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 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(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)
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.

(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)
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.

(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(X1, X2)
PLUS(N, 0) → U111(isNat(N), N)
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 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)  =  x2
isNat(x1)  =  isNat(x1)
tt  =  tt
n__isNat(x1)  =  n__isNat(x1)
ISNAT(x1)  =  ISNAT(x1)
AND(x1, x2)  =  AND(x1, x2)
activate(x1)  =  x1
n__s(x1)  =  n__s(x1)
s(x1)  =  s(x1)
U211(x1, x2, x3)  =  U211(x1, x2, x3)
and(x1, x2)  =  x2
U11(x1, x2)  =  x2
U21(x1, x2, x3)  =  U21(x1, x2, x3)
plus(x1, x2)  =  plus(x1, x2)
n__0  =  n__0

Lexicographic path order with status [LPO].
Quasi-Precedence:
[0, tt, n0] > [nplus2, U213, plus2] > [ns1, s1] > [PLUS2, isNat1, nisNat1, ISNAT1, U21^13] > AND2

Status:
nplus2: [2,1]
PLUS2: [1,2]
0: []
isNat1: [1]
tt: []
nisNat1: [1]
ISNAT1: [1]
AND2: [1,2]
ns1: [1]
s1: [1]
U21^13: [3,2,1]
U213: [2,3,1]
plus2: [2,1]
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)
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

(6) Obligation:

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

U111(tt, N) → ACTIVATE(N)
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)
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.

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(8) TRUE