Termination w.r.t. Q proof of /tmp/tmpylqdKA/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:

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

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:

MARK(tt) → ACTIVE(tt)
MARK(s(X)) → MARK(X)
PLUS(active(X1), X2) → PLUS(X1, X2)
PLUS(mark(X1), X2) → PLUS(X1, X2)
AND(X1, mark(X2)) → AND(X1, X2)
MARK(plus(X1, X2)) → MARK(X2)
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
MARK(plus(X1, X2)) → ACTIVE(plus(mark(X1), mark(X2)))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(and(X1, X2)) → MARK(X1)
ACTIVE(plus(N, s(M))) → PLUS(N, M)
PLUS(X1, mark(X2)) → PLUS(X1, X2)
PLUS(X1, active(X2)) → PLUS(X1, X2)
ACTIVE(plus(N, s(M))) → S(plus(N, M))
S(active(X)) → S(X)
S(mark(X)) → S(X)
MARK(s(X)) → S(mark(X))
MARK(and(X1, X2)) → AND(mark(X1), X2)
ACTIVE(plus(N, 0)) → MARK(N)
MARK(plus(X1, X2)) → MARK(X1)
ACTIVE(plus(N, s(M))) → MARK(s(plus(N, M)))
AND(X1, active(X2)) → AND(X1, X2)
AND(mark(X1), X2) → AND(X1, X2)
MARK(0) → ACTIVE(0)
MARK(plus(X1, X2)) → PLUS(mark(X1), mark(X2))
AND(active(X1), X2) → AND(X1, X2)
ACTIVE(and(tt, X)) → MARK(X)

The TRS R consists of the following rules:

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

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:

MARK(tt) → ACTIVE(tt)
MARK(s(X)) → MARK(X)
PLUS(active(X1), X2) → PLUS(X1, X2)
PLUS(mark(X1), X2) → PLUS(X1, X2)
AND(X1, mark(X2)) → AND(X1, X2)
MARK(plus(X1, X2)) → MARK(X2)
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
MARK(plus(X1, X2)) → ACTIVE(plus(mark(X1), mark(X2)))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(and(X1, X2)) → MARK(X1)
ACTIVE(plus(N, s(M))) → PLUS(N, M)
PLUS(X1, mark(X2)) → PLUS(X1, X2)
PLUS(X1, active(X2)) → PLUS(X1, X2)
ACTIVE(plus(N, s(M))) → S(plus(N, M))
S(active(X)) → S(X)
S(mark(X)) → S(X)
MARK(s(X)) → S(mark(X))
MARK(and(X1, X2)) → AND(mark(X1), X2)
ACTIVE(plus(N, 0)) → MARK(N)
MARK(plus(X1, X2)) → MARK(X1)
ACTIVE(plus(N, s(M))) → MARK(s(plus(N, M)))
AND(X1, active(X2)) → AND(X1, X2)
AND(mark(X1), X2) → AND(X1, X2)
MARK(0) → ACTIVE(0)
MARK(plus(X1, X2)) → PLUS(mark(X1), mark(X2))
AND(active(X1), X2) → AND(X1, X2)
ACTIVE(and(tt, X)) → MARK(X)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 7 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

S(mark(X)) → S(X)
S(active(X)) → S(X)

The TRS R consists of the following rules:

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

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.

S(mark(X)) → S(X)
S(active(X)) → S(X)

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

POL(active(x₁)) = 4 + x_1
POL(mark(x₁)) = 4 + (4)x_1
POL(S(x₁)) = (4)x_1

The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

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

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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

PLUS(active(X1), X2) → PLUS(X1, X2)
PLUS(mark(X1), X2) → PLUS(X1, X2)
PLUS(X1, mark(X2)) → PLUS(X1, X2)
PLUS(X1, active(X2)) → PLUS(X1, X2)

The TRS R consists of the following rules:

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

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(active(X1), X2) → PLUS(X1, X2)
PLUS(mark(X1), X2) → PLUS(X1, X2)
PLUS(X1, mark(X2)) → PLUS(X1, X2)
PLUS(X1, active(X2)) → PLUS(X1, X2)

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

POL(PLUS(x₁, x₂)) = x_1 + (4)x_2
POL(active(x₁)) = 3 + (4)x_1
POL(mark(x₁)) = 4 + (4)x_1

The value of delta used in the strict ordering is 3.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

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

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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

AND(mark(X1), X2) → AND(X1, X2)
AND(active(X1), X2) → AND(X1, X2)
AND(X1, mark(X2)) → AND(X1, X2)
AND(X1, active(X2)) → AND(X1, X2)

The TRS R consists of the following rules:

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

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.

AND(mark(X1), X2) → AND(X1, X2)
AND(active(X1), X2) → AND(X1, X2)
AND(X1, mark(X2)) → AND(X1, X2)
AND(X1, active(X2)) → AND(X1, X2)

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

POL(active(x₁)) = 4 + (4)x_1
POL(AND(x₁, x₂)) = (3)x_1 + (4)x_2
POL(mark(x₁)) = 4 + (4)x_1

The value of delta used in the strict ordering is 12.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

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

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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

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)) → ACTIVE(s(mark(X)))

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

POL(plus(x₁, x₂)) = 4
POL(active(x₁)) = 2
POL(MARK(x₁)) = 4
POL(tt) = 1
POL(mark(x₁)) = 3
POL(s(x₁)) = 0
POL(and(x₁, x₂)) = 4
POL(0) = 2
POL(ACTIVE(x₁)) = x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented:

and(X1, mark(X2)) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
s(active(X)) → s(X)
s(mark(X)) → s(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

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.

ACTIVE(plus(N, 0)) → MARK(N)
MARK(s(X)) → MARK(X)
MARK(plus(X1, X2)) → MARK(X1)
ACTIVE(plus(N, s(M))) → MARK(s(plus(N, M)))
MARK(plus(X1, X2)) → MARK(X2)

The remaining pairs can at least be oriented weakly.

MARK(plus(X1, X2)) → ACTIVE(plus(mark(X1), mark(X2)))
MARK(and(X1, X2)) → MARK(X1)
ACTIVE(and(tt, X)) → MARK(X)
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))

Used ordering: Polynomial interpretation [25,35]:

POL(plus(x₁, x₂)) = 2 + (4)x_1 + (4)x_2
POL(active(x₁)) = x_1
POL(MARK(x₁)) = (4)x_1
POL(tt) = 0
POL(mark(x₁)) = x_1
POL(s(x₁)) = 2 + x_1
POL(and(x₁, x₂)) = (4)x_1 + (4)x_2
POL(0) = 2
POL(ACTIVE(x₁)) = (4)x_1

The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

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

MARK(plus(X1, X2)) → ACTIVE(plus(mark(X1), mark(X2)))
MARK(and(X1, X2)) → MARK(X1)
ACTIVE(and(tt, X)) → MARK(X)
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))

The TRS R consists of the following rules:

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

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(and(X1, X2)) → MARK(X1)
ACTIVE(and(tt, X)) → MARK(X)

The remaining pairs can at least be oriented weakly.

MARK(plus(X1, X2)) → ACTIVE(plus(mark(X1), mark(X2)))
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))

Used ordering: Polynomial interpretation [25,35]:

POL(plus(x₁, x₂)) = (2)x_1
POL(active(x₁)) = x_1
POL(MARK(x₁)) = x_1
POL(tt) = 1
POL(mark(x₁)) = x_1
POL(s(x₁)) = x_1
POL(and(x₁, x₂)) = 1 + (4)x_1 + x_2
POL(0) = 1
POL(ACTIVE(x₁)) = x_1

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

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

MARK(plus(X1, X2)) → ACTIVE(plus(mark(X1), mark(X2)))
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))

The TRS R consists of the following rules:

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

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 2 less nodes.