(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(append, nil), ys) → ys
a(a(append, a(a(cons, x), xs)), ys) → a(a(cons, x), a(a(append, xs), ys))
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(le, 0), y) → true
a(a(le, a(s, x)), 0) → false
a(a(le, a(s, x)), a(s, y)) → a(a(le, x), y)
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), b) → a(not2, a(f, b))
a(not2, true) → false
a(not2, false) → true
a(qs, nil) → nil
a(qs, a(a(cons, x), xs)) → a(a(append, a(qs, a(a(filter, a(le, x)), xs))), a(a(cons, x), a(qs, a(a(filter, a(not, a(le, 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:
a(a(append, nil), ys) → ys
a(a(append, a(a(cons, x), xs)), ys) → a(a(cons, x), a(a(append, xs), ys))
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(le, 0), y) → true
a(a(le, a(s, x)), 0) → false
a(a(le, a(s, x)), a(s, y)) → a(a(le, x), y)
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), b) → a(not2, a(f, b))
a(not2, true) → false
a(not2, false) → true
a(qs, nil) → nil
a(qs, a(a(cons, x), xs)) → a(a(append, a(qs, a(a(filter, a(le, x)), xs))), a(a(cons, x), a(qs, a(a(filter, a(not, a(le, x))), xs))))
The set Q consists of the following terms:
a(a(append, nil), x0)
a(a(append, a(a(cons, x0), x1)), x2)
a(a(filter, x0), nil)
a(a(filter, x0), a(a(cons, x1), x2))
a(a(le, 0), x0)
a(a(le, a(s, x0)), 0)
a(a(le, a(s, x0)), a(s, x1))
a(a(a(if, true), x0), x1)
a(a(a(if, false), x0), x1)
a(a(not, x0), x1)
a(not2, true)
a(not2, false)
a(qs, nil)
a(qs, a(a(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:
A(a(append, a(a(cons, x), xs)), ys) → A(a(cons, x), a(a(append, xs), ys))
A(a(append, a(a(cons, x), xs)), ys) → A(a(append, xs), ys)
A(a(append, a(a(cons, x), xs)), ys) → A(append, xs)
A(a(filter, f), a(a(cons, x), xs)) → A(a(a(if, a(f, x)), x), a(a(filter, f), xs))
A(a(filter, f), a(a(cons, x), xs)) → A(a(if, a(f, x)), x)
A(a(filter, f), a(a(cons, x), xs)) → A(if, a(f, x))
A(a(filter, f), a(a(cons, x), xs)) → A(f, x)
A(a(filter, f), a(a(cons, x), xs)) → A(a(filter, f), xs)
A(a(le, a(s, x)), a(s, y)) → A(a(le, x), y)
A(a(le, a(s, x)), a(s, y)) → A(le, x)
A(a(a(if, true), x), xs) → A(a(cons, x), xs)
A(a(a(if, true), x), xs) → A(cons, x)
A(a(not, f), b) → A(not2, a(f, b))
A(a(not, f), b) → A(f, b)
A(qs, a(a(cons, x), xs)) → A(a(append, a(qs, a(a(filter, a(le, x)), xs))), a(a(cons, x), a(qs, a(a(filter, a(not, a(le, x))), xs))))
A(qs, a(a(cons, x), xs)) → A(append, a(qs, a(a(filter, a(le, x)), xs)))
A(qs, a(a(cons, x), xs)) → A(qs, a(a(filter, a(le, x)), xs))
A(qs, a(a(cons, x), xs)) → A(a(filter, a(le, x)), xs)
A(qs, a(a(cons, x), xs)) → A(filter, a(le, x))
A(qs, a(a(cons, x), xs)) → A(le, x)
A(qs, a(a(cons, x), xs)) → A(a(cons, x), a(qs, a(a(filter, a(not, a(le, x))), xs)))
A(qs, a(a(cons, x), xs)) → A(qs, a(a(filter, a(not, a(le, x))), xs))
A(qs, a(a(cons, x), xs)) → A(a(filter, a(not, a(le, x))), xs)
A(qs, a(a(cons, x), xs)) → A(filter, a(not, a(le, x)))
A(qs, a(a(cons, x), xs)) → A(not, a(le, x))
The TRS R consists of the following rules:
a(a(append, nil), ys) → ys
a(a(append, a(a(cons, x), xs)), ys) → a(a(cons, x), a(a(append, xs), ys))
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(le, 0), y) → true
a(a(le, a(s, x)), 0) → false
a(a(le, a(s, x)), a(s, y)) → a(a(le, x), y)
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), b) → a(not2, a(f, b))
a(not2, true) → false
a(not2, false) → true
a(qs, nil) → nil
a(qs, a(a(cons, x), xs)) → a(a(append, a(qs, a(a(filter, a(le, x)), xs))), a(a(cons, x), a(qs, a(a(filter, a(not, a(le, x))), xs))))
The set Q consists of the following terms:
a(a(append, nil), x0)
a(a(append, a(a(cons, x0), x1)), x2)
a(a(filter, x0), nil)
a(a(filter, x0), a(a(cons, x1), x2))
a(a(le, 0), x0)
a(a(le, a(s, x0)), 0)
a(a(le, a(s, x0)), a(s, x1))
a(a(a(if, true), x0), x1)
a(a(a(if, false), x0), x1)
a(a(not, x0), x1)
a(not2, true)
a(not2, false)
a(qs, nil)
a(qs, a(a(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 3 SCCs with 16 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(a(le, a(s, x)), a(s, y)) → A(a(le, x), y)
The TRS R consists of the following rules:
a(a(append, nil), ys) → ys
a(a(append, a(a(cons, x), xs)), ys) → a(a(cons, x), a(a(append, xs), ys))
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(le, 0), y) → true
a(a(le, a(s, x)), 0) → false
a(a(le, a(s, x)), a(s, y)) → a(a(le, x), y)
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), b) → a(not2, a(f, b))
a(not2, true) → false
a(not2, false) → true
a(qs, nil) → nil
a(qs, a(a(cons, x), xs)) → a(a(append, a(qs, a(a(filter, a(le, x)), xs))), a(a(cons, x), a(qs, a(a(filter, a(not, a(le, x))), xs))))
The set Q consists of the following terms:
a(a(append, nil), x0)
a(a(append, a(a(cons, x0), x1)), x2)
a(a(filter, x0), nil)
a(a(filter, x0), a(a(cons, x1), x2))
a(a(le, 0), x0)
a(a(le, a(s, x0)), 0)
a(a(le, a(s, x0)), a(s, x1))
a(a(a(if, true), x0), x1)
a(a(a(if, false), x0), x1)
a(a(not, x0), x1)
a(not2, true)
a(not2, false)
a(qs, nil)
a(qs, a(a(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:
A(a(append, a(a(cons, x), xs)), ys) → A(a(append, xs), ys)
The TRS R consists of the following rules:
a(a(append, nil), ys) → ys
a(a(append, a(a(cons, x), xs)), ys) → a(a(cons, x), a(a(append, xs), ys))
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(le, 0), y) → true
a(a(le, a(s, x)), 0) → false
a(a(le, a(s, x)), a(s, y)) → a(a(le, x), y)
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), b) → a(not2, a(f, b))
a(not2, true) → false
a(not2, false) → true
a(qs, nil) → nil
a(qs, a(a(cons, x), xs)) → a(a(append, a(qs, a(a(filter, a(le, x)), xs))), a(a(cons, x), a(qs, a(a(filter, a(not, a(le, x))), xs))))
The set Q consists of the following terms:
a(a(append, nil), x0)
a(a(append, a(a(cons, x0), x1)), x2)
a(a(filter, x0), nil)
a(a(filter, x0), a(a(cons, x1), x2))
a(a(le, 0), x0)
a(a(le, a(s, x0)), 0)
a(a(le, a(s, x0)), a(s, x1))
a(a(a(if, true), x0), x1)
a(a(a(if, false), x0), x1)
a(a(not, x0), x1)
a(not2, true)
a(not2, false)
a(qs, nil)
a(qs, a(a(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:
A(a(filter, f), a(a(cons, x), xs)) → A(a(filter, f), xs)
A(a(filter, f), a(a(cons, x), xs)) → A(f, x)
A(a(not, f), b) → A(f, b)
A(qs, a(a(cons, x), xs)) → A(qs, a(a(filter, a(le, x)), xs))
A(qs, a(a(cons, x), xs)) → A(a(filter, a(le, x)), xs)
A(qs, a(a(cons, x), xs)) → A(qs, a(a(filter, a(not, a(le, x))), xs))
A(qs, a(a(cons, x), xs)) → A(a(filter, a(not, a(le, x))), xs)
The TRS R consists of the following rules:
a(a(append, nil), ys) → ys
a(a(append, a(a(cons, x), xs)), ys) → a(a(cons, x), a(a(append, xs), ys))
a(a(filter, f), nil) → nil
a(a(filter, f), a(a(cons, x), xs)) → a(a(a(if, a(f, x)), x), a(a(filter, f), xs))
a(a(le, 0), y) → true
a(a(le, a(s, x)), 0) → false
a(a(le, a(s, x)), a(s, y)) → a(a(le, x), y)
a(a(a(if, true), x), xs) → a(a(cons, x), xs)
a(a(a(if, false), x), xs) → xs
a(a(not, f), b) → a(not2, a(f, b))
a(not2, true) → false
a(not2, false) → true
a(qs, nil) → nil
a(qs, a(a(cons, x), xs)) → a(a(append, a(qs, a(a(filter, a(le, x)), xs))), a(a(cons, x), a(qs, a(a(filter, a(not, a(le, x))), xs))))
The set Q consists of the following terms:
a(a(append, nil), x0)
a(a(append, a(a(cons, x0), x1)), x2)
a(a(filter, x0), nil)
a(a(filter, x0), a(a(cons, x1), x2))
a(a(le, 0), x0)
a(a(le, a(s, x0)), 0)
a(a(le, a(s, x0)), a(s, x1))
a(a(a(if, true), x0), x1)
a(a(a(if, false), x0), x1)
a(a(not, x0), x1)
a(not2, true)
a(not2, false)
a(qs, nil)
a(qs, a(a(cons, x0), x1))
We have to consider all minimal (P,Q,R)-chains.