProof

                                              ↳ QDP

                                                ↳ QReductionProof

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ RuleRemovalProof

                                                              ↳ QDP

                                                                ↳ RuleRemovalProof

                                                                  ↳ QDP

                                                                    ↳ DependencyGraphProof

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

LOG2(s(s(x0)), y1) → IF(false, s(s(x0)), inc(y1))
IF(false, s(s(x0)), y1) → LOG2(s(half(x0)), y1)

The TRS R consists of the following rules:

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))

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.