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

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

Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

  ↳ QTRS Reverse

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

Q is empty.

The following Q TRS is given: 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))
bits(0) → 0
bits(s(x)) → s(bits(half(s(x))))

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

bits(0) → 0

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(bits(x₁)) = 2 + x₁
POL(half(x₁)) = x₁
POL(s(x₁)) = x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ Overlay + Local Confluence

  ↳ QTRS Reverse

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))
bits(s(x)) → s(bits(half(s(x))))

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ Overlay + Local Confluence

        ↳ QTRS

          ↳ DependencyPairsProof

  ↳ QTRS Reverse

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))
bits(s(x)) → s(bits(half(s(x))))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
bits(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:

HALF(s(s(x))) → HALF(x)
BITS(s(x)) → BITS(half(s(x)))
BITS(s(x)) → HALF(s(x))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
bits(s(x0))

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ Overlay + Local Confluence

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

  ↳ QTRS Reverse

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

HALF(s(s(x))) → HALF(x)
BITS(s(x)) → BITS(half(s(x)))
BITS(s(x)) → HALF(s(x))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
bits(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ Overlay + Local Confluence

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                  ↳ QDP

  ↳ QTRS Reverse

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))
bits(s(x)) → s(bits(half(s(x))))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
bits(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

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ Overlay + Local Confluence

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ QReductionProof

                    ↳ UsableRulesProof

                  ↳ QDP

  ↳ QTRS Reverse

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)))
bits(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.

half(0)
half(s(0))
half(s(s(x0)))
bits(s(x0))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ Overlay + Local Confluence

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ QReductionProof

                          ↳ QDP

                    ↳ UsableRulesProof

                  ↳ QDP

  ↳ QTRS Reverse

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.
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

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ Overlay + Local Confluence

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ QReductionProof

                  ↳ QDP

  ↳ QTRS Reverse

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)))
bits(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.

half(0)
half(s(0))
half(s(s(x0)))
bits(s(x0))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ Overlay + Local Confluence

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ QReductionProof

                          ↳ QDP

                  ↳ QDP

  ↳ QTRS Reverse

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.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ Overlay + Local Confluence

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

  ↳ QTRS Reverse

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

BITS(s(x)) → BITS(half(s(x)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
bits(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We have reversed the following QTRS:
The set of rules R is

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

The set Q is empty.
We have obtained the following QTRS:

0'(half(x)) → 0'(x)
0'(s(half(x))) → 0'(x)
s(s(half(x))) → half(s(x))
0'(bits(x)) → 0'(x)
s(bits(x)) → s(half(bits(s(x))))

The set Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

  ↳ QTRS Reverse

    ↳ QTRS

      ↳ RFCMatchBoundsTRSProof

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

0'(half(x)) → 0'(x)
0'(s(half(x))) → 0'(x)
s(s(half(x))) → half(s(x))
0'(bits(x)) → 0'(x)
s(bits(x)) → s(half(bits(s(x))))

Q is empty.

Termination of the TRS R could be shown with a Match Bound [6,7] of 1. This implies Q-termination of R.
The following rules were used to construct the certificate:

0'(half(x)) → 0'(x)
0'(s(half(x))) → 0'(x)
s(s(half(x))) → half(s(x))
0'(bits(x)) → 0'(x)
s(bits(x)) → s(half(bits(s(x))))

The certificate found is represented by the following graph.

The certificate consists of the following enumerated nodes:

123, 124, 125, 128, 126, 127, 129, 132, 130, 131

Node 123 is start node and node 124 is final node.

Those nodes are connect through the following edges:

123 to 125 labelled half_1(0)
123 to 124 labelled 0'_1(0), 0'_1(1)
123 to 126 labelled s_1(0)
124 to 124 labelled #_1(0)
125 to 124 labelled s_1(0)
125 to 129 labelled half_1(1)
125 to 130 labelled s_1(1)
128 to 124 labelled s_1(0)
128 to 129 labelled half_1(1)
128 to 130 labelled s_1(1)
126 to 127 labelled half_1(0)
127 to 128 labelled bits_1(0)
129 to 124 labelled s_1(1)
129 to 129 labelled half_1(1)
129 to 130 labelled s_1(1)
132 to 124 labelled s_1(1)
132 to 129 labelled half_1(1)
132 to 130 labelled s_1(1)
130 to 131 labelled half_1(1)
131 to 132 labelled bits_1(1)