(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.