Termination w.r.t. Q proof of /tmp/tmpMnp2Mq/PEANO_nokinds

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 QDPOrderProof (⇔)

    ↳18 QDP

    ↳19 PisEmptyProof (⇔)

    ↳20 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.

PLUS(N, 0) → U11¹(isNat(N), N)
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)) → 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) = x1
tt = tt
n__isNat(x1) = x1
ISNAT(x1) = x1
AND(x1, x2) = AND(x1, 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) = x2
U21(x1, x2, x3) = U21(x2, x3)
plus(x1, x2) = plus(x1, x2)
n__0 = n__0

Recursive Path Order [RPO].
Precedence:

[n_{_plus₂}, PLUS₂, AND₂, U21^1₂, U21₂, plus₂]
[0, n_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(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)
U11¹(tt, N) → ACTIVATE(N)
ACTIVATE(n__isNat(X)) → ISNAT(X)
ISNAT(n__plus(V1, V2)) → AND(isNat(activate(V1)), n__isNat(activate(V2)))
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))
PLUS(N, s(M)) → AND(isNat(M), n__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)
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 2 SCCs with 4 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)
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.

(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))

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) = 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, n_{_plus₂}] > [U21^1₁, s₁, n_{_s₁}] > [tt, isNat, n_{_isNat}, n_₀, 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(X1, X2)
activate(n__isNat(X)) → isNat(X)
activate(n__s(X)) → s(X)
activate(X) → X

(11) Obligation:

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

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(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) 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:

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)
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.

(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)
ISNAT(n__s(V1)) → ACTIVATE(V1)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1) = 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

Recursive Path Order [RPO].
Precedence:

[n_{_isNat₁}, tt, isNat₁] > ISNAT₁ > [U21₂, plus₂, n_{_plus₂}]
[n_₀, 0] > [U21₂, plus₂, n_{_plus₂}]

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

(16) 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(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) 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(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, n_{_plus₂}] > [n_{_s₁}, s₁] > [tt, isNat, n_{_isNat}]
[U21₃, plus₂, n_{_plus₂}] > U11₁
[n_₀, 0] > U11₁
[n_₀, 0] > [tt, isNat, n_{_isNat}]

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

(18) 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(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) PisEmptyProof (EQUIVALENT transformation)

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

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Obligation:

(5) QDPOrderProof (EQUIVALENT transformation)

(6) Obligation:

(7) DependencyGraphProof (EQUIVALENT transformation)

(8) Complex Obligation (AND)

(9) Obligation:

(10) QDPOrderProof (EQUIVALENT transformation)

(11) Obligation:

(12) DependencyGraphProof (EQUIVALENT transformation)

(13) TRUE

(14) Obligation:

(15) QDPOrderProof (EQUIVALENT transformation)

(16) Obligation:

(17) QDPOrderProof (EQUIVALENT transformation)

(18) Obligation:

(19) PisEmptyProof (EQUIVALENT transformation)

(20) TRUE