Termination w.r.t. Q proof of /tmp/tmpKQJ5PL/4.14.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:

p(s(x)) → x
s(p(x)) → x
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

p(s(x)) → x
s(p(x)) → x
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

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:

+¹(p(x), y) → +¹(x, y)
+¹(p(x), y) → P(+(x, y))
+¹(s(x), y) → S(+(x, y))
*¹(p(x), y) → +¹(*(x, y), minus(y))
MINUS(p(x)) → MINUS(x)
*¹(p(x), y) → *¹(x, y)
MINUS(s(x)) → MINUS(x)
*¹(s(x), y) → +¹(*(x, y), y)
MINUS(s(x)) → P(minus(x))
+¹(s(x), y) → +¹(x, y)
MINUS(p(x)) → S(minus(x))
*¹(s(x), y) → *¹(x, y)
*¹(p(x), y) → MINUS(y)

The TRS R consists of the following rules:

p(s(x)) → x
s(p(x)) → x
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

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:

+¹(p(x), y) → +¹(x, y)
+¹(p(x), y) → P(+(x, y))
+¹(s(x), y) → S(+(x, y))
*¹(p(x), y) → +¹(*(x, y), minus(y))
MINUS(p(x)) → MINUS(x)
*¹(p(x), y) → *¹(x, y)
MINUS(s(x)) → MINUS(x)
*¹(s(x), y) → +¹(*(x, y), y)
MINUS(s(x)) → P(minus(x))
+¹(s(x), y) → +¹(x, y)
MINUS(p(x)) → S(minus(x))
*¹(s(x), y) → *¹(x, y)
*¹(p(x), y) → MINUS(y)

The TRS R consists of the following rules:

p(s(x)) → x
s(p(x)) → x
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

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 7 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

MINUS(p(x)) → MINUS(x)
MINUS(s(x)) → MINUS(x)

The TRS R consists of the following rules:

p(s(x)) → x
s(p(x)) → x
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

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:

MINUS(p(x)) → MINUS(x)
MINUS(s(x)) → MINUS(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:

MINUS(p(x)) → MINUS(x)
The graph contains the following edges 1 > 1
MINUS(s(x)) → MINUS(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:

+¹(p(x), y) → +¹(x, y)
+¹(s(x), y) → +¹(x, y)

The TRS R consists of the following rules:

p(s(x)) → x
s(p(x)) → x
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

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

+¹(p(x), y) → +¹(x, y)
+¹(s(x), y) → +¹(x, y)

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

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

*¹(p(x), y) → *¹(x, y)
*¹(s(x), y) → *¹(x, y)

The TRS R consists of the following rules:

p(s(x)) → x
s(p(x)) → x
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

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

*¹(p(x), y) → *¹(x, y)
*¹(s(x), y) → *¹(x, y)

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

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