Termination w.r.t. Q proof of /tmp/tmp2RJ5Ej/MYNAT_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)))
active(x(N, 0)) → mark(0)
active(x(N, s(M))) → mark(plus(x(N, M), N))
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)))
mark(x(X1, X2)) → active(x(mark(X1), mark(X2)))
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)
x(mark(X1), X2) → x(X1, X2)
x(X1, mark(X2)) → x(X1, X2)
x(active(X1), X2) → x(X1, X2)
x(X1, active(X2)) → x(X1, X2)

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)))
active(x(N, 0)) → mark(0)
active(x(N, s(M))) → mark(plus(x(N, M), N))
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)))
mark(x(X1, X2)) → active(x(mark(X1), mark(X2)))
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)
x(mark(X1), X2) → x(X1, X2)
x(X1, mark(X2)) → x(X1, X2)
x(active(X1), X2) → x(X1, X2)
x(X1, active(X2)) → x(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:

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

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(x(X1, X2)) → ACTIVE(x(mark(X1), mark(X2)))
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)
X(mark(X1), X2) → X(X1, X2)
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
MARK(x(X1, X2)) → MARK(X1)
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)
X(X1, mark(X2)) → X(X1, X2)
PLUS(X1, mark(X2)) → PLUS(X1, X2)
ACTIVE(x(N, s(M))) → X(N, M)
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(x(X1, X2)) → X(mark(X1), mark(X2))
X(X1, active(X2)) → X(X1, X2)
ACTIVE(x(N, s(M))) → PLUS(x(N, M), N)
ACTIVE(plus(N, s(M))) → MARK(s(plus(N, M)))
MARK(plus(X1, X2)) → MARK(X1)
ACTIVE(x(N, 0)) → MARK(0)
AND(X1, active(X2)) → AND(X1, X2)
X(active(X1), X2) → X(X1, X2)
MARK(x(X1, X2)) → MARK(X2)
AND(mark(X1), X2) → AND(X1, X2)
MARK(0) → ACTIVE(0)
MARK(plus(X1, X2)) → PLUS(mark(X1), mark(X2))
ACTIVE(x(N, s(M))) → MARK(plus(x(N, M), N))
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)))
active(x(N, 0)) → mark(0)
active(x(N, s(M))) → mark(plus(x(N, M), N))
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)))
mark(x(X1, X2)) → active(x(mark(X1), mark(X2)))
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)
x(mark(X1), X2) → x(X1, X2)
x(X1, mark(X2)) → x(X1, X2)
x(active(X1), X2) → x(X1, X2)
x(X1, active(X2)) → x(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 5 SCCs with 11 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

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

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

          ↳ QDP

          ↳ QDP

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

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

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:

X(X1, active(X2)) → X(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
X(X1, mark(X2)) → X(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
X(mark(X1), X2) → X(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
X(active(X1), X2) → X(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ 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)))
active(x(N, 0)) → mark(0)
active(x(N, s(M))) → mark(plus(x(N, M), N))
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)))
mark(x(X1, X2)) → active(x(mark(X1), mark(X2)))
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)
x(mark(X1), X2) → x(X1, X2)
x(X1, mark(X2)) → x(X1, X2)
x(active(X1), X2) → x(X1, X2)
x(X1, active(X2)) → x(X1, X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

S(active(X)) → S(X)
S(mark(X)) → S(X)

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

S(mark(X)) → S(X)
The graph contains the following edges 1 > 1
S(active(X)) → S(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ 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)))
active(x(N, 0)) → mark(0)
active(x(N, s(M))) → mark(plus(x(N, M), N))
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)))
mark(x(X1, X2)) → active(x(mark(X1), mark(X2)))
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)
x(mark(X1), X2) → x(X1, X2)
x(X1, mark(X2)) → x(X1, X2)
x(active(X1), X2) → x(X1, X2)
x(X1, active(X2)) → x(X1, X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ 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)

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

PLUS(active(X1), X2) → PLUS(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
PLUS(mark(X1), X2) → PLUS(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
PLUS(X1, mark(X2)) → PLUS(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
PLUS(X1, active(X2)) → PLUS(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ 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)))
active(x(N, 0)) → mark(0)
active(x(N, s(M))) → mark(plus(x(N, M), N))
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)))
mark(x(X1, X2)) → active(x(mark(X1), mark(X2)))
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)
x(mark(X1), X2) → x(X1, X2)
x(X1, mark(X2)) → x(X1, X2)
x(active(X1), X2) → x(X1, X2)
x(X1, active(X2)) → x(X1, X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ 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)

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

AND(mark(X1), X2) → AND(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
AND(active(X1), X2) → AND(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
AND(X1, mark(X2)) → AND(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
AND(X1, active(X2)) → AND(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(ACTIVE(x₁)) = x₁
POL(MARK(x₁)) = 1
POL(active(x₁)) = 0
POL(and(x₁, x₂)) = 1
POL(mark(x₁)) = 0
POL(plus(x₁, x₂)) = 1
POL(s(x₁)) = 0
POL(tt) = 0
POL(x(x₁, x₂)) = 1

The following usable rules [17] were oriented:

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)
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)
s(active(X)) → s(X)
s(mark(X)) → s(X)
x(X1, active(X2)) → x(X1, X2)
x(X1, mark(X2)) → x(X1, X2)
x(active(X1), X2) → x(X1, X2)
x(mark(X1), X2) → x(X1, X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

ACTIVE(plus(N, 0)) → MARK(N)
MARK(x(X1, X2)) → ACTIVE(x(mark(X1), mark(X2)))
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)
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
MARK(x(X1, X2)) → MARK(X1)
MARK(x(X1, X2)) → MARK(X2)
MARK(plus(X1, X2)) → ACTIVE(plus(mark(X1), mark(X2)))
MARK(and(X1, X2)) → MARK(X1)
ACTIVE(x(N, s(M))) → MARK(plus(x(N, M), N))
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)))
active(x(N, 0)) → mark(0)
active(x(N, s(M))) → mark(plus(x(N, M), N))
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)))
mark(x(X1, X2)) → active(x(mark(X1), mark(X2)))
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)
x(mark(X1), X2) → x(X1, X2)
x(X1, mark(X2)) → x(X1, X2)
x(active(X1), X2) → x(X1, X2)
x(X1, active(X2)) → x(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.

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = ACTIVE(x1)
plus(x1, x2) = plus(x1, x2)
0 = 0
MARK(x1) = MARK(x1)
x(x1, x2) = x(x1, x2)
mark(x1) = x1
s(x1) = s(x1)
and(x1, x2) = and(x1, x2)
tt = tt
active(x1) = x1

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

[ACTIVE₁, MARK₁, and₂] > x₂ > plus₂ > s₁
[ACTIVE₁, MARK₁, and₂] > x₂ > 0 > s₁
tt > s₁

Status:

plus₂: multiset
MARK₁: [1]
tt: multiset
s₁: [1]
x₂: [2,1]
and₂: multiset
0: multiset
ACTIVE₁: [1]

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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(x(X1, X2)) → ACTIVE(x(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)))
active(x(N, 0)) → mark(0)
active(x(N, s(M))) → mark(plus(x(N, M), N))
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)))
mark(x(X1, X2)) → active(x(mark(X1), mark(X2)))
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)
x(mark(X1), X2) → x(X1, X2)
x(X1, mark(X2)) → x(X1, X2)
x(active(X1), X2) → x(X1, X2)
x(X1, active(X2)) → x(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 0 SCCs with 3 less nodes.