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:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(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:

AND(true, X) → ACTIVATE(X)
IF(true, X, Y) → ACTIVATE(X)
IF(false, X, Y) → ACTIVATE(Y)
ADD(0, X) → ACTIVATE(X)
ADD(s(X), Y) → S(n__add(activate(X), activate(Y)))
ADD(s(X), Y) → ACTIVATE(X)
ADD(s(X), Y) → ACTIVATE(Y)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Y)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
FROM(X) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
ACTIVATE(n__from(X)) → FROM(X)
ACTIVATE(n__s(X)) → S(X)

The TRS R consists of the following rules:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(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 5 less nodes.

(4) Obligation:

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

ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
ADD(0, X) → ACTIVATE(X)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Y)
ACTIVATE(n__from(X)) → FROM(X)
FROM(X) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ADD(s(X), Y) → ACTIVATE(X)
ADD(s(X), Y) → ACTIVATE(Y)

The TRS R consists of the following rules:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(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.


ADD(0, X) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Y)
FROM(X) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ADD(s(X), Y) → ACTIVATE(X)
ADD(s(X), Y) → ACTIVATE(Y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  x1
n__add(x1, x2)  =  n__add(x1, x2)
ADD(x1, x2)  =  ADD(x1, x2)
0  =  0
n__first(x1, x2)  =  n__first(x1, x2)
FIRST(x1, x2)  =  FIRST(x1, x2)
s(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
n__from(x1)  =  n__from(x1)
FROM(x1)  =  FROM(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
0 > [nadd2, ADD2]
[nfirst2, FIRST2] > [nadd2, ADD2]
cons2 > [nadd2, ADD2]
[nfrom1, FROM1] > [nadd2, ADD2]

Status:
nfirst2: [2,1]
cons2: [1,2]
FIRST2: [2,1]
nadd2: [1,2]
nfrom1: [1]
0: []
ADD2: [1,2]
FROM1: [1]


The following usable rules [FROCOS05] were oriented: none

(6) Obligation:

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

ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
ACTIVATE(n__from(X)) → FROM(X)

The TRS R consists of the following rules:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(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 3 less nodes.

(8) TRUE