Termination w.r.t. Q proof of /tmp/tmp1ODIyj/PEANO_nosorts

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:

a__and(tt, X) → mark(X)
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → s(a__plus(mark(N), mark(M)))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(0) → 0
mark(s(X)) → s(mark(X))
a__and(X1, X2) → and(X1, X2)
a__plus(X1, X2) → plus(X1, X2)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__and(tt, X) → mark(X)
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → s(a__plus(mark(N), mark(M)))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(0) → 0
mark(s(X)) → s(mark(X))
a__and(X1, X2) → and(X1, X2)
a__plus(X1, X2) → plus(X1, X2)

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:

A__PLUS(N, s(M)) → A__PLUS(mark(N), mark(M))
A__PLUS(N, s(M)) → MARK(M)
MARK(s(X)) → MARK(X)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → MARK(X1)
A__PLUS(N, 0) → MARK(N)
MARK(plus(X1, X2)) → A__PLUS(mark(X1), mark(X2))
MARK(plus(X1, X2)) → MARK(X2)
A__PLUS(N, s(M)) → MARK(N)
A__AND(tt, X) → MARK(X)

The TRS R consists of the following rules:

a__and(tt, X) → mark(X)
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → s(a__plus(mark(N), mark(M)))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(0) → 0
mark(s(X)) → s(mark(X))
a__and(X1, X2) → and(X1, X2)
a__plus(X1, X2) → plus(X1, X2)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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

A__PLUS(N, s(M)) → A__PLUS(mark(N), mark(M))
A__PLUS(N, s(M)) → MARK(M)
MARK(s(X)) → MARK(X)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → MARK(X1)
A__PLUS(N, 0) → MARK(N)
MARK(plus(X1, X2)) → A__PLUS(mark(X1), mark(X2))
MARK(plus(X1, X2)) → MARK(X2)
A__PLUS(N, s(M)) → MARK(N)
A__AND(tt, X) → MARK(X)

The TRS R consists of the following rules:

a__and(tt, X) → mark(X)
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → s(a__plus(mark(N), mark(M)))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(0) → 0
mark(s(X)) → s(mark(X))
a__and(X1, X2) → and(X1, X2)
a__plus(X1, X2) → plus(X1, X2)

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.

A__PLUS(N, s(M)) → MARK(M)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → MARK(X1)
A__PLUS(N, 0) → MARK(N)
MARK(plus(X1, X2)) → A__PLUS(mark(X1), mark(X2))
MARK(plus(X1, X2)) → MARK(X2)
A__PLUS(N, s(M)) → MARK(N)

The remaining pairs can at least be oriented weakly.

A__PLUS(N, s(M)) → A__PLUS(mark(N), mark(M))
MARK(s(X)) → MARK(X)
A__AND(tt, X) → MARK(X)

Used ordering: Polynomial interpretation [25,35]:

POL(a__plus(x₁, x₂)) = 2 + x_1 + (4)x_2
POL(A__AND(x₁, x₂)) = (2)x_2
POL(plus(x₁, x₂)) = 2 + x_1 + (4)x_2
POL(A__PLUS(x₁, x₂)) = 1 + x_1 + (4)x_2
POL(MARK(x₁)) = x_1
POL(tt) = 0
POL(a__and(x₁, x₂)) = 4 + (4)x_1 + (2)x_2
POL(mark(x₁)) = x_1
POL(s(x₁)) = x_1
POL(and(x₁, x₂)) = 4 + (4)x_1 + (2)x_2
POL(0) = 2

The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

a__plus(N, s(M)) → s(a__plus(mark(N), mark(M)))
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
a__plus(N, 0) → mark(N)
a__and(tt, X) → mark(X)
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__plus(X1, X2) → plus(X1, X2)
a__and(X1, X2) → and(X1, X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

A__PLUS(N, s(M)) → A__PLUS(mark(N), mark(M))
MARK(s(X)) → MARK(X)
A__AND(tt, X) → MARK(X)

The TRS R consists of the following rules:

a__and(tt, X) → mark(X)
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → s(a__plus(mark(N), mark(M)))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(0) → 0
mark(s(X)) → s(mark(X))
a__and(X1, X2) → and(X1, X2)
a__plus(X1, X2) → plus(X1, X2)

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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__and(tt, X) → mark(X)
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → s(a__plus(mark(N), mark(M)))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(0) → 0
mark(s(X)) → s(mark(X))
a__and(X1, X2) → and(X1, X2)
a__plus(X1, X2) → plus(X1, X2)

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.

MARK(s(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(MARK(x₁)) = (4)x_1
POL(s(x₁)) = 1 + (4)x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

a__and(tt, X) → mark(X)
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → s(a__plus(mark(N), mark(M)))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(0) → 0
mark(s(X)) → s(mark(X))
a__and(X1, X2) → and(X1, X2)
a__plus(X1, X2) → plus(X1, X2)

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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

A__PLUS(N, s(M)) → A__PLUS(mark(N), mark(M))

The TRS R consists of the following rules:

a__and(tt, X) → mark(X)
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → s(a__plus(mark(N), mark(M)))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(0) → 0
mark(s(X)) → s(mark(X))
a__and(X1, X2) → and(X1, X2)
a__plus(X1, X2) → plus(X1, X2)

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.

A__PLUS(N, s(M)) → A__PLUS(mark(N), mark(M))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(a__plus(x₁, x₂)) = (2)x_1 + x_2
POL(plus(x₁, x₂)) = (2)x_1 + x_2
POL(A__PLUS(x₁, x₂)) = (2)x_2
POL(tt) = 0
POL(a__and(x₁, x₂)) = x_1 + (4)x_2
POL(mark(x₁)) = x_1
POL(s(x₁)) = 1 + x_1
POL(and(x₁, x₂)) = x_1 + (4)x_2
POL(0) = 4

The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented:

mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__plus(X1, X2) → plus(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__plus(N, s(M)) → s(a__plus(mark(N), mark(M)))
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
a__plus(N, 0) → mark(N)
a__and(tt, X) → mark(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

a__and(tt, X) → mark(X)
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → s(a__plus(mark(N), mark(M)))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(0) → 0
mark(s(X)) → s(mark(X))
a__and(X1, X2) → and(X1, X2)
a__plus(X1, X2) → plus(X1, X2)

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.