(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(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:
FST(s(X), cons(Y, Z)) → ACTIVATE(X)
FST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ADD(s(X), Y) → S(n__add(activate(X), Y))
ADD(s(X), Y) → ACTIVATE(X)
LEN(cons(X, Z)) → S(n__len(activate(Z)))
LEN(cons(X, Z)) → ACTIVATE(Z)
ACTIVATE(n__fst(X1, X2)) → FST(activate(X1), activate(X2))
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__from(X)) → FROM(activate(X))
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(X)
ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), activate(X2))
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__len(X)) → LEN(activate(X))
ACTIVATE(n__len(X)) → ACTIVATE(X)
The TRS R consists of the following rules:
fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(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 4 less nodes.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE(n__fst(X1, X2)) → FST(activate(X1), activate(X2))
FST(s(X), cons(Y, Z)) → ACTIVATE(X)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), activate(X2))
ADD(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__len(X)) → LEN(activate(X))
LEN(cons(X, Z)) → ACTIVATE(Z)
ACTIVATE(n__len(X)) → ACTIVATE(X)
FST(s(X), cons(Y, Z)) → ACTIVATE(Z)
The TRS R consists of the following rules:
fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(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(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 0
POL(ACTIVATE(x1)) = x1
POL(ADD(x1, x2)) = 1 + x1
POL(FST(x1, x2)) = x1 + x2
POL(LEN(x1)) = x1
POL(activate(x1)) = x1
POL(add(x1, x2)) = 1 + x1 + x2
POL(cons(x1, x2)) = x2
POL(from(x1)) = x1
POL(fst(x1, x2)) = x1 + x2
POL(len(x1)) = x1
POL(n__add(x1, x2)) = 1 + x1 + x2
POL(n__from(x1)) = x1
POL(n__fst(x1, x2)) = x1 + x2
POL(n__len(x1)) = x1
POL(n__s(x1)) = x1
POL(nil) = 0
POL(s(x1)) = x1
The following usable rules [FROCOS05] were oriented:
activate(X) → X
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(X))
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
len(X) → n__len(X)
len(cons(X, Z)) → s(n__len(activate(Z)))
len(nil) → 0
from(X) → n__from(X)
fst(X1, X2) → n__fst(X1, X2)
from(X) → cons(X, n__from(n__s(X)))
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
add(s(X), Y) → s(n__add(activate(X), Y))
add(0, X) → X
fst(0, Z) → nil
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE(n__fst(X1, X2)) → FST(activate(X1), activate(X2))
FST(s(X), cons(Y, Z)) → ACTIVATE(X)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), activate(X2))
ACTIVATE(n__len(X)) → LEN(activate(X))
LEN(cons(X, Z)) → ACTIVATE(Z)
ACTIVATE(n__len(X)) → ACTIVATE(X)
FST(s(X), cons(Y, Z)) → ACTIVATE(Z)
The TRS R consists of the following rules:
fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(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 1 SCC with 1 less node.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FST(s(X), cons(Y, Z)) → ACTIVATE(X)
ACTIVATE(n__fst(X1, X2)) → FST(activate(X1), activate(X2))
FST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__len(X)) → LEN(activate(X))
LEN(cons(X, Z)) → ACTIVATE(Z)
ACTIVATE(n__len(X)) → ACTIVATE(X)
The TRS R consists of the following rules:
fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(X))
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
ACTIVATE(n__from(X)) → ACTIVATE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 0
POL(ACTIVATE(x1)) = x1
POL(FST(x1, x2)) = x1 + x2
POL(LEN(x1)) = x1
POL(activate(x1)) = x1
POL(add(x1, x2)) = x2
POL(cons(x1, x2)) = x2
POL(from(x1)) = 1 + x1
POL(fst(x1, x2)) = x1 + x2
POL(len(x1)) = x1
POL(n__add(x1, x2)) = x2
POL(n__from(x1)) = 1 + x1
POL(n__fst(x1, x2)) = x1 + x2
POL(n__len(x1)) = x1
POL(n__s(x1)) = x1
POL(nil) = 0
POL(s(x1)) = x1
The following usable rules [FROCOS05] were oriented:
activate(X) → X
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(X))
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
len(X) → n__len(X)
len(cons(X, Z)) → s(n__len(activate(Z)))
len(nil) → 0
from(X) → n__from(X)
fst(X1, X2) → n__fst(X1, X2)
from(X) → cons(X, n__from(n__s(X)))
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
add(s(X), Y) → s(n__add(activate(X), Y))
add(0, X) → X
fst(0, Z) → nil
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FST(s(X), cons(Y, Z)) → ACTIVATE(X)
ACTIVATE(n__fst(X1, X2)) → FST(activate(X1), activate(X2))
FST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__len(X)) → LEN(activate(X))
LEN(cons(X, Z)) → ACTIVATE(Z)
ACTIVATE(n__len(X)) → ACTIVATE(X)
The TRS R consists of the following rules:
fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(X))
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
LEN(cons(X, Z)) → ACTIVATE(Z)
ACTIVATE(n__len(X)) → ACTIVATE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 0
POL(ACTIVATE(x1)) = x1
POL(FST(x1, x2)) = x1 + x2
POL(LEN(x1)) = 1 + x1
POL(activate(x1)) = x1
POL(add(x1, x2)) = x2
POL(cons(x1, x2)) = x2
POL(from(x1)) = 0
POL(fst(x1, x2)) = x1 + x2
POL(len(x1)) = 1 + x1
POL(n__add(x1, x2)) = x2
POL(n__from(x1)) = 0
POL(n__fst(x1, x2)) = x1 + x2
POL(n__len(x1)) = 1 + x1
POL(n__s(x1)) = x1
POL(nil) = 0
POL(s(x1)) = x1
The following usable rules [FROCOS05] were oriented:
activate(X) → X
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(X))
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
len(X) → n__len(X)
len(cons(X, Z)) → s(n__len(activate(Z)))
len(nil) → 0
from(X) → n__from(X)
fst(X1, X2) → n__fst(X1, X2)
from(X) → cons(X, n__from(n__s(X)))
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
add(s(X), Y) → s(n__add(activate(X), Y))
add(0, X) → X
fst(0, Z) → nil
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FST(s(X), cons(Y, Z)) → ACTIVATE(X)
ACTIVATE(n__fst(X1, X2)) → FST(activate(X1), activate(X2))
FST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__len(X)) → LEN(activate(X))
The TRS R consists of the following rules:
fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(X))
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE(n__fst(X1, X2)) → FST(activate(X1), activate(X2))
FST(s(X), cons(Y, Z)) → ACTIVATE(X)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X2)
FST(s(X), cons(Y, Z)) → ACTIVATE(Z)
The TRS R consists of the following rules:
fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(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__fst(X1, X2)) → FST(activate(X1), activate(X2))
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:
POL(ACTIVATE(x1)) = | | + | | · | x1 |
POL(n__fst(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(FST(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(activate(x1)) = | | + | | · | x1 |
POL(cons(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(n__add(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(add(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(fst(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
The following usable rules [FROCOS05] were oriented:
activate(X) → X
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(X))
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
len(X) → n__len(X)
len(cons(X, Z)) → s(n__len(activate(Z)))
len(nil) → 0
from(X) → n__from(X)
fst(X1, X2) → n__fst(X1, X2)
from(X) → cons(X, n__from(n__s(X)))
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
add(s(X), Y) → s(n__add(activate(X), Y))
add(0, X) → X
fst(0, Z) → nil
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FST(s(X), cons(Y, Z)) → ACTIVATE(X)
FST(s(X), cons(Y, Z)) → ACTIVATE(Z)
The TRS R consists of the following rules:
fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(X))
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
(18) TRUE