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

(0) Obligation:

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

and(tt, X) → activate(X)
plus(N, 0) → N
plus(N, s(M)) → s(plus(N, M))
x(N, 0) → 0
x(N, s(M)) → plus(x(N, M), N)
activate(X) → X

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

and(tt, X) → activate(X)
plus(N, 0) → N
plus(N, s(M)) → s(plus(N, M))
x(N, 0) → 0
x(N, s(M)) → plus(x(N, M), N)
activate(X) → X

The set Q consists of the following terms:

and(tt, x0)
plus(x0, 0)
plus(x0, s(x1))
x(x0, 0)
x(x0, s(x1))
activate(x0)

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

AND(tt, X) → ACTIVATE(X)
PLUS(N, s(M)) → PLUS(N, M)
X(N, s(M)) → PLUS(x(N, M), N)
X(N, s(M)) → X(N, M)

The TRS R consists of the following rules:

and(tt, X) → activate(X)
plus(N, 0) → N
plus(N, s(M)) → s(plus(N, M))
x(N, 0) → 0
x(N, s(M)) → plus(x(N, M), N)
activate(X) → X

The set Q consists of the following terms:

and(tt, x0)
plus(x0, 0)
plus(x0, s(x1))
x(x0, 0)
x(x0, s(x1))
activate(x0)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) Obligation:

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

PLUS(N, s(M)) → PLUS(N, M)

The TRS R consists of the following rules:

and(tt, X) → activate(X)
plus(N, 0) → N
plus(N, s(M)) → s(plus(N, M))
x(N, 0) → 0
x(N, s(M)) → plus(x(N, M), N)
activate(X) → X

The set Q consists of the following terms:

and(tt, x0)
plus(x0, 0)
plus(x0, s(x1))
x(x0, 0)
x(x0, s(x1))
activate(x0)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PLUS(N, s(M)) → PLUS(N, M)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2)  =  PLUS(x2)
s(x1)  =  s(x1)
and(x1, x2)  =  x2
tt  =  tt
activate(x1)  =  x1
plus(x1, x2)  =  plus(x1, x2)
0  =  0
x(x1, x2)  =  x(x1, x2)

Recursive path order with status [RPO].
Precedence:

tt > PLUS₁
0 > PLUS₁
x₂ > plus₂ > s₁ > PLUS₁

Status:

PLUS₁: multiset
s₁: multiset
tt: multiset
plus₂: [2,1]
0: multiset
x₂: [2,1]

The following usable rules [FROCOS05] were oriented:

and(tt, X) → activate(X)
plus(N, 0) → N
plus(N, s(M)) → s(plus(N, M))
x(N, 0) → 0
x(N, s(M)) → plus(x(N, M), N)
activate(X) → X

(9) Obligation:

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

and(tt, X) → activate(X)
plus(N, 0) → N
plus(N, s(M)) → s(plus(N, M))
x(N, 0) → 0
x(N, s(M)) → plus(x(N, M), N)
activate(X) → X

The set Q consists of the following terms:

and(tt, x0)
plus(x0, 0)
plus(x0, s(x1))
x(x0, 0)
x(x0, s(x1))
activate(x0)

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

X(N, s(M)) → X(N, M)

The TRS R consists of the following rules:

and(tt, X) → activate(X)
plus(N, 0) → N
plus(N, s(M)) → s(plus(N, M))
x(N, 0) → 0
x(N, s(M)) → plus(x(N, M), N)
activate(X) → X

The set Q consists of the following terms:

and(tt, x0)
plus(x0, 0)
plus(x0, s(x1))
x(x0, 0)
x(x0, s(x1))
activate(x0)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

X(N, s(M)) → X(N, M)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
X(x1, x2)  =  X(x2)
s(x1)  =  s(x1)
and(x1, x2)  =  x2
tt  =  tt
activate(x1)  =  x1
plus(x1, x2)  =  plus(x1, x2)
0  =  0
x(x1, x2)  =  x(x1, x2)

Recursive path order with status [RPO].
Precedence:

tt > X₁
0 > X₁
x₂ > plus₂ > s₁ > X₁

Status:

X₁: multiset
s₁: multiset
tt: multiset
plus₂: [2,1]
0: multiset
x₂: [2,1]

The following usable rules [FROCOS05] were oriented:

and(tt, X) → activate(X)
plus(N, 0) → N
plus(N, s(M)) → s(plus(N, M))
x(N, 0) → 0
x(N, s(M)) → plus(x(N, M), N)
activate(X) → X

(14) Obligation:

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

and(tt, X) → activate(X)
plus(N, 0) → N
plus(N, s(M)) → s(plus(N, M))
x(N, 0) → 0
x(N, s(M)) → plus(x(N, M), N)
activate(X) → X

The set Q consists of the following terms:

and(tt, x0)
plus(x0, 0)
plus(x0, s(x1))
x(x0, 0)
x(x0, s(x1))
activate(x0)

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE