Termination w.r.t. Q proof of /tmp/tmpkA0T53/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:

U11(tt, V2) → U12(isNat(activate(V2)))
U12(tt) → tt
U21(tt) → tt
U31(tt, N) → activate(N)
U41(tt, M, N) → U42(isNat(activate(N)), activate(M), activate(N))
U42(tt, M, N) → s(plus(activate(N), activate(M)))
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → U11(isNat(activate(V1)), activate(V2))
isNat(n__s(V1)) → U21(isNat(activate(V1)))
plus(N, 0) → U31(isNat(N), N)
plus(N, s(M)) → U41(isNat(M), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

U11(tt, V2) → U12(isNat(activate(V2)))
U12(tt) → tt
U21(tt) → tt
U31(tt, N) → activate(N)
U41(tt, M, N) → U42(isNat(activate(N)), activate(M), activate(N))
U42(tt, M, N) → s(plus(activate(N), activate(M)))
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → U11(isNat(activate(V1)), activate(V2))
isNat(n__s(V1)) → U21(isNat(activate(V1)))
plus(N, 0) → U31(isNat(N), N)
plus(N, s(M)) → U41(isNat(M), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
activate(n__s(X)) → s(X)
activate(X) → 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:

U11¹(tt, V2) → U12¹(isNat(activate(V2)))
ACTIVATE(n__plus(X1, X2)) → PLUS(X1, X2)
PLUS(N, s(M)) → ISNAT(M)
U42¹(tt, M, N) → PLUS(activate(N), activate(M))
U42¹(tt, M, N) → S(plus(activate(N), activate(M)))
U31¹(tt, N) → ACTIVATE(N)
ACTIVATE(n__0) → 0¹
U41¹(tt, M, N) → ISNAT(activate(N))
ISNAT(n__plus(V1, V2)) → U11¹(isNat(activate(V1)), activate(V2))
U41¹(tt, M, N) → ACTIVATE(N)
ISNAT(n__s(V1)) → U21¹(isNat(activate(V1)))
U11¹(tt, V2) → ACTIVATE(V2)
PLUS(N, 0) → ISNAT(N)
PLUS(N, 0) → U31¹(isNat(N), N)
U41¹(tt, M, N) → ACTIVATE(M)
U42¹(tt, M, N) → ACTIVATE(M)
ISNAT(n__s(V1)) → ISNAT(activate(V1))
ACTIVATE(n__s(X)) → S(X)
ISNAT(n__plus(V1, V2)) → ACTIVATE(V2)
ISNAT(n__s(V1)) → ACTIVATE(V1)
U41¹(tt, M, N) → U42¹(isNat(activate(N)), activate(M), activate(N))
ISNAT(n__plus(V1, V2)) → ACTIVATE(V1)
PLUS(N, s(M)) → U41¹(isNat(M), M, N)
U11¹(tt, V2) → ISNAT(activate(V2))
ISNAT(n__plus(V1, V2)) → ISNAT(activate(V1))
U42¹(tt, M, N) → ACTIVATE(N)

The TRS R consists of the following rules:

U11(tt, V2) → U12(isNat(activate(V2)))
U12(tt) → tt
U21(tt) → tt
U31(tt, N) → activate(N)
U41(tt, M, N) → U42(isNat(activate(N)), activate(M), activate(N))
U42(tt, M, N) → s(plus(activate(N), activate(M)))
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → U11(isNat(activate(V1)), activate(V2))
isNat(n__s(V1)) → U21(isNat(activate(V1)))
plus(N, 0) → U31(isNat(N), N)
plus(N, s(M)) → U41(isNat(M), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
activate(n__s(X)) → s(X)
activate(X) → 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:

U11¹(tt, V2) → U12¹(isNat(activate(V2)))
ACTIVATE(n__plus(X1, X2)) → PLUS(X1, X2)
PLUS(N, s(M)) → ISNAT(M)
U42¹(tt, M, N) → PLUS(activate(N), activate(M))
U42¹(tt, M, N) → S(plus(activate(N), activate(M)))
U31¹(tt, N) → ACTIVATE(N)
ACTIVATE(n__0) → 0¹
U41¹(tt, M, N) → ISNAT(activate(N))
ISNAT(n__plus(V1, V2)) → U11¹(isNat(activate(V1)), activate(V2))
U41¹(tt, M, N) → ACTIVATE(N)
ISNAT(n__s(V1)) → U21¹(isNat(activate(V1)))
U11¹(tt, V2) → ACTIVATE(V2)
PLUS(N, 0) → ISNAT(N)
PLUS(N, 0) → U31¹(isNat(N), N)
U41¹(tt, M, N) → ACTIVATE(M)
U42¹(tt, M, N) → ACTIVATE(M)
ISNAT(n__s(V1)) → ISNAT(activate(V1))
ACTIVATE(n__s(X)) → S(X)
ISNAT(n__plus(V1, V2)) → ACTIVATE(V2)
ISNAT(n__s(V1)) → ACTIVATE(V1)
U41¹(tt, M, N) → U42¹(isNat(activate(N)), activate(M), activate(N))
ISNAT(n__plus(V1, V2)) → ACTIVATE(V1)
PLUS(N, s(M)) → U41¹(isNat(M), M, N)
U11¹(tt, V2) → ISNAT(activate(V2))
ISNAT(n__plus(V1, V2)) → ISNAT(activate(V1))
U42¹(tt, M, N) → ACTIVATE(N)

The TRS R consists of the following rules:

U11(tt, V2) → U12(isNat(activate(V2)))
U12(tt) → tt
U21(tt) → tt
U31(tt, N) → activate(N)
U41(tt, M, N) → U42(isNat(activate(N)), activate(M), activate(N))
U42(tt, M, N) → s(plus(activate(N), activate(M)))
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → U11(isNat(activate(V1)), activate(V2))
isNat(n__s(V1)) → U21(isNat(activate(V1)))
plus(N, 0) → U31(isNat(N), N)
plus(N, s(M)) → U41(isNat(M), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
activate(n__s(X)) → s(X)
activate(X) → X

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 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

PLUS(N, 0) → U31¹(isNat(N), N)
U41¹(tt, M, N) → ACTIVATE(M)
ACTIVATE(n__plus(X1, X2)) → PLUS(X1, X2)
U42¹(tt, M, N) → ACTIVATE(M)
PLUS(N, s(M)) → ISNAT(M)
ISNAT(n__s(V1)) → ISNAT(activate(V1))
ISNAT(n__plus(V1, V2)) → ACTIVATE(V2)
ISNAT(n__s(V1)) → ACTIVATE(V1)
U42¹(tt, M, N) → PLUS(activate(N), activate(M))
U41¹(tt, M, N) → U42¹(isNat(activate(N)), activate(M), activate(N))
U31¹(tt, N) → ACTIVATE(N)
ISNAT(n__plus(V1, V2)) → ACTIVATE(V1)
PLUS(N, s(M)) → U41¹(isNat(M), M, N)
U11¹(tt, V2) → ISNAT(activate(V2))
U41¹(tt, M, N) → ISNAT(activate(N))
U41¹(tt, M, N) → ACTIVATE(N)
ISNAT(n__plus(V1, V2)) → U11¹(isNat(activate(V1)), activate(V2))
U11¹(tt, V2) → ACTIVATE(V2)
PLUS(N, 0) → ISNAT(N)
ISNAT(n__plus(V1, V2)) → ISNAT(activate(V1))
U42¹(tt, M, N) → ACTIVATE(N)

The TRS R consists of the following rules:

U11(tt, V2) → U12(isNat(activate(V2)))
U12(tt) → tt
U21(tt) → tt
U31(tt, N) → activate(N)
U41(tt, M, N) → U42(isNat(activate(N)), activate(M), activate(N))
U42(tt, M, N) → s(plus(activate(N), activate(M)))
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → U11(isNat(activate(V1)), activate(V2))
isNat(n__s(V1)) → U21(isNat(activate(V1)))
plus(N, 0) → U31(isNat(N), N)
plus(N, s(M)) → U41(isNat(M), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
activate(n__s(X)) → s(X)
activate(X) → 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.

U41¹(tt, M, N) → ACTIVATE(M)
ACTIVATE(n__plus(X1, X2)) → PLUS(X1, X2)
U42¹(tt, M, N) → ACTIVATE(M)
PLUS(N, s(M)) → ISNAT(M)
ISNAT(n__plus(V1, V2)) → ACTIVATE(V2)
U31¹(tt, N) → ACTIVATE(N)
ISNAT(n__plus(V1, V2)) → ACTIVATE(V1)
U11¹(tt, V2) → ISNAT(activate(V2))
U41¹(tt, M, N) → ISNAT(activate(N))
U41¹(tt, M, N) → ACTIVATE(N)
U11¹(tt, V2) → ACTIVATE(V2)
PLUS(N, 0) → ISNAT(N)
ISNAT(n__plus(V1, V2)) → ISNAT(activate(V1))
U42¹(tt, M, N) → ACTIVATE(N)

The remaining pairs can at least be oriented weakly.

PLUS(N, 0) → U31¹(isNat(N), N)
ISNAT(n__s(V1)) → ISNAT(activate(V1))
ISNAT(n__s(V1)) → ACTIVATE(V1)
U42¹(tt, M, N) → PLUS(activate(N), activate(M))
U41¹(tt, M, N) → U42¹(isNat(activate(N)), activate(M), activate(N))
PLUS(N, s(M)) → U41¹(isNat(M), M, N)
ISNAT(n__plus(V1, V2)) → U11¹(isNat(activate(V1)), activate(V2))

Used ordering: Polynomial interpretation [25,35]:

POL(n__plus(x₁, x₂)) = 2 + x_1 + (2)x_2
POL(plus(x₁, x₂)) = 2 + x_1 + (2)x_2
POL(U42¹(x₁, x₂, x₃)) = 1 + x_2 + x_3
POL(U31(x₁, x₂)) = 2 + x_2
POL(U41(x₁, x₂, x₃)) = 2 + (2)x_2 + x_3
POL(activate(x₁)) = x_1
POL(U11(x₁, x₂)) = 0
POL(n__s(x₁)) = x_1
POL(0) = 0
POL(U41¹(x₁, x₂, x₃)) = 1 + x_2 + x_3
POL(ISNAT(x₁)) = x_1
POL(PLUS(x₁, x₂)) = 1 + x_1 + x_2
POL(tt) = 0
POL(U11¹(x₁, x₂)) = 2 + (2)x_2
POL(U42(x₁, x₂, x₃)) = 2 + (2)x_2 + x_3
POL(U31¹(x₁, x₂)) = 1 + x_2
POL(n__0) = 0
POL(U12(x₁)) = 0
POL(s(x₁)) = x_1
POL(isNat(x₁)) = 0
POL(ACTIVATE(x₁)) = x_1
POL(U21(x₁)) = 0

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

isNat(n__plus(V1, V2)) → U11(isNat(activate(V1)), activate(V2))
isNat(n__s(V1)) → U21(isNat(activate(V1)))
U42(tt, M, N) → s(plus(activate(N), activate(M)))
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
plus(N, s(M)) → U41(isNat(M), M, N)
activate(n__plus(X1, X2)) → plus(X1, X2)
plus(N, 0) → U31(isNat(N), N)
U31(tt, N) → activate(N)
activate(n__s(X)) → s(X)
s(X) → n__s(X)
activate(n__0) → 0
activate(X) → X
U11(tt, V2) → U12(isNat(activate(V2)))
U21(tt) → tt
U12(tt) → tt
U41(tt, M, N) → U42(isNat(activate(N)), activate(M), activate(N))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

PLUS(N, 0) → U31¹(isNat(N), N)
PLUS(N, s(M)) → U41¹(isNat(M), M, N)
ISNAT(n__s(V1)) → ISNAT(activate(V1))
ISNAT(n__plus(V1, V2)) → U11¹(isNat(activate(V1)), activate(V2))
ISNAT(n__s(V1)) → ACTIVATE(V1)
U41¹(tt, M, N) → U42¹(isNat(activate(N)), activate(M), activate(N))
U42¹(tt, M, N) → PLUS(activate(N), activate(M))

The TRS R consists of the following rules:

U11(tt, V2) → U12(isNat(activate(V2)))
U12(tt) → tt
U21(tt) → tt
U31(tt, N) → activate(N)
U41(tt, M, N) → U42(isNat(activate(N)), activate(M), activate(N))
U42(tt, M, N) → s(plus(activate(N), activate(M)))
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → U11(isNat(activate(V1)), activate(V2))
isNat(n__s(V1)) → U21(isNat(activate(V1)))
plus(N, 0) → U31(isNat(N), N)
plus(N, s(M)) → U41(isNat(M), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
activate(n__s(X)) → s(X)
activate(X) → X

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

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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

ISNAT(n__s(V1)) → ISNAT(activate(V1))

The TRS R consists of the following rules:

U11(tt, V2) → U12(isNat(activate(V2)))
U12(tt) → tt
U21(tt) → tt
U31(tt, N) → activate(N)
U41(tt, M, N) → U42(isNat(activate(N)), activate(M), activate(N))
U42(tt, M, N) → s(plus(activate(N), activate(M)))
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → U11(isNat(activate(V1)), activate(V2))
isNat(n__s(V1)) → U21(isNat(activate(V1)))
plus(N, 0) → U31(isNat(N), N)
plus(N, s(M)) → U41(isNat(M), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
activate(n__s(X)) → s(X)
activate(X) → 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.

ISNAT(n__s(V1)) → ISNAT(activate(V1))

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

POL(n__plus(x₁, x₂)) = x_1 + (4)x_2
POL(plus(x₁, x₂)) = x_1 + (4)x_2
POL(U31(x₁, x₂)) = x_2
POL(U41(x₁, x₂, x₃)) = 1 + (4)x_2 + x_3
POL(activate(x₁)) = x_1
POL(n__s(x₁)) = 1 + x_1
POL(U11(x₁, x₂)) = 4
POL(0) = 0
POL(ISNAT(x₁)) = (4)x_1
POL(tt) = 2
POL(U42(x₁, x₂, x₃)) = 1 + (4)x_2 + x_3
POL(n__0) = 0
POL(U12(x₁)) = 4
POL(s(x₁)) = 1 + x_1
POL(isNat(x₁)) = 4
POL(U21(x₁)) = 4

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

isNat(n__plus(V1, V2)) → U11(isNat(activate(V1)), activate(V2))
isNat(n__s(V1)) → U21(isNat(activate(V1)))
U42(tt, M, N) → s(plus(activate(N), activate(M)))
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
plus(N, s(M)) → U41(isNat(M), M, N)
activate(n__plus(X1, X2)) → plus(X1, X2)
plus(N, 0) → U31(isNat(N), N)
U31(tt, N) → activate(N)
activate(n__s(X)) → s(X)
s(X) → n__s(X)
activate(n__0) → 0
activate(X) → X
U11(tt, V2) → U12(isNat(activate(V2)))
U21(tt) → tt
U12(tt) → tt
U41(tt, M, N) → U42(isNat(activate(N)), activate(M), activate(N))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

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

U11(tt, V2) → U12(isNat(activate(V2)))
U12(tt) → tt
U21(tt) → tt
U31(tt, N) → activate(N)
U41(tt, M, N) → U42(isNat(activate(N)), activate(M), activate(N))
U42(tt, M, N) → s(plus(activate(N), activate(M)))
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → U11(isNat(activate(V1)), activate(V2))
isNat(n__s(V1)) → U21(isNat(activate(V1)))
plus(N, 0) → U31(isNat(N), N)
plus(N, s(M)) → U41(isNat(M), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
activate(n__s(X)) → s(X)
activate(X) → 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

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

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

PLUS(N, s(M)) → U41¹(isNat(M), M, N)
U42¹(tt, M, N) → PLUS(activate(N), activate(M))
U41¹(tt, M, N) → U42¹(isNat(activate(N)), activate(M), activate(N))

The TRS R consists of the following rules:

U11(tt, V2) → U12(isNat(activate(V2)))
U12(tt) → tt
U21(tt) → tt
U31(tt, N) → activate(N)
U41(tt, M, N) → U42(isNat(activate(N)), activate(M), activate(N))
U42(tt, M, N) → s(plus(activate(N), activate(M)))
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → U11(isNat(activate(V1)), activate(V2))
isNat(n__s(V1)) → U21(isNat(activate(V1)))
plus(N, 0) → U31(isNat(N), N)
plus(N, s(M)) → U41(isNat(M), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
activate(n__s(X)) → s(X)
activate(X) → 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(N, s(M)) → U41¹(isNat(M), M, N)

The remaining pairs can at least be oriented weakly.

U42¹(tt, M, N) → PLUS(activate(N), activate(M))
U41¹(tt, M, N) → U42¹(isNat(activate(N)), activate(M), activate(N))

Used ordering: Polynomial interpretation [25,35]:

POL(n__plus(x₁, x₂)) = (4)x_1 + x_2
POL(plus(x₁, x₂)) = (4)x_1 + x_2
POL(U42¹(x₁, x₂, x₃)) = 4 + (3)x_2
POL(U31(x₁, x₂)) = (4)x_2
POL(U41(x₁, x₂, x₃)) = 4 + x_2 + (4)x_3
POL(activate(x₁)) = x_1
POL(U11(x₁, x₂)) = 0
POL(n__s(x₁)) = 4 + x_1
POL(0) = 0
POL(U41¹(x₁, x₂, x₃)) = 4 + (3)x_2
POL(PLUS(x₁, x₂)) = 4 + (3)x_2
POL(tt) = 0
POL(U42(x₁, x₂, x₃)) = 4 + x_2 + (4)x_3
POL(n__0) = 0
POL(U12(x₁)) = 0
POL(s(x₁)) = 4 + x_1
POL(isNat(x₁)) = 0
POL(U21(x₁)) = 0

The value of delta used in the strict ordering is 12.
The following usable rules [17] were oriented:

isNat(n__plus(V1, V2)) → U11(isNat(activate(V1)), activate(V2))
isNat(n__s(V1)) → U21(isNat(activate(V1)))
U42(tt, M, N) → s(plus(activate(N), activate(M)))
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
plus(N, s(M)) → U41(isNat(M), M, N)
activate(n__plus(X1, X2)) → plus(X1, X2)
plus(N, 0) → U31(isNat(N), N)
U31(tt, N) → activate(N)
activate(n__s(X)) → s(X)
s(X) → n__s(X)
activate(n__0) → 0
activate(X) → X
U11(tt, V2) → U12(isNat(activate(V2)))
U21(tt) → tt
U12(tt) → tt
U41(tt, M, N) → U42(isNat(activate(N)), activate(M), activate(N))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

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

U41¹(tt, M, N) → U42¹(isNat(activate(N)), activate(M), activate(N))
U42¹(tt, M, N) → PLUS(activate(N), activate(M))

The TRS R consists of the following rules:

U11(tt, V2) → U12(isNat(activate(V2)))
U12(tt) → tt
U21(tt) → tt
U31(tt, N) → activate(N)
U41(tt, M, N) → U42(isNat(activate(N)), activate(M), activate(N))
U42(tt, M, N) → s(plus(activate(N), activate(M)))
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → U11(isNat(activate(V1)), activate(V2))
isNat(n__s(V1)) → U21(isNat(activate(V1)))
plus(N, 0) → U31(isNat(N), N)
plus(N, s(M)) → U41(isNat(M), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
activate(n__s(X)) → s(X)
activate(X) → 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.