Termination w.r.t. Q proof of /tmp/tmpwBSZQ0/PEANO_nokinds_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, V2) → a__U12(a__isNat(V2))
a__U12(tt) → tt
a__U21(tt) → tt
a__U31(tt, N) → mark(N)
a__U41(tt, M, N) → a__U42(a__isNat(N), M, N)
a__U42(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__U11(a__isNat(V1), V2)
a__isNat(s(V1)) → a__U21(a__isNat(V1))
a__plus(N, 0) → a__U31(a__isNat(N), N)
a__plus(N, s(M)) → a__U41(a__isNat(M), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X)) → a__U12(mark(X))
mark(isNat(X)) → a__isNat(X)
mark(U21(X)) → a__U21(mark(X))
mark(U31(X1, X2)) → a__U31(mark(X1), X2)
mark(U41(X1, X2, X3)) → a__U41(mark(X1), X2, X3)
mark(U42(X1, X2, X3)) → a__U42(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) → U11(X1, X2)
a__U12(X) → U12(X)
a__isNat(X) → isNat(X)
a__U21(X) → U21(X)
a__U31(X1, X2) → U31(X1, X2)
a__U41(X1, X2, X3) → U41(X1, X2, X3)
a__U42(X1, X2, X3) → U42(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, V2) → a__U12(a__isNat(V2))
a__U12(tt) → tt
a__U21(tt) → tt
a__U31(tt, N) → mark(N)
a__U41(tt, M, N) → a__U42(a__isNat(N), M, N)
a__U42(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__U11(a__isNat(V1), V2)
a__isNat(s(V1)) → a__U21(a__isNat(V1))
a__plus(N, 0) → a__U31(a__isNat(N), N)
a__plus(N, s(M)) → a__U41(a__isNat(M), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X)) → a__U12(mark(X))
mark(isNat(X)) → a__isNat(X)
mark(U21(X)) → a__U21(mark(X))
mark(U31(X1, X2)) → a__U31(mark(X1), X2)
mark(U41(X1, X2, X3)) → a__U41(mark(X1), X2, X3)
mark(U42(X1, X2, X3)) → a__U42(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) → U11(X1, X2)
a__U12(X) → U12(X)
a__isNat(X) → isNat(X)
a__U21(X) → U21(X)
a__U31(X1, X2) → U31(X1, X2)
a__U41(X1, X2, X3) → U41(X1, X2, X3)
a__U42(X1, X2, X3) → U42(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:

A__U31(tt, N) → MARK(N)
A__U42(tt, M, N) → A__PLUS(mark(N), mark(M))
MARK(U12(X)) → MARK(X)
MARK(U42(X1, X2, X3)) → A__U42(mark(X1), X2, X3)
A__ISNAT(plus(V1, V2)) → A__U11(a__isNat(V1), V2)
MARK(U11(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(U12(X)) → A__U12(mark(X))
A__ISNAT(s(V1)) → A__U21(a__isNat(V1))
MARK(U31(X1, X2)) → MARK(X1)
A__U41(tt, M, N) → A__U42(a__isNat(N), M, N)
MARK(plus(X1, X2)) → MARK(X2)
A__U42(tt, M, N) → MARK(M)
MARK(U21(X)) → A__U21(mark(X))
MARK(U11(X1, X2)) → A__U11(mark(X1), X2)
MARK(U41(X1, X2, X3)) → A__U41(mark(X1), X2, X3)
A__PLUS(N, s(M)) → A__ISNAT(M)
A__U11(tt, V2) → A__U12(a__isNat(V2))
A__PLUS(N, s(M)) → A__U41(a__isNat(M), M, N)
A__ISNAT(plus(V1, V2)) → A__ISNAT(V1)
MARK(U41(X1, X2, X3)) → MARK(X1)
A__PLUS(N, 0) → A__U31(a__isNat(N), N)
MARK(plus(X1, X2)) → MARK(X1)
A__ISNAT(s(V1)) → A__ISNAT(V1)
A__PLUS(N, 0) → A__ISNAT(N)
MARK(U21(X)) → MARK(X)
MARK(U42(X1, X2, X3)) → MARK(X1)
A__U42(tt, M, N) → MARK(N)
MARK(U31(X1, X2)) → A__U31(mark(X1), X2)
MARK(plus(X1, X2)) → A__PLUS(mark(X1), mark(X2))
MARK(isNat(X)) → A__ISNAT(X)
A__U41(tt, M, N) → A__ISNAT(N)
A__U11(tt, V2) → A__ISNAT(V2)

The TRS R consists of the following rules:

a__U11(tt, V2) → a__U12(a__isNat(V2))
a__U12(tt) → tt
a__U21(tt) → tt
a__U31(tt, N) → mark(N)
a__U41(tt, M, N) → a__U42(a__isNat(N), M, N)
a__U42(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__U11(a__isNat(V1), V2)
a__isNat(s(V1)) → a__U21(a__isNat(V1))
a__plus(N, 0) → a__U31(a__isNat(N), N)
a__plus(N, s(M)) → a__U41(a__isNat(M), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X)) → a__U12(mark(X))
mark(isNat(X)) → a__isNat(X)
mark(U21(X)) → a__U21(mark(X))
mark(U31(X1, X2)) → a__U31(mark(X1), X2)
mark(U41(X1, X2, X3)) → a__U41(mark(X1), X2, X3)
mark(U42(X1, X2, X3)) → a__U42(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) → U11(X1, X2)
a__U12(X) → U12(X)
a__isNat(X) → isNat(X)
a__U21(X) → U21(X)
a__U31(X1, X2) → U31(X1, X2)
a__U41(X1, X2, X3) → U41(X1, X2, X3)
a__U42(X1, X2, X3) → U42(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

      ↳ DependencyGraphProof

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

A__U31(tt, N) → MARK(N)
A__U42(tt, M, N) → A__PLUS(mark(N), mark(M))
MARK(U12(X)) → MARK(X)
MARK(U42(X1, X2, X3)) → A__U42(mark(X1), X2, X3)
A__ISNAT(plus(V1, V2)) → A__U11(a__isNat(V1), V2)
MARK(U11(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(U12(X)) → A__U12(mark(X))
A__ISNAT(s(V1)) → A__U21(a__isNat(V1))
MARK(U31(X1, X2)) → MARK(X1)
A__U41(tt, M, N) → A__U42(a__isNat(N), M, N)
MARK(plus(X1, X2)) → MARK(X2)
A__U42(tt, M, N) → MARK(M)
MARK(U21(X)) → A__U21(mark(X))
MARK(U11(X1, X2)) → A__U11(mark(X1), X2)
MARK(U41(X1, X2, X3)) → A__U41(mark(X1), X2, X3)
A__PLUS(N, s(M)) → A__ISNAT(M)
A__U11(tt, V2) → A__U12(a__isNat(V2))
A__PLUS(N, s(M)) → A__U41(a__isNat(M), M, N)
A__ISNAT(plus(V1, V2)) → A__ISNAT(V1)
MARK(U41(X1, X2, X3)) → MARK(X1)
A__PLUS(N, 0) → A__U31(a__isNat(N), N)
MARK(plus(X1, X2)) → MARK(X1)
A__ISNAT(s(V1)) → A__ISNAT(V1)
A__PLUS(N, 0) → A__ISNAT(N)
MARK(U21(X)) → MARK(X)
MARK(U42(X1, X2, X3)) → MARK(X1)
A__U42(tt, M, N) → MARK(N)
MARK(U31(X1, X2)) → A__U31(mark(X1), X2)
MARK(plus(X1, X2)) → A__PLUS(mark(X1), mark(X2))
MARK(isNat(X)) → A__ISNAT(X)
A__U41(tt, M, N) → A__ISNAT(N)
A__U11(tt, V2) → A__ISNAT(V2)

The TRS R consists of the following rules:

a__U11(tt, V2) → a__U12(a__isNat(V2))
a__U12(tt) → tt
a__U21(tt) → tt
a__U31(tt, N) → mark(N)
a__U41(tt, M, N) → a__U42(a__isNat(N), M, N)
a__U42(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__U11(a__isNat(V1), V2)
a__isNat(s(V1)) → a__U21(a__isNat(V1))
a__plus(N, 0) → a__U31(a__isNat(N), N)
a__plus(N, s(M)) → a__U41(a__isNat(M), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X)) → a__U12(mark(X))
mark(isNat(X)) → a__isNat(X)
mark(U21(X)) → a__U21(mark(X))
mark(U31(X1, X2)) → a__U31(mark(X1), X2)
mark(U41(X1, X2, X3)) → a__U41(mark(X1), X2, X3)
mark(U42(X1, X2, X3)) → a__U42(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) → U11(X1, X2)
a__U12(X) → U12(X)
a__isNat(X) → isNat(X)
a__U21(X) → U21(X)
a__U31(X1, X2) → U31(X1, X2)
a__U41(X1, X2, X3) → U41(X1, X2, X3)
a__U42(X1, X2, X3) → U42(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 2 SCCs with 9 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

A__ISNAT(plus(V1, V2)) → A__ISNAT(V1)
A__ISNAT(plus(V1, V2)) → A__U11(a__isNat(V1), V2)
A__ISNAT(s(V1)) → A__ISNAT(V1)
A__U11(tt, V2) → A__ISNAT(V2)

The TRS R consists of the following rules:

a__U11(tt, V2) → a__U12(a__isNat(V2))
a__U12(tt) → tt
a__U21(tt) → tt
a__U31(tt, N) → mark(N)
a__U41(tt, M, N) → a__U42(a__isNat(N), M, N)
a__U42(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__U11(a__isNat(V1), V2)
a__isNat(s(V1)) → a__U21(a__isNat(V1))
a__plus(N, 0) → a__U31(a__isNat(N), N)
a__plus(N, s(M)) → a__U41(a__isNat(M), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X)) → a__U12(mark(X))
mark(isNat(X)) → a__isNat(X)
mark(U21(X)) → a__U21(mark(X))
mark(U31(X1, X2)) → a__U31(mark(X1), X2)
mark(U41(X1, X2, X3)) → a__U41(mark(X1), X2, X3)
mark(U42(X1, X2, X3)) → a__U42(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) → U11(X1, X2)
a__U12(X) → U12(X)
a__isNat(X) → isNat(X)
a__U21(X) → U21(X)
a__U31(X1, X2) → U31(X1, X2)
a__U41(X1, X2, X3) → U41(X1, X2, X3)
a__U42(X1, X2, X3) → U42(X1, X2, X3)
a__plus(X1, X2) → plus(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

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

A__ISNAT(plus(V1, V2)) → A__ISNAT(V1)
A__ISNAT(plus(V1, V2)) → A__U11(a__isNat(V1), V2)
A__ISNAT(s(V1)) → A__ISNAT(V1)
A__U11(tt, V2) → A__ISNAT(V2)

The TRS R consists of the following rules:

a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__U11(a__isNat(V1), V2)
a__isNat(s(V1)) → a__U21(a__isNat(V1))
a__isNat(X) → isNat(X)
a__U21(tt) → tt
a__U21(X) → U21(X)
a__U11(tt, V2) → a__U12(a__isNat(V2))
a__U11(X1, X2) → U11(X1, X2)
a__U12(tt) → tt
a__U12(X) → U12(X)

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:

A__ISNAT(plus(V1, V2)) → A__U11(a__isNat(V1), V2)
The graph contains the following edges 1 > 2
A__U11(tt, V2) → A__ISNAT(V2)
The graph contains the following edges 2 >= 1
A__ISNAT(plus(V1, V2)) → A__ISNAT(V1)
The graph contains the following edges 1 > 1
A__ISNAT(s(V1)) → A__ISNAT(V1)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

A__U31(tt, N) → MARK(N)
MARK(U12(X)) → MARK(X)
A__U42(tt, M, N) → A__PLUS(mark(N), mark(M))
MARK(U42(X1, X2, X3)) → A__U42(mark(X1), X2, X3)
MARK(U41(X1, X2, X3)) → MARK(X1)
MARK(U11(X1, X2)) → MARK(X1)
A__PLUS(N, 0) → A__U31(a__isNat(N), N)
MARK(s(X)) → MARK(X)
MARK(plus(X1, X2)) → MARK(X1)
MARK(U31(X1, X2)) → MARK(X1)
A__U41(tt, M, N) → A__U42(a__isNat(N), M, N)
MARK(plus(X1, X2)) → MARK(X2)
MARK(U21(X)) → MARK(X)
A__U42(tt, M, N) → MARK(M)
MARK(U42(X1, X2, X3)) → MARK(X1)
MARK(U41(X1, X2, X3)) → A__U41(mark(X1), X2, X3)
A__U42(tt, M, N) → MARK(N)
MARK(U31(X1, X2)) → A__U31(mark(X1), X2)
MARK(plus(X1, X2)) → A__PLUS(mark(X1), mark(X2))
A__PLUS(N, s(M)) → A__U41(a__isNat(M), M, N)

The TRS R consists of the following rules:

a__U11(tt, V2) → a__U12(a__isNat(V2))
a__U12(tt) → tt
a__U21(tt) → tt
a__U31(tt, N) → mark(N)
a__U41(tt, M, N) → a__U42(a__isNat(N), M, N)
a__U42(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__U11(a__isNat(V1), V2)
a__isNat(s(V1)) → a__U21(a__isNat(V1))
a__plus(N, 0) → a__U31(a__isNat(N), N)
a__plus(N, s(M)) → a__U41(a__isNat(M), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X)) → a__U12(mark(X))
mark(isNat(X)) → a__isNat(X)
mark(U21(X)) → a__U21(mark(X))
mark(U31(X1, X2)) → a__U31(mark(X1), X2)
mark(U41(X1, X2, X3)) → a__U41(mark(X1), X2, X3)
mark(U42(X1, X2, X3)) → a__U42(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) → U11(X1, X2)
a__U12(X) → U12(X)
a__isNat(X) → isNat(X)
a__U21(X) → U21(X)
a__U31(X1, X2) → U31(X1, X2)
a__U41(X1, X2, X3) → U41(X1, X2, X3)
a__U42(X1, X2, X3) → U42(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(U42(X1, X2, X3)) → A__U42(mark(X1), X2, X3)
MARK(U41(X1, X2, X3)) → MARK(X1)
MARK(plus(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → MARK(X2)
MARK(U42(X1, X2, X3)) → MARK(X1)
MARK(U41(X1, X2, X3)) → A__U41(mark(X1), X2, X3)
MARK(plus(X1, X2)) → A__PLUS(mark(X1), mark(X2))

The remaining pairs can at least be oriented weakly.

A__U31(tt, N) → MARK(N)
MARK(U12(X)) → MARK(X)
A__U42(tt, M, N) → A__PLUS(mark(N), mark(M))
MARK(U11(X1, X2)) → MARK(X1)
A__PLUS(N, 0) → A__U31(a__isNat(N), N)
MARK(s(X)) → MARK(X)
MARK(U31(X1, X2)) → MARK(X1)
A__U41(tt, M, N) → A__U42(a__isNat(N), M, N)
MARK(U21(X)) → MARK(X)
A__U42(tt, M, N) → MARK(M)
A__U42(tt, M, N) → MARK(N)
MARK(U31(X1, X2)) → A__U31(mark(X1), X2)
A__PLUS(N, s(M)) → A__U41(a__isNat(M), M, N)

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(A__PLUS(x₁, x₂)) = x₁ + x₂
POL(A__U31(x₁, x₂)) = x₂
POL(A__U41(x₁, x₂, x₃)) = x₂ + x₃
POL(A__U42(x₁, x₂, x₃)) = x₂ + x₃
POL(MARK(x₁)) = x₁
POL(U11(x₁, x₂)) = x₁
POL(U12(x₁)) = x₁
POL(U21(x₁)) = x₁
POL(U31(x₁, x₂)) = x₁ + x₂
POL(U41(x₁, x₂, x₃)) = 1 + x₁ + x₂ + x₃
POL(U42(x₁, x₂, x₃)) = 1 + x₁ + x₂ + x₃
POL(a__U11(x₁, x₂)) = x₁
POL(a__U12(x₁)) = x₁
POL(a__U21(x₁)) = x₁
POL(a__U31(x₁, x₂)) = x₁ + x₂
POL(a__U41(x₁, x₂, x₃)) = 1 + x₁ + x₂ + x₃
POL(a__U42(x₁, x₂, x₃)) = 1 + x₁ + x₂ + x₃
POL(a__isNat(x₁)) = 0
POL(a__plus(x₁, x₂)) = 1 + x₁ + x₂
POL(isNat(x₁)) = 0
POL(mark(x₁)) = x₁
POL(plus(x₁, x₂)) = 1 + x₁ + x₂
POL(s(x₁)) = x₁
POL(tt) = 0

The following usable rules [17] were oriented:

a__U11(tt, V2) → a__U12(a__isNat(V2))
a__U21(tt) → tt
a__U12(tt) → tt
a__isNat(s(V1)) → a__U21(a__isNat(V1))
a__isNat(plus(V1, V2)) → a__U11(a__isNat(V1), V2)
a__plus(N, s(M)) → a__U41(a__isNat(M), M, N)
a__U41(tt, M, N) → a__U42(a__isNat(N), M, N)
a__isNat(0) → tt
a__U42(tt, M, N) → s(a__plus(mark(N), mark(M)))
mark(U42(X1, X2, X3)) → a__U42(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
a__U31(tt, N) → mark(N)
a__plus(N, 0) → a__U31(a__isNat(N), N)
mark(U31(X1, X2)) → a__U31(mark(X1), X2)
mark(U41(X1, X2, X3)) → a__U41(mark(X1), X2, X3)
mark(isNat(X)) → a__isNat(X)
mark(U21(X)) → a__U21(mark(X))
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X)) → a__U12(mark(X))
a__U21(X) → U21(X)
a__U31(X1, X2) → U31(X1, X2)
a__U12(X) → U12(X)
a__isNat(X) → isNat(X)
mark(0) → 0
a__U11(X1, X2) → U11(X1, X2)
mark(tt) → tt
mark(s(X)) → s(mark(X))
a__plus(X1, X2) → plus(X1, X2)
a__U42(X1, X2, X3) → U42(X1, X2, X3)
a__U41(X1, X2, X3) → U41(X1, X2, X3)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

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

A__U31(tt, N) → MARK(N)
MARK(U12(X)) → MARK(X)
A__U42(tt, M, N) → A__PLUS(mark(N), mark(M))
MARK(U11(X1, X2)) → MARK(X1)
A__PLUS(N, 0) → A__U31(a__isNat(N), N)
MARK(s(X)) → MARK(X)
MARK(U31(X1, X2)) → MARK(X1)
A__U41(tt, M, N) → A__U42(a__isNat(N), M, N)
MARK(U21(X)) → MARK(X)
A__U42(tt, M, N) → MARK(M)
A__U42(tt, M, N) → MARK(N)
MARK(U31(X1, X2)) → A__U31(mark(X1), X2)
A__PLUS(N, s(M)) → A__U41(a__isNat(M), M, N)

The TRS R consists of the following rules:

a__U11(tt, V2) → a__U12(a__isNat(V2))
a__U12(tt) → tt
a__U21(tt) → tt
a__U31(tt, N) → mark(N)
a__U41(tt, M, N) → a__U42(a__isNat(N), M, N)
a__U42(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__U11(a__isNat(V1), V2)
a__isNat(s(V1)) → a__U21(a__isNat(V1))
a__plus(N, 0) → a__U31(a__isNat(N), N)
a__plus(N, s(M)) → a__U41(a__isNat(M), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X)) → a__U12(mark(X))
mark(isNat(X)) → a__isNat(X)
mark(U21(X)) → a__U21(mark(X))
mark(U31(X1, X2)) → a__U31(mark(X1), X2)
mark(U41(X1, X2, X3)) → a__U41(mark(X1), X2, X3)
mark(U42(X1, X2, X3)) → a__U42(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) → U11(X1, X2)
a__U12(X) → U12(X)
a__isNat(X) → isNat(X)
a__U21(X) → U21(X)
a__U31(X1, X2) → U31(X1, X2)
a__U41(X1, X2, X3) → U41(X1, X2, X3)
a__U42(X1, X2, X3) → U42(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 2 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ AND

                    ↳ QDP

                      ↳ QDPSizeChangeProof

                    ↳ QDP

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

A__U31(tt, N) → MARK(N)
MARK(U12(X)) → MARK(X)
MARK(U11(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(U31(X1, X2)) → A__U31(mark(X1), X2)
MARK(U31(X1, X2)) → MARK(X1)
MARK(U21(X)) → MARK(X)

The TRS R consists of the following rules:

a__U11(tt, V2) → a__U12(a__isNat(V2))
a__U12(tt) → tt
a__U21(tt) → tt
a__U31(tt, N) → mark(N)
a__U41(tt, M, N) → a__U42(a__isNat(N), M, N)
a__U42(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__U11(a__isNat(V1), V2)
a__isNat(s(V1)) → a__U21(a__isNat(V1))
a__plus(N, 0) → a__U31(a__isNat(N), N)
a__plus(N, s(M)) → a__U41(a__isNat(M), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X)) → a__U12(mark(X))
mark(isNat(X)) → a__isNat(X)
mark(U21(X)) → a__U21(mark(X))
mark(U31(X1, X2)) → a__U31(mark(X1), X2)
mark(U41(X1, X2, X3)) → a__U41(mark(X1), X2, X3)
mark(U42(X1, X2, X3)) → a__U42(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) → U11(X1, X2)
a__U12(X) → U12(X)
a__isNat(X) → isNat(X)
a__U21(X) → U21(X)
a__U31(X1, X2) → U31(X1, X2)
a__U41(X1, X2, X3) → U41(X1, X2, X3)
a__U42(X1, X2, X3) → U42(X1, X2, X3)
a__plus(X1, X2) → plus(X1, X2)

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

MARK(U31(X1, X2)) → A__U31(mark(X1), X2)
The graph contains the following edges 1 > 2
A__U31(tt, N) → MARK(N)
The graph contains the following edges 2 >= 1
MARK(U11(X1, X2)) → MARK(X1)
The graph contains the following edges 1 > 1
MARK(U12(X)) → MARK(X)
The graph contains the following edges 1 > 1
MARK(U21(X)) → MARK(X)
The graph contains the following edges 1 > 1
MARK(U31(X1, X2)) → MARK(X1)
The graph contains the following edges 1 > 1
MARK(s(X)) → MARK(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ AND

                    ↳ QDP

                    ↳ QDP

                      ↳ QDPOrderProof

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

A__U42(tt, M, N) → A__PLUS(mark(N), mark(M))
A__U41(tt, M, N) → A__U42(a__isNat(N), M, N)
A__PLUS(N, s(M)) → A__U41(a__isNat(M), M, N)

The TRS R consists of the following rules:

a__U11(tt, V2) → a__U12(a__isNat(V2))
a__U12(tt) → tt
a__U21(tt) → tt
a__U31(tt, N) → mark(N)
a__U41(tt, M, N) → a__U42(a__isNat(N), M, N)
a__U42(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__U11(a__isNat(V1), V2)
a__isNat(s(V1)) → a__U21(a__isNat(V1))
a__plus(N, 0) → a__U31(a__isNat(N), N)
a__plus(N, s(M)) → a__U41(a__isNat(M), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X)) → a__U12(mark(X))
mark(isNat(X)) → a__isNat(X)
mark(U21(X)) → a__U21(mark(X))
mark(U31(X1, X2)) → a__U31(mark(X1), X2)
mark(U41(X1, X2, X3)) → a__U41(mark(X1), X2, X3)
mark(U42(X1, X2, X3)) → a__U42(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) → U11(X1, X2)
a__U12(X) → U12(X)
a__isNat(X) → isNat(X)
a__U21(X) → U21(X)
a__U31(X1, X2) → U31(X1, X2)
a__U41(X1, X2, X3) → U41(X1, X2, X3)
a__U42(X1, X2, X3) → U42(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.

A__U41(tt, M, N) → A__U42(a__isNat(N), M, N)

The remaining pairs can at least be oriented weakly.

A__U42(tt, M, N) → A__PLUS(mark(N), mark(M))
A__PLUS(N, s(M)) → A__U41(a__isNat(M), M, N)

Used ordering: Polynomial interpretation [25]:

POL(0) = 1
POL(A__PLUS(x₁, x₂)) = 1 + x₂
POL(A__U41(x₁, x₂, x₃)) = 1 + x₁ + x₂
POL(A__U42(x₁, x₂, x₃)) = 1 + x₂
POL(U11(x₁, x₂)) = 1
POL(U12(x₁)) = x₁
POL(U21(x₁)) = 1
POL(U31(x₁, x₂)) = x₂
POL(U41(x₁, x₂, x₃)) = 1 + x₂ + x₃
POL(U42(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(a__U11(x₁, x₂)) = 1
POL(a__U12(x₁)) = x₁
POL(a__U21(x₁)) = 1
POL(a__U31(x₁, x₂)) = x₂
POL(a__U41(x₁, x₂, x₃)) = 1 + x₂ + x₃
POL(a__U42(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(a__isNat(x₁)) = 1
POL(a__plus(x₁, x₂)) = x₁ + x₂
POL(isNat(x₁)) = 1
POL(mark(x₁)) = x₁
POL(plus(x₁, x₂)) = x₁ + x₂
POL(s(x₁)) = 1 + x₁
POL(tt) = 1

The following usable rules [17] were oriented:

a__U11(tt, V2) → a__U12(a__isNat(V2))
a__U21(tt) → tt
a__U12(tt) → tt
a__isNat(s(V1)) → a__U21(a__isNat(V1))
a__isNat(plus(V1, V2)) → a__U11(a__isNat(V1), V2)
a__plus(N, s(M)) → a__U41(a__isNat(M), M, N)
a__U41(tt, M, N) → a__U42(a__isNat(N), M, N)
a__isNat(0) → tt
a__U42(tt, M, N) → s(a__plus(mark(N), mark(M)))
mark(U42(X1, X2, X3)) → a__U42(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
a__U31(tt, N) → mark(N)
a__plus(N, 0) → a__U31(a__isNat(N), N)
mark(U31(X1, X2)) → a__U31(mark(X1), X2)
mark(U41(X1, X2, X3)) → a__U41(mark(X1), X2, X3)
mark(isNat(X)) → a__isNat(X)
mark(U21(X)) → a__U21(mark(X))
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X)) → a__U12(mark(X))
a__U21(X) → U21(X)
a__U31(X1, X2) → U31(X1, X2)
a__U12(X) → U12(X)
a__isNat(X) → isNat(X)
mark(0) → 0
a__U11(X1, X2) → U11(X1, X2)
mark(tt) → tt
mark(s(X)) → s(mark(X))
a__plus(X1, X2) → plus(X1, X2)
a__U42(X1, X2, X3) → U42(X1, X2, X3)
a__U41(X1, X2, X3) → U41(X1, X2, X3)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ AND

                    ↳ QDP

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ DependencyGraphProof

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

A__U42(tt, M, N) → A__PLUS(mark(N), mark(M))
A__PLUS(N, s(M)) → A__U41(a__isNat(M), M, N)

The TRS R consists of the following rules:

a__U11(tt, V2) → a__U12(a__isNat(V2))
a__U12(tt) → tt
a__U21(tt) → tt
a__U31(tt, N) → mark(N)
a__U41(tt, M, N) → a__U42(a__isNat(N), M, N)
a__U42(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__U11(a__isNat(V1), V2)
a__isNat(s(V1)) → a__U21(a__isNat(V1))
a__plus(N, 0) → a__U31(a__isNat(N), N)
a__plus(N, s(M)) → a__U41(a__isNat(M), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X)) → a__U12(mark(X))
mark(isNat(X)) → a__isNat(X)
mark(U21(X)) → a__U21(mark(X))
mark(U31(X1, X2)) → a__U31(mark(X1), X2)
mark(U41(X1, X2, X3)) → a__U41(mark(X1), X2, X3)
mark(U42(X1, X2, X3)) → a__U42(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) → U11(X1, X2)
a__U12(X) → U12(X)
a__isNat(X) → isNat(X)
a__U21(X) → U21(X)
a__U31(X1, X2) → U31(X1, X2)
a__U41(X1, X2, X3) → U41(X1, X2, X3)
a__U42(X1, X2, X3) → U42(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 0 SCCs with 2 less nodes.