Termination w.r.t. Q proof of /tmp/tmpl4sBBh/#3.19.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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) → s(plus(plus(x, y), z))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) → s(plus(plus(x, y), z))

Q is empty.

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

PLUS(s(x), y) → PLUS(minus(x, y), double(y))
MINUS(s(x), s(y)) → MINUS(x, y)
DOUBLE(s(x)) → DOUBLE(x)
PLUS(s(x), y) → DOUBLE(y)
PLUS(s(x), y) → PLUS(x, s(y))
PLUS(s(plus(x, y)), z) → PLUS(plus(x, y), z)
PLUS(s(x), y) → MINUS(x, y)
PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) → s(plus(plus(x, y), z))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

PLUS(s(x), y) → PLUS(minus(x, y), double(y))
MINUS(s(x), s(y)) → MINUS(x, y)
DOUBLE(s(x)) → DOUBLE(x)
PLUS(s(x), y) → DOUBLE(y)
PLUS(s(x), y) → PLUS(x, s(y))
PLUS(s(plus(x, y)), z) → PLUS(plus(x, y), z)
PLUS(s(x), y) → MINUS(x, y)
PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) → s(plus(plus(x, y), z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

DOUBLE(s(x)) → DOUBLE(x)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) → s(plus(plus(x, y), z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

DOUBLE(s(x)) → DOUBLE(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:

DOUBLE(s(x)) → DOUBLE(x)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) → s(plus(plus(x, y), z))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(x, y)

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

PLUS(s(x), y) → PLUS(minus(x, y), double(y))
PLUS(s(x), y) → PLUS(x, s(y))
PLUS(s(plus(x, y)), z) → PLUS(plus(x, y), z)
PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) → s(plus(plus(x, y), z))

Q is empty.
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.

PLUS(s(x), y) → PLUS(minus(x, y), double(y))
PLUS(s(x), y) → PLUS(x, s(y))
PLUS(s(plus(x, y)), z) → PLUS(plus(x, y), z)
PLUS(s(x), y) → PLUS(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2) = PLUS(x1, x2)
s(x1) = s(x1)
minus(x1, x2) = x1
double(x1) = double(x1)
plus(x1, x2) = plus(x1, x2)
0 = 0

Recursive path order with status [2].
Quasi-Precedence:

PLUS₂ > plus₂ > double₁ > s₁
0 > s₁

Status:

PLUS₂: [1,2]
plus₂: [1,2]
s₁: [1]
double₁: [1]
0: multiset

The following usable rules [17] were oriented:

minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
double(s(x)) → s(s(double(x)))
double(0) → 0
plus(s(x), y) → s(plus(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(minus(x, y), double(y)))
plus(s(x), y) → plus(x, s(y))
plus(s(plus(x, y)), z) → s(plus(plus(x, y), z))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(y)))
plus(s(plus(x, y)), z) → s(plus(plus(x, y), z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.