Termination w.r.t. Q proof of /tmp/tmpPKoykW/#3.22.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:

times(x, plus(y, s(z))) → plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

times(x, plus(y, s(z))) → plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, 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:

TIMES(x, plus(y, s(z))) → TIMES(x, s(z))
TIMES(x, s(y)) → TIMES(x, y)
TIMES(x, s(y)) → PLUS(times(x, y), x)
TIMES(x, plus(y, s(z))) → TIMES(x, plus(y, times(s(z), 0)))
TIMES(x, plus(y, s(z))) → PLUS(y, times(s(z), 0))
PLUS(x, s(y)) → PLUS(x, y)
TIMES(x, plus(y, s(z))) → PLUS(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
TIMES(x, plus(y, s(z))) → TIMES(s(z), 0)

The TRS R consists of the following rules:

times(x, plus(y, s(z))) → plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, 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:

TIMES(x, plus(y, s(z))) → TIMES(x, s(z))
TIMES(x, s(y)) → TIMES(x, y)
TIMES(x, s(y)) → PLUS(times(x, y), x)
TIMES(x, plus(y, s(z))) → TIMES(x, plus(y, times(s(z), 0)))
TIMES(x, plus(y, s(z))) → PLUS(y, times(s(z), 0))
PLUS(x, s(y)) → PLUS(x, y)
TIMES(x, plus(y, s(z))) → PLUS(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
TIMES(x, plus(y, s(z))) → TIMES(s(z), 0)

The TRS R consists of the following rules:

times(x, plus(y, s(z))) → plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

PLUS(x, s(y)) → PLUS(x, y)

The TRS R consists of the following rules:

times(x, plus(y, s(z))) → plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, 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

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

PLUS(x, s(y)) → PLUS(x, y)

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:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

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

TIMES(x, plus(y, s(z))) → TIMES(x, s(z))
TIMES(x, s(y)) → TIMES(x, y)
TIMES(x, plus(y, s(z))) → TIMES(x, plus(y, times(s(z), 0)))

The TRS R consists of the following rules:

times(x, plus(y, s(z))) → plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

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

TIMES(x, plus(y, s(z))) → TIMES(x, s(z))
TIMES(x, s(y)) → TIMES(x, y)
TIMES(x, plus(y, s(z))) → TIMES(x, plus(y, times(s(z), 0)))

The TRS R consists of the following rules:

times(x, 0) → 0
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

TIMES(x, plus(y, s(z))) → TIMES(x, s(z))

Strictly oriented rules of the TRS R:

plus(x, 0) → x

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(TIMES(x₁, x₂)) = x₁ + 2·x₂
POL(plus(x₁, x₂)) = 1 + x₁ + 2·x₂
POL(s(x₁)) = x₁
POL(times(x₁, x₂)) = x₁ + 2·x₂

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ RuleRemovalProof

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

TIMES(x, s(y)) → TIMES(x, y)
TIMES(x, plus(y, s(z))) → TIMES(x, plus(y, times(s(z), 0)))

The TRS R consists of the following rules:

times(x, 0) → 0
plus(x, s(y)) → s(plus(x, y))

TIMES(x, s(y)) → TIMES(x, y)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(TIMES(x₁, x₂)) = x₁ + 2·x₂
POL(plus(x₁, x₂)) = 1 + 2·x₁ + x₂
POL(s(x₁)) = 1 + x₁
POL(times(x₁, x₂)) = x₁ + x₂

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

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

TIMES(x, plus(y, s(z))) → TIMES(x, plus(y, times(s(z), 0)))

The TRS R consists of the following rules:

times(x, 0) → 0
plus(x, s(y)) → s(plus(x, y))

plus(x, s(y)) → s(plus(x, y))

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(TIMES(x₁, x₂)) = x₁ + 2·x₂
POL(plus(x₁, x₂)) = 2·x₁ + 2·x₂
POL(s(x₁)) = 2 + x₁
POL(times(x₁, x₂)) = x₁ + x₂

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ RuleRemovalProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ DependencyGraphProof

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

TIMES(x, plus(y, s(z))) → TIMES(x, plus(y, times(s(z), 0)))

The TRS R consists of the following rules:

times(x, 0) → 0

Q is empty.
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.