Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
↳5 QDPOrderProof (⇔)
↳6 QDP
↳7 DependencyGraphProof (⇔)
↳8 AND
  ↳9 QDP
    ↳10 QDPOrderProof (⇔)
    ↳11 QDP
    ↳12 DependencyGraphProof (⇔)
    ↳13 TRUE
  ↳14 QDP
    ↳15 QDPOrderProof (⇔)
    ↳16 QDP
    ↳17 DependencyGraphProof (⇔)
    ↳18 AND
      ↳19 QDP
        ↳20 QDPOrderProof (⇔)
        ↳21 QDP
        ↳22 PisEmptyProof (⇔)
        ↳23 TRUE
      ↳24 QDP
        ↳25 QDPOrderProof (⇔)
        ↳26 QDP
        ↳27 PisEmptyProof (⇔)
        ↳28 TRUE

(0) Obligation:

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(activate(X1), activate(X2))
activate(n__isNat(X)) → isNat(X)
activate(n__s(X)) → s(activate(X))
activate(X) → X

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(4) Obligation:

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

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVATE(n__plus(X1, X2)) → PLUS(activate(X1), activate(X2))
PLUS(N, 0) → U111(isNat(N), N)
U111(tt, N) → ACTIVATE(N)
ACTIVATE(n__plus(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__plus(X1, X2)) → ACTIVATE(X2)
ISNAT(n__plus(V1, V2)) → AND(isNat(activate(V1)), n__isNat(activate(V2)))
AND(tt, X) → ACTIVATE(X)
ISNAT(n__plus(V1, V2)) → ISNAT(activate(V1))
ISNAT(n__plus(V1, V2)) → ACTIVATE(V1)
ISNAT(n__plus(V1, V2)) → ACTIVATE(V2)
PLUS(N, 0) → ISNAT(N)
PLUS(N, s(M)) → AND(isNat(M), n__isNat(N))
PLUS(N, s(M)) → ISNAT(M)
U211(tt, M, N) → ACTIVATE(N)
U211(tt, M, N) → ACTIVATE(M)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  x1
n__plus(x1, x2)  =  n__plus(x1, x2)
PLUS(x1, x2)  =  PLUS(x1, x2)
activate(x1)  =  x1
0  =  0
U111(x1, x2)  =  U111(x1, x2)
isNat(x1)  =  x1
tt  =  tt
n__isNat(x1)  =  x1
ISNAT(x1)  =  x1
AND(x1, x2)  =  AND(x2)
n__s(x1)  =  x1
s(x1)  =  x1
U211(x1, x2, x3)  =  U211(x2, x3)
and(x1, x2)  =  and(x1, x2)
U11(x1, x2)  =  x2
U21(x1, x2, x3)  =  U21(x2, x3)
plus(x1, x2)  =  plus(x1, x2)
n__0  =  n__0

Recursive Path Order [RPO].
Precedence:
[nplus2, U212, plus2] > [PLUS2, tt, U21^12] > U11^12
[nplus2, U212, plus2] > [PLUS2, tt, U21^12] > AND1
[nplus2, U212, plus2] > and2
[0, n0] > [PLUS2, tt, U21^12] > U11^12
[0, n0] > [PLUS2, tt, U21^12] > AND1


The following usable rules [FROCOS05] were oriented:

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(activate(X1), activate(X2))
activate(n__isNat(X)) → isNat(X)
activate(n__s(X)) → s(activate(X))
activate(X) → X

(6) Obligation:

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

ACTIVATE(n__isNat(X)) → ISNAT(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)
ISNAT(n__s(V1)) → ISNAT(activate(V1))
ISNAT(n__s(V1)) → ACTIVATE(V1)
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(activate(X1), activate(X2))
activate(n__isNat(X)) → isNat(X)
activate(n__s(X)) → s(activate(X))
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.

(8) Complex Obligation (AND)

(9) Obligation:

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

U211(tt, M, N) → PLUS(activate(N), activate(M))
PLUS(N, s(M)) → U211(and(isNat(M), n__isNat(N)), M, 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(activate(X1), activate(X2))
activate(n__isNat(X)) → isNat(X)
activate(n__s(X)) → s(activate(X))
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(10) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


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.
Used ordering: Combined order from the following AFS and order.
U211(x1, x2, x3)  =  U211(x2)
tt  =  tt
PLUS(x1, x2)  =  PLUS(x2)
activate(x1)  =  x1
s(x1)  =  s(x1)
and(x1, x2)  =  x2
isNat(x1)  =  isNat
n__isNat(x1)  =  n__isNat
U11(x1, x2)  =  x2
U21(x1, x2, x3)  =  U21(x1, x2, x3)
plus(x1, x2)  =  plus(x1, x2)
n__0  =  n__0
n__plus(x1, x2)  =  n__plus(x1, x2)
n__s(x1)  =  n__s(x1)
0  =  0

Recursive Path Order [RPO].
Precedence:
[U213, plus2, nplus2] > [U21^11, tt, PLUS1, s1, isNat, nisNat, ns1]
[n0, 0]


The following usable rules [FROCOS05] were oriented:

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(activate(X1), activate(X2))
activate(n__isNat(X)) → isNat(X)
activate(n__s(X)) → s(activate(X))
activate(X) → X

(11) Obligation:

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(activate(X1), activate(X2))
activate(n__isNat(X)) → isNat(X)
activate(n__s(X)) → s(activate(X))
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(12) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(13) TRUE

(14) Obligation:

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

ISNAT(n__s(V1)) → ISNAT(activate(V1))
ISNAT(n__s(V1)) → ACTIVATE(V1)
ACTIVATE(n__isNat(X)) → ISNAT(X)
ACTIVATE(n__s(X)) → ACTIVATE(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)
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(activate(X1), activate(X2))
activate(n__isNat(X)) → isNat(X)
activate(n__s(X)) → s(activate(X))
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


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.
Used ordering: Combined order from the following AFS and order.
ISNAT(x1)  =  ISNAT(x1)
n__s(x1)  =  x1
activate(x1)  =  x1
ACTIVATE(x1)  =  ACTIVATE(x1)
n__isNat(x1)  =  n__isNat(x1)
U11(x1, x2)  =  x2
tt  =  tt
U21(x1, x2, x3)  =  U21(x2, x3)
s(x1)  =  x1
plus(x1, x2)  =  plus(x1, x2)
and(x1, x2)  =  and(x2)
isNat(x1)  =  isNat(x1)
n__0  =  n__0
n__plus(x1, x2)  =  n__plus(x1, x2)
0  =  0

Recursive Path Order [RPO].
Precedence:
[ISNAT1, ACTIVATE1, nisNat1, isNat1] > tt
[ISNAT1, ACTIVATE1, nisNat1, isNat1] > and1
[U212, plus2, nplus2]
[n0, 0]


The following usable rules [FROCOS05] were oriented:

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(activate(X1), activate(X2))
activate(n__isNat(X)) → isNat(X)
activate(n__s(X)) → s(activate(X))
activate(X) → X

(16) Obligation:

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

ISNAT(n__s(V1)) → ISNAT(activate(V1))
ISNAT(n__s(V1)) → ACTIVATE(V1)
ACTIVATE(n__s(X)) → ACTIVATE(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)
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(activate(X1), activate(X2))
activate(n__isNat(X)) → isNat(X)
activate(n__s(X)) → s(activate(X))
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(17) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

(18) Complex Obligation (AND)

(19) Obligation:

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

ACTIVATE(n__s(X)) → ACTIVATE(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)
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(activate(X1), activate(X2))
activate(n__isNat(X)) → isNat(X)
activate(n__s(X)) → s(activate(X))
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVATE(n__s(X)) → ACTIVATE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  ACTIVATE(x1)
n__s(x1)  =  n__s(x1)
U11(x1, x2)  =  U11(x1, x2)
tt  =  tt
activate(x1)  =  x1
U21(x1, x2, x3)  =  U21(x1, x2, x3)
s(x1)  =  s(x1)
plus(x1, x2)  =  plus(x1, x2)
and(x1, x2)  =  x2
isNat(x1)  =  isNat
n__0  =  n__0
n__plus(x1, x2)  =  n__plus(x1, x2)
n__isNat(x1)  =  n__isNat
0  =  0

Recursive Path Order [RPO].
Precedence:
[U112, U213, plus2, nplus2] > [ACTIVATE1, ns1, s1, isNat, n0, nisNat, 0] > tt


The following usable rules [FROCOS05] were oriented:

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(activate(X1), activate(X2))
activate(n__isNat(X)) → isNat(X)
activate(n__s(X)) → s(activate(X))
activate(X) → X

(21) Obligation:

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(activate(X1), activate(X2))
activate(n__isNat(X)) → isNat(X)
activate(n__s(X)) → s(activate(X))
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(22) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(23) TRUE

(24) Obligation:

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(activate(X1), activate(X2))
activate(n__isNat(X)) → isNat(X)
activate(n__s(X)) → s(activate(X))
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(25) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


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.
Used ordering: Combined order from the following AFS and order.
ISNAT(x1)  =  ISNAT(x1)
n__s(x1)  =  n__s(x1)
activate(x1)  =  x1
U11(x1, x2)  =  U11(x1, x2)
tt  =  tt
U21(x1, x2, x3)  =  U21(x1, x2, x3)
s(x1)  =  s(x1)
plus(x1, x2)  =  plus(x1, x2)
and(x1, x2)  =  x2
isNat(x1)  =  isNat
n__0  =  n__0
n__plus(x1, x2)  =  n__plus(x1, x2)
n__isNat(x1)  =  n__isNat
0  =  0

Recursive Path Order [RPO].
Precedence:
[U213, plus2, nplus2] > [ns1, s1, isNat, nisNat] > ISNAT1
[U213, plus2, nplus2] > [ns1, s1, isNat, nisNat] > [tt, n0, 0]
[U213, plus2, nplus2] > U112


The following usable rules [FROCOS05] were oriented:

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(activate(X1), activate(X2))
activate(n__isNat(X)) → isNat(X)
activate(n__s(X)) → s(activate(X))
activate(X) → X

(26) Obligation:

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(activate(X1), activate(X2))
activate(n__isNat(X)) → isNat(X)
activate(n__s(X)) → s(activate(X))
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(27) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(28) TRUE