Termination w.r.t. Q proof of /tmp/tmpa6nv9

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:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
logarithm(x) → ifa(lt(0, x), x)
ifa(true, x) → help(x, 1)
ifa(false, x) → logZeroError
help(x, y) → ifb(lt(y, x), x, y)
ifb(true, x, y) → help(half(x), s(y))
ifb(false, x, y) → y
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
logarithm(x) → ifa(lt(0, x), x)
ifa(true, x) → help(x, 1)
ifa(false, x) → logZeroError
help(x, y) → ifb(lt(y, x), x, y)
ifb(true, x, y) → help(half(x), s(y))
ifb(false, x, y) → y
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))

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:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
logarithm(x) → ifa(lt(0, x), x)
ifa(true, x) → help(x, 1)
ifa(false, x) → logZeroError
help(x, y) → ifb(lt(y, x), x, y)
ifb(true, x, y) → help(half(x), s(y))
ifb(false, x, y) → y
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(s(s(x0)))

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

LT(s(x), s(y)) → LT(x, y)
HALF(s(s(x))) → HALF(x)
HELP(x, y) → LT(y, x)
LOGARITHM(x) → LT(0, x)
IFA(true, x) → HELP(x, 1)
IFB(true, x, y) → HALF(x)
IFB(true, x, y) → HELP(half(x), s(y))
HELP(x, y) → IFB(lt(y, x), x, y)
LOGARITHM(x) → IFA(lt(0, x), x)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
logarithm(x) → ifa(lt(0, x), x)
ifa(true, x) → help(x, 1)
ifa(false, x) → logZeroError
help(x, y) → ifb(lt(y, x), x, y)
ifb(true, x, y) → help(half(x), s(y))
ifb(false, x, y) → y
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(s(s(x0)))

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:

LT(s(x), s(y)) → LT(x, y)
HALF(s(s(x))) → HALF(x)
HELP(x, y) → LT(y, x)
LOGARITHM(x) → LT(0, x)
IFA(true, x) → HELP(x, 1)
IFB(true, x, y) → HALF(x)
IFB(true, x, y) → HELP(half(x), s(y))
HELP(x, y) → IFB(lt(y, x), x, y)
LOGARITHM(x) → IFA(lt(0, x), x)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
logarithm(x) → ifa(lt(0, x), x)
ifa(true, x) → help(x, 1)
ifa(false, x) → logZeroError
help(x, y) → ifb(lt(y, x), x, y)
ifb(true, x, y) → help(half(x), s(y))
ifb(false, x, y) → y
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(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 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:

HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
logarithm(x) → ifa(lt(0, x), x)
ifa(true, x) → help(x, 1)
ifa(false, x) → logZeroError
help(x, y) → ifb(lt(y, x), x, y)
ifb(true, x, y) → help(half(x), s(y))
ifb(false, x, y) → y
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(s(s(x0)))

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:

HALF(s(s(x))) → HALF(x)

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

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(s(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.

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(s(s(x0)))

↳ 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:

HALF(s(s(x))) → HALF(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:

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

                ↳ UsableRulesProof

              ↳ QDP

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

LT(s(x), s(y)) → LT(x, y)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
logarithm(x) → ifa(lt(0, x), x)
ifa(true, x) → help(x, 1)
ifa(false, x) → logZeroError
help(x, y) → ifb(lt(y, x), x, y)
ifb(true, x, y) → help(half(x), s(y))
ifb(false, x, y) → y
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(s(s(x0)))

↳ 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:

LT(s(x), s(y)) → LT(x, y)

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

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(s(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.

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(s(s(x0)))

↳ 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:

LT(s(x), s(y)) → LT(x, y)

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

LT(s(x), s(y)) → LT(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

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

IFB(true, x, y) → HELP(half(x), s(y))
HELP(x, y) → IFB(lt(y, x), x, y)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
logarithm(x) → ifa(lt(0, x), x)
ifa(true, x) → help(x, 1)
ifa(false, x) → logZeroError
help(x, y) → ifb(lt(y, x), x, y)
ifb(true, x, y) → help(half(x), s(y))
ifb(false, x, y) → y
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(s(s(x0)))

↳ 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:

IFB(true, x, y) → HELP(half(x), s(y))
HELP(x, y) → IFB(lt(y, x), x, y)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)
half(0)
half(s(0))
half(s(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.

logarithm(x0)
ifa(true, x0)
ifa(false, x0)
help(x0, x1)
ifb(true, x0, x1)
ifb(false, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

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

HELP(x, y) → IFB(lt(y, x), x, y)
IFB(true, x, y) → HELP(half(x), s(y))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
half(0)
half(s(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.

IFB(true, x, y) → HELP(half(x), s(y))

The remaining pairs can at least be oriented weakly.

HELP(x, y) → IFB(lt(y, x), x, y)

Used ordering: Polynomial interpretation [25,35]:

POL(half(x₁)) = (1/2)x_1
POL(true) = 1/4
POL(false) = 0
POL(s(x₁)) = 4 + (2)x_1
POL(IFB(x₁, x₂, x₃)) = (1/4)x_1 + (1/2)x_2
POL(lt(x₁, x₂)) = (1/4)x_2
POL(HELP(x₁, x₂)) = x_1
POL(0) = 0

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

lt(s(x), s(y)) → lt(x, y)
half(0) → 0
lt(0, s(x)) → true
lt(x, 0) → false
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

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ DependencyGraphProof

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

HELP(x, y) → IFB(lt(y, x), x, y)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
half(0)
half(s(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.