Termination w.r.t. Q proof of /tmp/tmp64ILqR/PEANO_nokinds

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, N) → activate(N)
U21(tt, M, N) → s(plus(activate(N), activate(M)))
and(tt, X) → activate(X)
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → and(isNat(activate(V1)), n__isNat(activate(V2)))
isNat(n__s(V1)) → isNat(activate(V1))
plus(N, 0) → U11(isNat(N), N)
plus(N, s(M)) → U21(and(isNat(M), n__isNat(N)), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
isNat(X) → n__isNat(X)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
activate(n__isNat(X)) → isNat(X)
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, N) → activate(N)
U21(tt, M, N) → s(plus(activate(N), activate(M)))
and(tt, X) → activate(X)
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → and(isNat(activate(V1)), n__isNat(activate(V2)))
isNat(n__s(V1)) → isNat(activate(V1))
plus(N, 0) → U11(isNat(N), N)
plus(N, s(M)) → U21(and(isNat(M), n__isNat(N)), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
isNat(X) → n__isNat(X)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
activate(n__isNat(X)) → isNat(X)
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:

ACTIVATE(n__plus(X1, X2)) → PLUS(X1, X2)
U21¹(tt, M, N) → S(plus(activate(N), activate(M)))
PLUS(N, s(M)) → ISNAT(M)
ISNAT(n__s(V1)) → ISNAT(activate(V1))
ACTIVATE(n__s(X)) → S(X)
ISNAT(n__plus(V1, V2)) → AND(isNat(activate(V1)), n__isNat(activate(V2)))
AND(tt, X) → ACTIVATE(X)
ISNAT(n__plus(V1, V2)) → ACTIVATE(V2)
ISNAT(n__s(V1)) → ACTIVATE(V1)
PLUS(N, 0) → U11¹(isNat(N), N)
U21¹(tt, M, N) → PLUS(activate(N), activate(M))
U21¹(tt, M, N) → ACTIVATE(M)
PLUS(N, s(M)) → U21¹(and(isNat(M), n__isNat(N)), M, N)
ISNAT(n__plus(V1, V2)) → ACTIVATE(V1)
ACTIVATE(n__0) → 0¹
U21¹(tt, M, N) → ACTIVATE(N)
ACTIVATE(n__isNat(X)) → ISNAT(X)
PLUS(N, s(M)) → AND(isNat(M), n__isNat(N))
PLUS(N, 0) → ISNAT(N)
U11¹(tt, N) → ACTIVATE(N)
ISNAT(n__plus(V1, V2)) → ISNAT(activate(V1))

The TRS R consists of the following rules:

U11(tt, N) → activate(N)
U21(tt, M, N) → s(plus(activate(N), activate(M)))
and(tt, X) → activate(X)
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → and(isNat(activate(V1)), n__isNat(activate(V2)))
isNat(n__s(V1)) → isNat(activate(V1))
plus(N, 0) → U11(isNat(N), N)
plus(N, s(M)) → U21(and(isNat(M), n__isNat(N)), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
isNat(X) → n__isNat(X)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
activate(n__isNat(X)) → isNat(X)
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:

ACTIVATE(n__plus(X1, X2)) → PLUS(X1, X2)
U21¹(tt, M, N) → S(plus(activate(N), activate(M)))
PLUS(N, s(M)) → ISNAT(M)
ISNAT(n__s(V1)) → ISNAT(activate(V1))
ACTIVATE(n__s(X)) → S(X)
ISNAT(n__plus(V1, V2)) → AND(isNat(activate(V1)), n__isNat(activate(V2)))
AND(tt, X) → ACTIVATE(X)
ISNAT(n__plus(V1, V2)) → ACTIVATE(V2)
ISNAT(n__s(V1)) → ACTIVATE(V1)
PLUS(N, 0) → U11¹(isNat(N), N)
U21¹(tt, M, N) → PLUS(activate(N), activate(M))
U21¹(tt, M, N) → ACTIVATE(M)
PLUS(N, s(M)) → U21¹(and(isNat(M), n__isNat(N)), M, N)
ISNAT(n__plus(V1, V2)) → ACTIVATE(V1)
ACTIVATE(n__0) → 0¹
U21¹(tt, M, N) → ACTIVATE(N)
ACTIVATE(n__isNat(X)) → ISNAT(X)
PLUS(N, s(M)) → AND(isNat(M), n__isNat(N))
PLUS(N, 0) → ISNAT(N)
U11¹(tt, N) → ACTIVATE(N)
ISNAT(n__plus(V1, V2)) → ISNAT(activate(V1))

The TRS R consists of the following rules:

U11(tt, N) → activate(N)
U21(tt, M, N) → s(plus(activate(N), activate(M)))
and(tt, X) → activate(X)
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → and(isNat(activate(V1)), n__isNat(activate(V2)))
isNat(n__s(V1)) → isNat(activate(V1))
plus(N, 0) → U11(isNat(N), N)
plus(N, s(M)) → U21(and(isNat(M), n__isNat(N)), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
isNat(X) → n__isNat(X)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
activate(n__isNat(X)) → isNat(X)
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 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

ACTIVATE(n__plus(X1, X2)) → PLUS(X1, X2)
PLUS(N, s(M)) → ISNAT(M)
ISNAT(n__s(V1)) → ISNAT(activate(V1))
ISNAT(n__plus(V1, V2)) → AND(isNat(activate(V1)), n__isNat(activate(V2)))
PLUS(N, 0) → U11¹(isNat(N), N)
AND(tt, X) → ACTIVATE(X)
ISNAT(n__plus(V1, V2)) → ACTIVATE(V2)
ISNAT(n__s(V1)) → ACTIVATE(V1)
U21¹(tt, M, N) → ACTIVATE(M)
U21¹(tt, M, N) → PLUS(activate(N), activate(M))
PLUS(N, s(M)) → U21¹(and(isNat(M), n__isNat(N)), M, N)
ISNAT(n__plus(V1, V2)) → ACTIVATE(V1)
U21¹(tt, M, N) → ACTIVATE(N)
ACTIVATE(n__isNat(X)) → ISNAT(X)
PLUS(N, s(M)) → AND(isNat(M), n__isNat(N))
U11¹(tt, N) → ACTIVATE(N)
ISNAT(n__plus(V1, V2)) → ISNAT(activate(V1))
PLUS(N, 0) → ISNAT(N)

The TRS R consists of the following rules:

U11(tt, N) → activate(N)
U21(tt, M, N) → s(plus(activate(N), activate(M)))
and(tt, X) → activate(X)
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → and(isNat(activate(V1)), n__isNat(activate(V2)))
isNat(n__s(V1)) → isNat(activate(V1))
plus(N, 0) → U11(isNat(N), N)
plus(N, s(M)) → U21(and(isNat(M), n__isNat(N)), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
isNat(X) → n__isNat(X)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
activate(n__isNat(X)) → isNat(X)
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.

ACTIVATE(n__plus(X1, X2)) → PLUS(X1, X2)
ISNAT(n__plus(V1, V2)) → AND(isNat(activate(V1)), n__isNat(activate(V2)))
PLUS(N, 0) → U11¹(isNat(N), N)
ISNAT(n__plus(V1, V2)) → ACTIVATE(V2)
ISNAT(n__plus(V1, V2)) → ACTIVATE(V1)
ISNAT(n__plus(V1, V2)) → ISNAT(activate(V1))
PLUS(N, 0) → ISNAT(N)

The remaining pairs can at least be oriented weakly.

PLUS(N, s(M)) → ISNAT(M)
ISNAT(n__s(V1)) → ISNAT(activate(V1))
AND(tt, X) → ACTIVATE(X)
ISNAT(n__s(V1)) → ACTIVATE(V1)
U21¹(tt, M, N) → ACTIVATE(M)
U21¹(tt, M, N) → PLUS(activate(N), activate(M))
PLUS(N, s(M)) → U21¹(and(isNat(M), n__isNat(N)), M, N)
U21¹(tt, M, N) → ACTIVATE(N)
ACTIVATE(n__isNat(X)) → ISNAT(X)
PLUS(N, s(M)) → AND(isNat(M), n__isNat(N))
U11¹(tt, N) → ACTIVATE(N)

Used ordering: Polynomial interpretation [25]:

POL(0) = 1
POL(ACTIVATE(x₁)) = x₁
POL(AND(x₁, x₂)) = x₂
POL(ISNAT(x₁)) = x₁
POL(PLUS(x₁, x₂)) = x₁ + x₂
POL(U11(x₁, x₂)) = x₂
POL(U11¹(x₁, x₂)) = x₂
POL(U21(x₁, x₂, x₃)) = 1 + x₂ + x₃
POL(U21¹(x₁, x₂, x₃)) = x₂ + x₃
POL(activate(x₁)) = x₁
POL(and(x₁, x₂)) = x₂
POL(isNat(x₁)) = x₁
POL(n__0) = 1
POL(n__isNat(x₁)) = x₁
POL(n__plus(x₁, x₂)) = 1 + x₁ + x₂
POL(n__s(x₁)) = x₁
POL(plus(x₁, x₂)) = 1 + x₁ + x₂
POL(s(x₁)) = x₁
POL(tt) = 0

The following usable rules [17] were oriented:

activate(n__s(X)) → s(X)
activate(X) → X
U21(tt, M, N) → s(plus(activate(N), activate(M)))
isNat(n__0) → tt
isNat(X) → n__isNat(X)
plus(X1, X2) → n__plus(X1, X2)
0 → n__0
plus(N, s(M)) → U21(and(isNat(M), n__isNat(N)), M, N)
activate(n__isNat(X)) → isNat(X)
isNat(n__s(V1)) → isNat(activate(V1))
and(tt, X) → activate(X)
isNat(n__plus(V1, V2)) → and(isNat(activate(V1)), n__isNat(activate(V2)))
activate(n__plus(X1, X2)) → plus(X1, X2)
plus(N, 0) → U11(isNat(N), N)
U11(tt, N) → activate(N)
activate(n__0) → 0
s(X) → n__s(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

PLUS(N, s(M)) → U21¹(and(isNat(M), n__isNat(N)), M, N)
U21¹(tt, M, N) → ACTIVATE(N)
ACTIVATE(n__isNat(X)) → ISNAT(X)
ISNAT(n__s(V1)) → ISNAT(activate(V1))
PLUS(N, s(M)) → ISNAT(M)
ISNAT(n__s(V1)) → ACTIVATE(V1)
AND(tt, X) → ACTIVATE(X)
PLUS(N, s(M)) → AND(isNat(M), n__isNat(N))
U11¹(tt, N) → ACTIVATE(N)
U21¹(tt, M, N) → PLUS(activate(N), activate(M))
U21¹(tt, M, N) → ACTIVATE(M)

The TRS R consists of the following rules:

U11(tt, N) → activate(N)
U21(tt, M, N) → s(plus(activate(N), activate(M)))
and(tt, X) → activate(X)
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → and(isNat(activate(V1)), n__isNat(activate(V2)))
isNat(n__s(V1)) → isNat(activate(V1))
plus(N, 0) → U11(isNat(N), N)
plus(N, s(M)) → U21(and(isNat(M), n__isNat(N)), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
isNat(X) → n__isNat(X)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
activate(n__isNat(X)) → isNat(X)
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 6 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:

ACTIVATE(n__isNat(X)) → ISNAT(X)
ISNAT(n__s(V1)) → ISNAT(activate(V1))
ISNAT(n__s(V1)) → ACTIVATE(V1)

The TRS R consists of the following rules:

U11(tt, N) → activate(N)
U21(tt, M, N) → s(plus(activate(N), activate(M)))
and(tt, X) → activate(X)
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → and(isNat(activate(V1)), n__isNat(activate(V2)))
isNat(n__s(V1)) → isNat(activate(V1))
plus(N, 0) → U11(isNat(N), N)
plus(N, s(M)) → U21(and(isNat(M), n__isNat(N)), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
isNat(X) → n__isNat(X)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
activate(n__isNat(X)) → isNat(X)
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.

ACTIVATE(n__isNat(X)) → ISNAT(X)

The remaining pairs can at least be oriented weakly.

ISNAT(n__s(V1)) → ISNAT(activate(V1))
ISNAT(n__s(V1)) → ACTIVATE(V1)

Used ordering: Matrix interpretation [3]:
Non-tuple symbols:

M( n__plus(x₁, x₂) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

/	0	0	\
\	0	1	/

x₂

M( plus(x₁, x₂) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

/	0	0	\
\	0	1	/

x₂

M( n__isNat(x₁) ) =

/	0	\
\	1	/

/	0	0	\
\	0	1	/

x₁

M( activate(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

M( n__s(x₁) ) =

/	0	\
\	0	/

/	0	0	\
\	0	1	/

x₁

M( and(x₁, x₂) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

/	1	0	\
\	0	1	/

x₂

M( U11(x₁, x₂) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

/	1	0	\
\	0	1	/

x₂

M( 0 ) =

/	0	\
\	0	/

M( tt ) =

/	0	\
\	0	/

M( n__0 ) =

/	0	\
\	0	/

M( s(x₁) ) =

/	0	\
\	0	/

/	0	0	\
\	0	1	/

x₁

M( isNat(x₁) ) =

/	0	\
\	1	/

/	0	0	\
\	0	1	/

x₁

M( U21(x₁, ..., x₃) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

/	0	0	\
\	0	1	/

x₂

/	0	0	\
\	0	1	/

x₃

Tuple symbols:

M( ACTIVATE(x₁) ) =

[

]

x₁

M( ISNAT(x₁) ) =

[

]

x₁

Matrix type:
We used a basic matrix type which is not further parametrizeable.

As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

U21(tt, M, N) → s(plus(activate(N), activate(M)))
activate(n__s(X)) → s(X)
activate(X) → X
isNat(n__0) → tt
isNat(X) → n__isNat(X)
plus(X1, X2) → n__plus(X1, X2)
0 → n__0
plus(N, s(M)) → U21(and(isNat(M), n__isNat(N)), M, N)
activate(n__isNat(X)) → isNat(X)
isNat(n__s(V1)) → isNat(activate(V1))
and(tt, X) → activate(X)
isNat(n__plus(V1, V2)) → and(isNat(activate(V1)), n__isNat(activate(V2)))
activate(n__plus(X1, X2)) → plus(X1, X2)
plus(N, 0) → U11(isNat(N), N)
U11(tt, N) → activate(N)
activate(n__0) → 0
s(X) → n__s(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                  ↳ QDP

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

ISNAT(n__s(V1)) → ISNAT(activate(V1))
ISNAT(n__s(V1)) → ACTIVATE(V1)

The TRS R consists of the following rules:

U11(tt, N) → activate(N)
U21(tt, M, N) → s(plus(activate(N), activate(M)))
and(tt, X) → activate(X)
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → and(isNat(activate(V1)), n__isNat(activate(V2)))
isNat(n__s(V1)) → isNat(activate(V1))
plus(N, 0) → U11(isNat(N), N)
plus(N, s(M)) → U21(and(isNat(M), n__isNat(N)), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
isNat(X) → n__isNat(X)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
activate(n__isNat(X)) → isNat(X)
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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ 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, N) → activate(N)
U21(tt, M, N) → s(plus(activate(N), activate(M)))
and(tt, X) → activate(X)
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → and(isNat(activate(V1)), n__isNat(activate(V2)))
isNat(n__s(V1)) → isNat(activate(V1))
plus(N, 0) → U11(isNat(N), N)
plus(N, s(M)) → U21(and(isNat(M), n__isNat(N)), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
isNat(X) → n__isNat(X)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
activate(n__isNat(X)) → isNat(X)
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]:

POL(0) = 0
POL(ISNAT(x₁)) = x₁
POL(U11(x₁, x₂)) = x₂
POL(U21(x₁, x₂, x₃)) = 1 + x₂ + x₃
POL(activate(x₁)) = x₁
POL(and(x₁, x₂)) = x₂
POL(isNat(x₁)) = 0
POL(n__0) = 0
POL(n__isNat(x₁)) = 0
POL(n__plus(x₁, x₂)) = x₁ + x₂
POL(n__s(x₁)) = 1 + x₁
POL(plus(x₁, x₂)) = x₁ + x₂
POL(s(x₁)) = 1 + x₁
POL(tt) = 0

The following usable rules [17] were oriented:

U21(tt, M, N) → s(plus(activate(N), activate(M)))
activate(n__s(X)) → s(X)
activate(X) → X
isNat(n__0) → tt
isNat(X) → n__isNat(X)
plus(X1, X2) → n__plus(X1, X2)
0 → n__0
plus(N, s(M)) → U21(and(isNat(M), n__isNat(N)), M, N)
activate(n__isNat(X)) → isNat(X)
isNat(n__s(V1)) → isNat(activate(V1))
and(tt, X) → activate(X)
isNat(n__plus(V1, V2)) → and(isNat(activate(V1)), n__isNat(activate(V2)))
activate(n__plus(X1, X2)) → plus(X1, X2)
plus(N, 0) → U11(isNat(N), N)
U11(tt, N) → activate(N)
activate(n__0) → 0
s(X) → n__s(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ PisEmptyProof

                  ↳ QDP

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

U11(tt, N) → activate(N)
U21(tt, M, N) → s(plus(activate(N), activate(M)))
and(tt, X) → activate(X)
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → and(isNat(activate(V1)), n__isNat(activate(V2)))
isNat(n__s(V1)) → isNat(activate(V1))
plus(N, 0) → U11(isNat(N), N)
plus(N, s(M)) → U21(and(isNat(M), n__isNat(N)), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
isNat(X) → n__isNat(X)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
activate(n__isNat(X)) → isNat(X)
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)) → U21¹(and(isNat(M), n__isNat(N)), M, N)
U21¹(tt, M, N) → PLUS(activate(N), activate(M))

The TRS R consists of the following rules:

U11(tt, N) → activate(N)
U21(tt, M, N) → s(plus(activate(N), activate(M)))
and(tt, X) → activate(X)
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → and(isNat(activate(V1)), n__isNat(activate(V2)))
isNat(n__s(V1)) → isNat(activate(V1))
plus(N, 0) → U11(isNat(N), N)
plus(N, s(M)) → U21(and(isNat(M), n__isNat(N)), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
isNat(X) → n__isNat(X)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
activate(n__isNat(X)) → isNat(X)
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)) → U21¹(and(isNat(M), n__isNat(N)), M, N)

The remaining pairs can at least be oriented weakly.

U21¹(tt, M, N) → PLUS(activate(N), activate(M))

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(PLUS(x₁, x₂)) = x₂
POL(U11(x₁, x₂)) = x₂
POL(U21(x₁, x₂, x₃)) = 1 + x₂ + x₃
POL(U21¹(x₁, x₂, x₃)) = x₂
POL(activate(x₁)) = x₁
POL(and(x₁, x₂)) = x₂
POL(isNat(x₁)) = 0
POL(n__0) = 0
POL(n__isNat(x₁)) = 0
POL(n__plus(x₁, x₂)) = x₁ + x₂
POL(n__s(x₁)) = 1 + x₁
POL(plus(x₁, x₂)) = x₁ + x₂
POL(s(x₁)) = 1 + x₁
POL(tt) = 0

The following usable rules [17] were oriented:

U21(tt, M, N) → s(plus(activate(N), activate(M)))
activate(n__s(X)) → s(X)
activate(X) → X
isNat(n__0) → tt
isNat(X) → n__isNat(X)
plus(X1, X2) → n__plus(X1, X2)
0 → n__0
plus(N, s(M)) → U21(and(isNat(M), n__isNat(N)), M, N)
activate(n__isNat(X)) → isNat(X)
isNat(n__s(V1)) → isNat(activate(V1))
and(tt, X) → activate(X)
isNat(n__plus(V1, V2)) → and(isNat(activate(V1)), n__isNat(activate(V2)))
activate(n__plus(X1, X2)) → plus(X1, X2)
plus(N, 0) → U11(isNat(N), N)
U11(tt, N) → activate(N)
activate(n__0) → 0
s(X) → n__s(X)

↳ 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:

U21¹(tt, M, N) → PLUS(activate(N), activate(M))

The TRS R consists of the following rules:

U11(tt, N) → activate(N)
U21(tt, M, N) → s(plus(activate(N), activate(M)))
and(tt, X) → activate(X)
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → and(isNat(activate(V1)), n__isNat(activate(V2)))
isNat(n__s(V1)) → isNat(activate(V1))
plus(N, 0) → U11(isNat(N), N)
plus(N, s(M)) → U21(and(isNat(M), n__isNat(N)), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
isNat(X) → n__isNat(X)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
activate(n__isNat(X)) → isNat(X)
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 1 less node.