(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, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))
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:
fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))
The set Q consists of the following terms:
fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))
(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:
FST(s(X), cons(Y, Z)) → FST(X, Z)
FROM(X) → FROM(s(X))
ADD(s(X), Y) → ADD(X, Y)
LEN(cons(X, Z)) → LEN(Z)
The TRS R consists of the following rules:
fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))
The set Q consists of the following terms:
fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LEN(cons(X, Z)) → LEN(Z)
The TRS R consists of the following rules:
fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))
The set Q consists of the following terms:
fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ADD(s(X), Y) → ADD(X, Y)
The TRS R consists of the following rules:
fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))
The set Q consists of the following terms:
fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FROM(X) → FROM(s(X))
The TRS R consists of the following rules:
fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))
The set Q consists of the following terms:
fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FST(s(X), cons(Y, Z)) → FST(X, Z)
The TRS R consists of the following rules:
fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, fst(X, Z))
from(X) → cons(X, from(s(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
len(nil) → 0
len(cons(X, Z)) → s(len(Z))
The set Q consists of the following terms:
fst(0, x0)
fst(s(x0), cons(x1, x2))
from(x0)
add(0, x0)
add(s(x0), x1)
len(nil)
len(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.