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)
0n__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)
0n__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)
0n__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)
0n__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)
0n__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)
0n__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)
0n__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)
0n__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))
ISNAT(n__s(V1)) → ACTIVATE(V1)
The remaining pairs can at least be oriented weakly.

ACTIVATE(n__isNat(X)) → ISNAT(X)
Used ordering: Polynomial interpretation [25]:

POL(0) = 0   
POL(ACTIVATE(x1)) = 1 + x1   
POL(ISNAT(x1)) = 1 + 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)) = x1   
POL(n__0) = 0   
POL(n__isNat(x1)) = x1   
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)
0n__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:

ACTIVATE(n__isNat(X)) → ISNAT(X)

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)
0n__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.

↳ 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)
0n__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)
0n__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)
0n__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.