Termination w.r.t. Q proof of /tmp/tmpk4mI4u/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:

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

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

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)))
a__x(N, 0) → 0
a__x(N, s(M)) → a__plus(a__x(mark(N), mark(M)), mark(N))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(x(X1, X2)) → a__x(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)
a__x(X1, X2) → x(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))
MARK(s(X)) → MARK(X)
MARK(plus(X1, X2)) → MARK(X1)
A__X(N, s(M)) → A__X(mark(N), mark(M))
MARK(plus(X1, X2)) → MARK(X2)
A__X(N, s(M)) → MARK(M)
A__PLUS(N, s(M)) → MARK(N)
A__AND(tt, X) → MARK(X)
MARK(x(X1, X2)) → MARK(X1)
MARK(x(X1, X2)) → MARK(X2)
A__PLUS(N, s(M)) → MARK(M)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
A__X(N, s(M)) → A__PLUS(a__x(mark(N), mark(M)), mark(N))
MARK(and(X1, X2)) → MARK(X1)
MARK(x(X1, X2)) → A__X(mark(X1), mark(X2))
A__PLUS(N, 0) → MARK(N)
A__X(N, s(M)) → MARK(N)
MARK(plus(X1, X2)) → A__PLUS(mark(X1), mark(X2))

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)))
a__x(N, 0) → 0
a__x(N, s(M)) → a__plus(a__x(mark(N), mark(M)), mark(N))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(x(X1, X2)) → a__x(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)
a__x(X1, 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.

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

The remaining pairs can at least be oriented weakly.

A__AND(tt, X) → MARK(X)
MARK(x(X1, X2)) → A__X(mark(X1), mark(X2))

Used ordering: Combined order from the following AFS and order.
A__PLUS(x1, x2) = A__PLUS(x1, x2)
s(x1) = s(x1)
mark(x1) = x1
MARK(x1) = x1
plus(x1, x2) = plus(x1, x2)
A__X(x1, x2) = A__X(x1, x2)
A__AND(x1, x2) = x2
tt = tt
x(x1, x2) = x(x1, x2)
and(x1, x2) = and(x1, x2)
a__x(x1, x2) = a__x(x1, x2)
0 = 0
a__plus(x1, x2) = a__plus(x1, x2)
a__and(x1, x2) = a__and(x1, x2)

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

[A_{_X₂}, x₂, a_{_x₂}] > [plus₂, a_{_plus₂}] > s₁ > A_{_PLUS₂}
[A_{_X₂}, x₂, a_{_x₂}] > 0
[and₂, a_{_and₂}]

Status:

a_{_x₂}: multiset
a_{_plus₂}: multiset
plus₂: multiset
A_{_PLUS₂}: multiset
tt: multiset
a_{_and₂}: multiset
A_{_X₂}: multiset
s₁: multiset
x₂: multiset
and₂: multiset
0: multiset

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

MARK(x(X1, X2)) → A__X(mark(X1), mark(X2))
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)))
a__x(N, 0) → 0
a__x(N, s(M)) → a__plus(a__x(mark(N), mark(M)), mark(N))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(x(X1, X2)) → a__x(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)
a__x(X1, 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 2 less nodes.