(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
double(0) → 0
double(s(x)) → s(s(double(x)))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
first(nil) → 0
first(cons(x, xs)) → x
doublelist(nil) → nil
doublelist(cons(x, xs)) → cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))
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:
double(0) → 0
double(s(x)) → s(s(double(x)))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
first(nil) → 0
first(cons(x, xs)) → x
doublelist(nil) → nil
doublelist(cons(x, xs)) → cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))
The set Q consists of the following terms:
double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(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:
DOUBLE(s(x)) → DOUBLE(x)
DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)
DEL(x, cons(y, xs)) → EQ(x, y)
IF(false, x, y, xs) → DEL(x, xs)
EQ(s(x), s(y)) → EQ(x, y)
DOUBLELIST(cons(x, xs)) → DOUBLE(x)
DOUBLELIST(cons(x, xs)) → DOUBLELIST(del(first(cons(x, xs)), cons(x, xs)))
DOUBLELIST(cons(x, xs)) → DEL(first(cons(x, xs)), cons(x, xs))
DOUBLELIST(cons(x, xs)) → FIRST(cons(x, xs))
The TRS R consists of the following rules:
double(0) → 0
double(s(x)) → s(s(double(x)))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
first(nil) → 0
first(cons(x, xs)) → x
doublelist(nil) → nil
doublelist(cons(x, xs)) → cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))
The set Q consists of the following terms:
double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(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 with 4 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
EQ(s(x), s(y)) → EQ(x, y)
The TRS R consists of the following rules:
double(0) → 0
double(s(x)) → s(s(double(x)))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
first(nil) → 0
first(cons(x, xs)) → x
doublelist(nil) → nil
doublelist(cons(x, xs)) → cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))
The set Q consists of the following terms:
double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(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:
IF(false, x, y, xs) → DEL(x, xs)
DEL(x, cons(y, xs)) → IF(eq(x, y), x, y, xs)
The TRS R consists of the following rules:
double(0) → 0
double(s(x)) → s(s(double(x)))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
first(nil) → 0
first(cons(x, xs)) → x
doublelist(nil) → nil
doublelist(cons(x, xs)) → cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))
The set Q consists of the following terms:
double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(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:
DOUBLE(s(x)) → DOUBLE(x)
The TRS R consists of the following rules:
double(0) → 0
double(s(x)) → s(s(double(x)))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
first(nil) → 0
first(cons(x, xs)) → x
doublelist(nil) → nil
doublelist(cons(x, xs)) → cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))
The set Q consists of the following terms:
double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(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:
DOUBLELIST(cons(x, xs)) → DOUBLELIST(del(first(cons(x, xs)), cons(x, xs)))
The TRS R consists of the following rules:
double(0) → 0
double(s(x)) → s(s(double(x)))
del(x, nil) → nil
del(x, cons(y, xs)) → if(eq(x, y), x, y, xs)
if(true, x, y, xs) → xs
if(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
first(nil) → 0
first(cons(x, xs)) → x
doublelist(nil) → nil
doublelist(cons(x, xs)) → cons(double(x), doublelist(del(first(cons(x, xs)), cons(x, xs))))
The set Q consists of the following terms:
double(0)
double(s(x0))
del(x0, nil)
del(x0, cons(x1, x2))
if(true, x0, x1, x2)
if(false, x0, x1, x2)
eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
first(nil)
first(cons(x0, x1))
doublelist(nil)
doublelist(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.