(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)
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.

(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(X1, X2)
ACTIVATE(n__isNat(X)) → ISNAT(X)
ACTIVATE(n__s(X)) → S(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.

(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(X1, X2)
PLUS(N, 0) → U111(isNat(N), N)
U111(tt, N) → ACTIVATE(N)
ACTIVATE(n__isNat(X)) → ISNAT(X)
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)
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)
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.

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic path order with status [LPO].
Quasi-Precedence:
[nplus2, PLUS2, U21^12, U212, plus2] > [U11^11, U112]
[0, n0] > tt > [U11^11, U112]

Status:
nplus2: [1,2]
PLUS2: [1,2]
0: []
U11^11: [1]
tt: []
U21^12: [2,1]
U112: [2,1]
U212: [2,1]
plus2: [1,2]
n0: []


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)
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

(6) Obligation:

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

ACTIVATE(n__plus(X1, X2)) → PLUS(X1, X2)
ACTIVATE(n__isNat(X)) → ISNAT(X)
AND(tt, 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)
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.

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(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)
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.

(10) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


U211(tt, M, N) → PLUS(activate(N), activate(M))
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)  =  x2
activate(x1)  =  x1
s(x1)  =  s(x1)
and(x1, x2)  =  and(x2)
isNat(x1)  =  isNat(x1)
n__isNat(x1)  =  n__isNat(x1)
U11(x1, x2)  =  U11(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

Lexicographic path order with status [LPO].
Quasi-Precedence:
[U213, plus2, nplus2] > [s1, ns1] > [U21^11, tt, isNat1, nisNat1]
[U213, plus2, nplus2] > and1 > [U21^11, tt, isNat1, nisNat1]
[U213, plus2, nplus2] > U111 > [U21^11, tt, isNat1, nisNat1]
[n0, 0] > [U21^11, tt, isNat1, nisNat1]

Status:
U21^11: [1]
tt: []
s1: [1]
and1: [1]
isNat1: [1]
nisNat1: [1]
U111: [1]
U213: [3,2,1]
plus2: [1,2]
n0: []
nplus2: [1,2]
ns1: [1]
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)
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

(11) 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)
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.

(12) PisEmptyProof (EQUIVALENT transformation)

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

(13) TRUE

(14) Obligation:

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.

(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.
ACTIVATE(x1)  =  ACTIVATE(x1)
n__isNat(x1)  =  n__isNat(x1)
ISNAT(x1)  =  ISNAT(x1)
n__s(x1)  =  x1
activate(x1)  =  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)  =  x2
isNat(x1)  =  isNat(x1)
n__0  =  n__0
n__plus(x1, x2)  =  n__plus(x1, x2)
0  =  0

Lexicographic path order with status [LPO].
Quasi-Precedence:
[n0, 0] > [ACTIVATE1, nisNat1, ISNAT1, tt, U212, plus2, isNat1, nplus2]

Status:
ACTIVATE1: [1]
nisNat1: [1]
ISNAT1: [1]
tt: []
U212: [2,1]
plus2: [1,2]
isNat1: [1]
n0: []
nplus2: [1,2]
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)
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

(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)

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.

(17) DependencyGraphProof (EQUIVALENT transformation)

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

(18) 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)
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.

(19) 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)  =  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(x1)
n__0  =  n__0
n__plus(x1, x2)  =  n__plus(x1, x2)
n__isNat(x1)  =  n__isNat(x1)
0  =  0

Lexicographic path order with status [LPO].
Quasi-Precedence:
ISNAT1 > [tt, isNat1, nisNat1]
[U213, plus2, nplus2] > [ns1, s1] > [tt, isNat1, nisNat1]
[n0, 0] > [tt, isNat1, nisNat1]

Status:
ISNAT1: [1]
ns1: [1]
tt: []
U213: [2,3,1]
s1: [1]
plus2: [2,1]
isNat1: [1]
n0: []
nplus2: [2,1]
nisNat1: [1]
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)
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

(20) 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)
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.

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) TRUE