Termination w.r.t. Q proof of /tmp/tmp0oC53T/PEANO_nosorts_noand

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__U11(tt, M, N) → a__U12(tt, M, N)
a__U12(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → a__U11(tt, M, N)
mark(U11(X1, X2, X3)) → a__U11(mark(X1), X2, X3)
mark(U12(X1, X2, X3)) → a__U12(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2, X3) → U11(X1, X2, X3)
a__U12(X1, X2, X3) → U12(X1, X2, X3)
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__U11(tt, M, N) → a__U12(tt, M, N)
a__U12(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → a__U11(tt, M, N)
mark(U11(X1, X2, X3)) → a__U11(mark(X1), X2, X3)
mark(U12(X1, X2, X3)) → a__U12(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2, X3) → U11(X1, X2, X3)
a__U12(X1, X2, X3) → U12(X1, X2, X3)
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:

MARK(U12(X1, X2, X3)) → MARK(X1)
MARK(U11(X1, X2, X3)) → A__U11(mark(X1), X2, X3)
MARK(U11(X1, X2, X3)) → MARK(X1)
MARK(s(X)) → MARK(X)
A__U12(tt, M, N) → A__PLUS(mark(N), mark(M))
MARK(plus(X1, X2)) → MARK(X1)
A__PLUS(N, s(M)) → A__U11(tt, M, N)
MARK(plus(X1, X2)) → MARK(X2)
A__U11(tt, M, N) → A__U12(tt, M, N)
A__PLUS(N, 0) → MARK(N)
MARK(plus(X1, X2)) → A__PLUS(mark(X1), mark(X2))
A__U12(tt, M, N) → MARK(N)
A__U12(tt, M, N) → MARK(M)
MARK(U12(X1, X2, X3)) → A__U12(mark(X1), X2, X3)

The TRS R consists of the following rules:

a__U11(tt, M, N) → a__U12(tt, M, N)
a__U12(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → a__U11(tt, M, N)
mark(U11(X1, X2, X3)) → a__U11(mark(X1), X2, X3)
mark(U12(X1, X2, X3)) → a__U12(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2, X3) → U11(X1, X2, X3)
a__U12(X1, X2, X3) → U12(X1, X2, X3)
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:

MARK(U12(X1, X2, X3)) → MARK(X1)
MARK(U11(X1, X2, X3)) → A__U11(mark(X1), X2, X3)
MARK(U11(X1, X2, X3)) → MARK(X1)
MARK(s(X)) → MARK(X)
A__U12(tt, M, N) → A__PLUS(mark(N), mark(M))
MARK(plus(X1, X2)) → MARK(X1)
A__PLUS(N, s(M)) → A__U11(tt, M, N)
MARK(plus(X1, X2)) → MARK(X2)
A__U11(tt, M, N) → A__U12(tt, M, N)
A__PLUS(N, 0) → MARK(N)
MARK(plus(X1, X2)) → A__PLUS(mark(X1), mark(X2))
A__U12(tt, M, N) → MARK(N)
A__U12(tt, M, N) → MARK(M)
MARK(U12(X1, X2, X3)) → A__U12(mark(X1), X2, X3)

The TRS R consists of the following rules:

a__U11(tt, M, N) → a__U12(tt, M, N)
a__U12(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → a__U11(tt, M, N)
mark(U11(X1, X2, X3)) → a__U11(mark(X1), X2, X3)
mark(U12(X1, X2, X3)) → a__U12(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2, X3) → U11(X1, X2, X3)
a__U12(X1, X2, X3) → U12(X1, X2, X3)
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(U11(X1, X2, X3)) → MARK(X1)
MARK(s(X)) → MARK(X)
A__U12(tt, M, N) → A__PLUS(mark(N), mark(M))
A__PLUS(N, s(M)) → A__U11(tt, M, N)
A__U11(tt, M, N) → A__U12(tt, M, N)
A__U12(tt, M, N) → MARK(N)
A__U12(tt, M, N) → MARK(M)
MARK(U12(X1, X2, X3)) → A__U12(mark(X1), X2, X3)

The remaining pairs can at least be oriented weakly.

MARK(U12(X1, X2, X3)) → MARK(X1)
MARK(U11(X1, X2, X3)) → A__U11(mark(X1), X2, X3)
MARK(plus(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → MARK(X2)
A__PLUS(N, 0) → MARK(N)
MARK(plus(X1, X2)) → A__PLUS(mark(X1), mark(X2))

Used ordering: Polynomial interpretation [25,35]:

POL(plus(x₁, x₂)) = (4)x_1 + (4)x_2
POL(A__PLUS(x₁, x₂)) = 3 + (4)x_1 + (4)x_2
POL(U11(x₁, x₂, x₃)) = 1 + (4)x_1 + (4)x_2 + (4)x_3
POL(mark(x₁)) = x_1
POL(a__U12(x₁, x₂, x₃)) = (4)x_1 + (4)x_2 + (4)x_3
POL(0) = 0
POL(a__plus(x₁, x₂)) = (4)x_1 + (4)x_2
POL(A__U12(x₁, x₂, x₃)) = (4)x_1 + (4)x_2 + (4)x_3
POL(MARK(x₁)) = 3 + x_1
POL(A__U11(x₁, x₂, x₃)) = 4 + (4)x_1 + (4)x_2 + (4)x_3
POL(tt) = 1
POL(a__U11(x₁, x₂, x₃)) = 1 + (4)x_1 + (4)x_2 + (4)x_3
POL(s(x₁)) = 2 + x_1
POL(U12(x₁, x₂, x₃)) = (4)x_1 + (4)x_2 + (4)x_3

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

a__U11(tt, M, N) → a__U12(tt, M, N)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
a__plus(N, 0) → mark(N)
a__U12(tt, M, N) → s(a__plus(mark(N), mark(M)))
mark(U11(X1, X2, X3)) → a__U11(mark(X1), X2, X3)
a__plus(N, s(M)) → a__U11(tt, M, N)
mark(U12(X1, X2, X3)) → a__U12(mark(X1), X2, X3)
mark(s(X)) → s(mark(X))
mark(tt) → tt
a__U11(X1, X2, X3) → U11(X1, X2, X3)
mark(0) → 0
a__plus(X1, X2) → plus(X1, X2)
a__U12(X1, X2, X3) → U12(X1, X2, X3)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

MARK(U12(X1, X2, X3)) → MARK(X1)
MARK(U11(X1, X2, X3)) → A__U11(mark(X1), X2, X3)
A__PLUS(N, 0) → MARK(N)
MARK(plus(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → A__PLUS(mark(X1), mark(X2))
MARK(plus(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

a__U11(tt, M, N) → a__U12(tt, M, N)
a__U12(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → a__U11(tt, M, N)
mark(U11(X1, X2, X3)) → a__U11(mark(X1), X2, X3)
mark(U12(X1, X2, X3)) → a__U12(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2, X3) → U11(X1, X2, X3)
a__U12(X1, X2, X3) → U12(X1, X2, X3)
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 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

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

MARK(U12(X1, X2, X3)) → 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)

The TRS R consists of the following rules:

a__U11(tt, M, N) → a__U12(tt, M, N)
a__U12(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → a__U11(tt, M, N)
mark(U11(X1, X2, X3)) → a__U11(mark(X1), X2, X3)
mark(U12(X1, X2, X3)) → a__U12(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2, X3) → U11(X1, X2, X3)
a__U12(X1, X2, X3) → U12(X1, X2, X3)
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(U12(X1, X2, X3)) → 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)

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

POL(a__plus(x₁, x₂)) = 3 + (2)x_1 + (4)x_2
POL(plus(x₁, x₂)) = 3 + (2)x_1 + (4)x_2
POL(MARK(x₁)) = 4 + (4)x_1
POL(A__PLUS(x₁, x₂)) = (4)x_1 + (4)x_2
POL(tt) = 1
POL(a__U11(x₁, x₂, x₃)) = 4 + (4)x_1 + x_3
POL(U11(x₁, x₂, x₃)) = 4 + (4)x_1 + x_3
POL(mark(x₁)) = (2)x_1
POL(s(x₁)) = 2
POL(U12(x₁, x₂, x₃)) = 2 + (2)x_1 + x_3
POL(a__U12(x₁, x₂, x₃)) = 3 + (2)x_1 + x_3
POL(0) = 2

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

mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
a__plus(N, 0) → mark(N)
a__U12(tt, M, N) → s(a__plus(mark(N), mark(M)))
mark(U11(X1, X2, X3)) → a__U11(mark(X1), X2, X3)
a__plus(N, s(M)) → a__U11(tt, M, N)
mark(U12(X1, X2, X3)) → a__U12(mark(X1), X2, X3)
mark(s(X)) → s(mark(X))
mark(tt) → tt
a__U11(X1, X2, X3) → U11(X1, X2, X3)
mark(0) → 0
a__plus(X1, X2) → plus(X1, X2)
a__U12(X1, X2, X3) → U12(X1, X2, X3)
a__U11(tt, M, N) → a__U12(tt, M, N)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ PisEmptyProof

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

a__U11(tt, M, N) → a__U12(tt, M, N)
a__U12(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → a__U11(tt, M, N)
mark(U11(X1, X2, X3)) → a__U11(mark(X1), X2, X3)
mark(U12(X1, X2, X3)) → a__U12(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2, X3) → U11(X1, X2, X3)
a__U12(X1, X2, X3) → U12(X1, X2, X3)
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.