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

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

  ↳ 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.
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:

121, 122, 123, 126, 124, 125, 127, 130, 128, 129

Node 121 is start node and node 122 is final node.

Those nodes are connect through the following edges:

121 to 123 labelled half_1(0)
121 to 122 labelled 0'_1(0), 0'_1(1)
121 to 124 labelled s_1(0)
122 to 122 labelled #_1(0)
123 to 122 labelled s_1(0)
123 to 127 labelled half_1(1)
123 to 128 labelled s_1(1)
126 to 122 labelled s_1(0)
126 to 127 labelled half_1(1)
126 to 128 labelled s_1(1)
124 to 125 labelled half_1(0)
125 to 126 labelled bits_1(0)
127 to 122 labelled s_1(1)
127 to 127 labelled half_1(1)
127 to 128 labelled s_1(1)
130 to 122 labelled s_1(1)
130 to 127 labelled half_1(1)
130 to 128 labelled s_1(1)
128 to 129 labelled half_1(1)
129 to 130 labelled bits_1(1)