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 PisEmptyProof (⇔)
        ↳18 TRUE

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

U11(tt, V2) → U12(isNat(activate(V2)))
U12(tt) → tt
U21(tt) → tt
U31(tt, N) → activate(N)
U41(tt, M, N) → U42(isNat(activate(N)), activate(M), activate(N))
U42(tt, M, N) → s(plus(activate(N), activate(M)))
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → U11(isNat(activate(V1)), activate(V2))
isNat(n__s(V1)) → U21(isNat(activate(V1)))
plus(N, 0) → U31(isNat(N), N)
plus(N, s(M)) → U41(isNat(M), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
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, V2) → U121(isNat(activate(V2)))
U111(tt, V2) → ISNAT(activate(V2))
U111(tt, V2) → ACTIVATE(V2)
U311(tt, N) → ACTIVATE(N)
U411(tt, M, N) → U421(isNat(activate(N)), activate(M), activate(N))
U411(tt, M, N) → ISNAT(activate(N))
U411(tt, M, N) → ACTIVATE(N)
U411(tt, M, N) → ACTIVATE(M)
U421(tt, M, N) → S(plus(activate(N), activate(M)))
U421(tt, M, N) → PLUS(activate(N), activate(M))
U421(tt, M, N) → ACTIVATE(N)
U421(tt, M, N) → ACTIVATE(M)
ISNAT(n__plus(V1, V2)) → U111(isNat(activate(V1)), 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)) → U211(isNat(activate(V1)))
ISNAT(n__s(V1)) → ISNAT(activate(V1))
ISNAT(n__s(V1)) → ACTIVATE(V1)
PLUS(N, 0) → U311(isNat(N), N)
PLUS(N, 0) → ISNAT(N)
PLUS(N, s(M)) → U411(isNat(M), M, N)
PLUS(N, s(M)) → ISNAT(M)
ACTIVATE(n__0) → 01
ACTIVATE(n__plus(X1, X2)) → PLUS(X1, X2)
ACTIVATE(n__s(X)) → S(X)

The TRS R consists of the following rules:

U11(tt, V2) → U12(isNat(activate(V2)))
U12(tt) → tt
U21(tt) → tt
U31(tt, N) → activate(N)
U41(tt, M, N) → U42(isNat(activate(N)), activate(M), activate(N))
U42(tt, M, N) → s(plus(activate(N), activate(M)))
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → U11(isNat(activate(V1)), activate(V2))
isNat(n__s(V1)) → U21(isNat(activate(V1)))
plus(N, 0) → U31(isNat(N), N)
plus(N, s(M)) → U41(isNat(M), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
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 5 less nodes.

(4) Obligation:

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

U111(tt, V2) → ISNAT(activate(V2))
ISNAT(n__plus(V1, V2)) → U111(isNat(activate(V1)), activate(V2))
U111(tt, V2) → ACTIVATE(V2)
ACTIVATE(n__plus(X1, X2)) → PLUS(X1, X2)
PLUS(N, 0) → U311(isNat(N), N)
U311(tt, N) → ACTIVATE(N)
PLUS(N, 0) → ISNAT(N)
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, s(M)) → U411(isNat(M), M, N)
U411(tt, M, N) → U421(isNat(activate(N)), activate(M), activate(N))
U421(tt, M, N) → PLUS(activate(N), activate(M))
PLUS(N, s(M)) → ISNAT(M)
U421(tt, M, N) → ACTIVATE(N)
U421(tt, M, N) → ACTIVATE(M)
U411(tt, M, N) → ISNAT(activate(N))
U411(tt, M, N) → ACTIVATE(N)
U411(tt, M, N) → ACTIVATE(M)

The TRS R consists of the following rules:

U11(tt, V2) → U12(isNat(activate(V2)))
U12(tt) → tt
U21(tt) → tt
U31(tt, N) → activate(N)
U41(tt, M, N) → U42(isNat(activate(N)), activate(M), activate(N))
U42(tt, M, N) → s(plus(activate(N), activate(M)))
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → U11(isNat(activate(V1)), activate(V2))
isNat(n__s(V1)) → U21(isNat(activate(V1)))
plus(N, 0) → U31(isNat(N), N)
plus(N, s(M)) → U41(isNat(M), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
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.


ISNAT(n__plus(V1, V2)) → U111(isNat(activate(V1)), activate(V2))
ACTIVATE(n__plus(X1, X2)) → PLUS(X1, X2)
PLUS(N, 0) → U311(isNat(N), N)
PLUS(N, 0) → ISNAT(N)
ISNAT(n__plus(V1, V2)) → ISNAT(activate(V1))
ISNAT(n__plus(V1, V2)) → ACTIVATE(V1)
ISNAT(n__plus(V1, V2)) → ACTIVATE(V2)
PLUS(N, s(M)) → ISNAT(M)
U421(tt, M, N) → ACTIVATE(N)
U421(tt, M, N) → ACTIVATE(M)
U411(tt, M, N) → ISNAT(activate(N))
U411(tt, M, N) → ACTIVATE(N)
U411(tt, M, N) → ACTIVATE(M)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
U111(x1, x2)  =  x2
tt  =  tt
ISNAT(x1)  =  x1
activate(x1)  =  x1
n__plus(x1, x2)  =  n__plus(x1, x2)
isNat(x1)  =  isNat(x1)
ACTIVATE(x1)  =  x1
PLUS(x1, x2)  =  PLUS(x1, x2)
0  =  0
U311(x1, x2)  =  x2
n__s(x1)  =  x1
s(x1)  =  x1
U411(x1, x2, x3)  =  U411(x2, x3)
U421(x1, x2, x3)  =  U421(x2, x3)
n__0  =  n__0
plus(x1, x2)  =  plus(x1, x2)
U31(x1, x2)  =  x2
U11(x1, x2)  =  U11(x1, x2)
U21(x1)  =  U21
U41(x1, x2, x3)  =  U41(x2, x3)
U12(x1)  =  U12(x1)
U42(x1, x2, x3)  =  U42(x2, x3)

Lexicographic path order with status [LPO].
Quasi-Precedence:
[nplus2, plus2, U412, U422] > [PLUS2, U41^12, U42^12] > U121
[0, n0] > U121
U112 > isNat1 > tt > U121
U21 > tt > U121

Status:
tt: []
nplus2: [1,2]
isNat1: [1]
PLUS2: [2,1]
0: []
U41^12: [1,2]
U42^12: [1,2]
n0: []
plus2: [1,2]
U112: [2,1]
U21: []
U412: [2,1]
U121: [1]
U422: [2,1]


The following usable rules [FROCOS05] were oriented:

activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
plus(N, 0) → U31(isNat(N), N)
U31(tt, N) → activate(N)
activate(n__s(X)) → s(X)
activate(X) → X
plus(N, s(M)) → U41(isNat(M), M, N)
U41(tt, M, N) → U42(isNat(activate(N)), activate(M), activate(N))
U42(tt, M, N) → s(plus(activate(N), activate(M)))
plus(X1, X2) → n__plus(X1, X2)
s(X) → n__s(X)
0 → n__0

(6) Obligation:

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

U111(tt, V2) → ISNAT(activate(V2))
U111(tt, V2) → ACTIVATE(V2)
U311(tt, N) → ACTIVATE(N)
ISNAT(n__s(V1)) → ISNAT(activate(V1))
ISNAT(n__s(V1)) → ACTIVATE(V1)
PLUS(N, s(M)) → U411(isNat(M), M, N)
U411(tt, M, N) → U421(isNat(activate(N)), activate(M), activate(N))
U421(tt, M, N) → PLUS(activate(N), activate(M))

The TRS R consists of the following rules:

U11(tt, V2) → U12(isNat(activate(V2)))
U12(tt) → tt
U21(tt) → tt
U31(tt, N) → activate(N)
U41(tt, M, N) → U42(isNat(activate(N)), activate(M), activate(N))
U42(tt, M, N) → s(plus(activate(N), activate(M)))
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → U11(isNat(activate(V1)), activate(V2))
isNat(n__s(V1)) → U21(isNat(activate(V1)))
plus(N, 0) → U31(isNat(N), N)
plus(N, s(M)) → U41(isNat(M), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
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 4 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

U411(tt, M, N) → U421(isNat(activate(N)), activate(M), activate(N))
U421(tt, M, N) → PLUS(activate(N), activate(M))
PLUS(N, s(M)) → U411(isNat(M), M, N)

The TRS R consists of the following rules:

U11(tt, V2) → U12(isNat(activate(V2)))
U12(tt) → tt
U21(tt) → tt
U31(tt, N) → activate(N)
U41(tt, M, N) → U42(isNat(activate(N)), activate(M), activate(N))
U42(tt, M, N) → s(plus(activate(N), activate(M)))
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → U11(isNat(activate(V1)), activate(V2))
isNat(n__s(V1)) → U21(isNat(activate(V1)))
plus(N, 0) → U31(isNat(N), N)
plus(N, s(M)) → U41(isNat(M), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
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.


U421(tt, M, N) → PLUS(activate(N), activate(M))
PLUS(N, s(M)) → U411(isNat(M), M, N)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
U411(x1, x2, x3)  =  U411(x1, x2)
tt  =  tt
U421(x1, x2, x3)  =  U421(x1, x2)
isNat(x1)  =  isNat
activate(x1)  =  x1
PLUS(x1, x2)  =  x2
s(x1)  =  s(x1)
n__0  =  n__0
0  =  0
n__plus(x1, x2)  =  n__plus(x1, x2)
plus(x1, x2)  =  plus(x1, x2)
U31(x1, x2)  =  x2
n__s(x1)  =  n__s(x1)
U11(x1, x2)  =  x1
U21(x1)  =  U21
U41(x1, x2, x3)  =  U41(x1, x2, x3)
U12(x1)  =  x1
U42(x1, x2, x3)  =  U42(x1, x2, x3)

Lexicographic path order with status [LPO].
Quasi-Precedence:
[n0, 0]
[nplus2, plus2, U413, U423] > [tt, isNat, s1, ns1, U21] > [U41^12, U42^12]

Status:
U41^12: [1,2]
tt: []
U42^12: [1,2]
isNat: []
s1: [1]
n0: []
0: []
nplus2: [1,2]
plus2: [1,2]
ns1: [1]
U21: []
U413: [3,2,1]
U423: [3,2,1]


The following usable rules [FROCOS05] were oriented:

activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
plus(N, 0) → U31(isNat(N), N)
U31(tt, N) → activate(N)
activate(n__s(X)) → s(X)
activate(X) → X
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → U11(isNat(activate(V1)), activate(V2))
isNat(n__s(V1)) → U21(isNat(activate(V1)))
plus(N, s(M)) → U41(isNat(M), M, N)
U11(tt, V2) → U12(isNat(activate(V2)))
U41(tt, M, N) → U42(isNat(activate(N)), activate(M), activate(N))
U42(tt, M, N) → s(plus(activate(N), activate(M)))
plus(X1, X2) → n__plus(X1, X2)
U21(tt) → tt
U12(tt) → tt
s(X) → n__s(X)
0 → n__0

(11) Obligation:

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

U411(tt, M, N) → U421(isNat(activate(N)), activate(M), activate(N))

The TRS R consists of the following rules:

U11(tt, V2) → U12(isNat(activate(V2)))
U12(tt) → tt
U21(tt) → tt
U31(tt, N) → activate(N)
U41(tt, M, N) → U42(isNat(activate(N)), activate(M), activate(N))
U42(tt, M, N) → s(plus(activate(N), activate(M)))
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → U11(isNat(activate(V1)), activate(V2))
isNat(n__s(V1)) → U21(isNat(activate(V1)))
plus(N, 0) → U31(isNat(N), N)
plus(N, s(M)) → U41(isNat(M), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
activate(n__s(X)) → s(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))

The TRS R consists of the following rules:

U11(tt, V2) → U12(isNat(activate(V2)))
U12(tt) → tt
U21(tt) → tt
U31(tt, N) → activate(N)
U41(tt, M, N) → U42(isNat(activate(N)), activate(M), activate(N))
U42(tt, M, N) → s(plus(activate(N), activate(M)))
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → U11(isNat(activate(V1)), activate(V2))
isNat(n__s(V1)) → U21(isNat(activate(V1)))
plus(N, 0) → U31(isNat(N), N)
plus(N, s(M)) → U41(isNat(M), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
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.


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)  =  x1
n__s(x1)  =  n__s(x1)
activate(x1)  =  x1
n__0  =  n__0
0  =  0
n__plus(x1, x2)  =  n__plus(x1, x2)
plus(x1, x2)  =  plus(x1, x2)
U31(x1, x2)  =  x2
isNat(x1)  =  isNat
tt  =  tt
s(x1)  =  s(x1)
U41(x1, x2, x3)  =  U41(x1, x2, x3)
U11(x1, x2)  =  x1
U21(x1)  =  x1
U42(x1, x2, x3)  =  U42(x1, x2, x3)
U12(x1)  =  x1

Lexicographic path order with status [LPO].
Quasi-Precedence:
[n0, 0]
[nplus2, plus2, isNat, tt, U413, U423] > [ns1, s1]

Status:
ns1: [1]
n0: []
0: []
nplus2: [1,2]
plus2: [1,2]
isNat: []
tt: []
s1: [1]
U413: [3,2,1]
U423: [3,2,1]


The following usable rules [FROCOS05] were oriented:

activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
plus(N, 0) → U31(isNat(N), N)
U31(tt, N) → activate(N)
activate(n__s(X)) → s(X)
activate(X) → X
plus(N, s(M)) → U41(isNat(M), M, N)
isNat(n__plus(V1, V2)) → U11(isNat(activate(V1)), activate(V2))
isNat(n__s(V1)) → U21(isNat(activate(V1)))
U41(tt, M, N) → U42(isNat(activate(N)), activate(M), activate(N))
U11(tt, V2) → U12(isNat(activate(V2)))
U42(tt, M, N) → s(plus(activate(N), activate(M)))
plus(X1, X2) → n__plus(X1, X2)
isNat(n__0) → tt
U21(tt) → tt
U12(tt) → tt
s(X) → n__s(X)
0 → n__0

(16) Obligation:

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

U11(tt, V2) → U12(isNat(activate(V2)))
U12(tt) → tt
U21(tt) → tt
U31(tt, N) → activate(N)
U41(tt, M, N) → U42(isNat(activate(N)), activate(M), activate(N))
U42(tt, M, N) → s(plus(activate(N), activate(M)))
isNat(n__0) → tt
isNat(n__plus(V1, V2)) → U11(isNat(activate(V1)), activate(V2))
isNat(n__s(V1)) → U21(isNat(activate(V1)))
plus(N, 0) → U31(isNat(N), N)
plus(N, s(M)) → U41(isNat(M), M, N)
0 → n__0
plus(X1, X2) → n__plus(X1, X2)
s(X) → n__s(X)
activate(n__0) → 0
activate(n__plus(X1, X2)) → plus(X1, X2)
activate(n__s(X)) → s(X)
activate(X) → X

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

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE