Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), xs))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), xs))
Q is empty.
The TRS is non-overlapping. Hence, we can switch to innermost.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), xs))
The set Q consists of the following terms:
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(sub, x0), 0)
app2(app2(sub, app2(s, x0)), app2(s, x1))
app2(app2(gtr, 0), x0)
app2(app2(gtr, app2(s, x0)), 0)
app2(app2(gtr, app2(s, x0)), app2(s, x1))
app2(app2(d, x0), 0)
app2(app2(d, app2(s, x0)), app2(s, x1))
app2(len, nil)
app2(len, app2(app2(cons, x0), x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), xs))
APP2(app2(gtr, app2(s, x)), app2(s, y)) -> APP2(app2(gtr, x), y)
APP2(app2(gtr, app2(s, x)), app2(s, y)) -> APP2(gtr, x)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
APP2(len, app2(app2(cons, x), xs)) -> APP2(len, xs)
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(if, app2(p, x))
APP2(app2(sub, app2(s, x)), app2(s, y)) -> APP2(app2(sub, x), y)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(app2(d, app2(s, x)), app2(app2(sub, y), x))
APP2(app2(sub, app2(s, x)), app2(s, y)) -> APP2(sub, x)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(gtr, x)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(sub, y)
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(p, x)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(app2(if, app2(app2(gtr, x), y)), false)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(if, app2(app2(gtr, x), y))
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(app2(filter, p), xs)
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(app2(cons, x), app2(app2(filter, p), xs))
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(app2(gtr, x), y)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(app2(sub, y), x)
APP2(len, app2(app2(cons, x), xs)) -> APP2(s, app2(len, xs))
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs)))
The TRS R consists of the following rules:
app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), xs))
The set Q consists of the following terms:
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(sub, x0), 0)
app2(app2(sub, app2(s, x0)), app2(s, x1))
app2(app2(gtr, 0), x0)
app2(app2(gtr, app2(s, x0)), 0)
app2(app2(gtr, app2(s, x0)), app2(s, x1))
app2(app2(d, x0), 0)
app2(app2(d, app2(s, x0)), app2(s, x1))
app2(len, nil)
app2(len, app2(app2(cons, x0), x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), xs))
APP2(app2(gtr, app2(s, x)), app2(s, y)) -> APP2(app2(gtr, x), y)
APP2(app2(gtr, app2(s, x)), app2(s, y)) -> APP2(gtr, x)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
APP2(len, app2(app2(cons, x), xs)) -> APP2(len, xs)
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(if, app2(p, x))
APP2(app2(sub, app2(s, x)), app2(s, y)) -> APP2(app2(sub, x), y)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(app2(d, app2(s, x)), app2(app2(sub, y), x))
APP2(app2(sub, app2(s, x)), app2(s, y)) -> APP2(sub, x)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(gtr, x)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(sub, y)
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(p, x)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(app2(if, app2(app2(gtr, x), y)), false)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(if, app2(app2(gtr, x), y))
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(app2(filter, p), xs)
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(app2(cons, x), app2(app2(filter, p), xs))
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(app2(gtr, x), y)
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(app2(sub, y), x)
APP2(len, app2(app2(cons, x), xs)) -> APP2(s, app2(len, xs))
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs)))
The TRS R consists of the following rules:
app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), xs))
The set Q consists of the following terms:
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(sub, x0), 0)
app2(app2(sub, app2(s, x0)), app2(s, x1))
app2(app2(gtr, 0), x0)
app2(app2(gtr, app2(s, x0)), 0)
app2(app2(gtr, app2(s, x0)), app2(s, x1))
app2(app2(d, x0), 0)
app2(app2(d, app2(s, x0)), app2(s, x1))
app2(len, nil)
app2(len, app2(app2(cons, x0), x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 5 SCCs with 14 less nodes.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP2(len, app2(app2(cons, x), xs)) -> APP2(len, xs)
The TRS R consists of the following rules:
app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), xs))
The set Q consists of the following terms:
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(sub, x0), 0)
app2(app2(sub, app2(s, x0)), app2(s, x1))
app2(app2(gtr, 0), x0)
app2(app2(gtr, app2(s, x0)), 0)
app2(app2(gtr, app2(s, x0)), app2(s, x1))
app2(app2(d, x0), 0)
app2(app2(d, app2(s, x0)), app2(s, x1))
app2(len, nil)
app2(len, app2(app2(cons, x0), x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
APP2(len, app2(app2(cons, x), xs)) -> APP2(len, xs)
Used argument filtering: APP2(x1, x2) = x2
app2(x1, x2) = app1(x2)
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), xs))
The set Q consists of the following terms:
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(sub, x0), 0)
app2(app2(sub, app2(s, x0)), app2(s, x1))
app2(app2(gtr, 0), x0)
app2(app2(gtr, app2(s, x0)), 0)
app2(app2(gtr, app2(s, x0)), app2(s, x1))
app2(app2(d, x0), 0)
app2(app2(d, app2(s, x0)), app2(s, x1))
app2(len, nil)
app2(len, app2(app2(cons, x0), x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(gtr, app2(s, x)), app2(s, y)) -> APP2(app2(gtr, x), y)
The TRS R consists of the following rules:
app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), xs))
The set Q consists of the following terms:
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(sub, x0), 0)
app2(app2(sub, app2(s, x0)), app2(s, x1))
app2(app2(gtr, 0), x0)
app2(app2(gtr, app2(s, x0)), 0)
app2(app2(gtr, app2(s, x0)), app2(s, x1))
app2(app2(d, x0), 0)
app2(app2(d, app2(s, x0)), app2(s, x1))
app2(len, nil)
app2(len, app2(app2(cons, x0), x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
APP2(app2(gtr, app2(s, x)), app2(s, y)) -> APP2(app2(gtr, x), y)
Used argument filtering: APP2(x1, x2) = x2
app2(x1, x2) = app1(x2)
s = s
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), xs))
The set Q consists of the following terms:
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(sub, x0), 0)
app2(app2(sub, app2(s, x0)), app2(s, x1))
app2(app2(gtr, 0), x0)
app2(app2(gtr, app2(s, x0)), 0)
app2(app2(gtr, app2(s, x0)), app2(s, x1))
app2(app2(d, x0), 0)
app2(app2(d, app2(s, x0)), app2(s, x1))
app2(len, nil)
app2(len, app2(app2(cons, x0), x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(sub, app2(s, x)), app2(s, y)) -> APP2(app2(sub, x), y)
The TRS R consists of the following rules:
app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), xs))
The set Q consists of the following terms:
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(sub, x0), 0)
app2(app2(sub, app2(s, x0)), app2(s, x1))
app2(app2(gtr, 0), x0)
app2(app2(gtr, app2(s, x0)), 0)
app2(app2(gtr, app2(s, x0)), app2(s, x1))
app2(app2(d, x0), 0)
app2(app2(d, app2(s, x0)), app2(s, x1))
app2(len, nil)
app2(len, app2(app2(cons, x0), x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
APP2(app2(sub, app2(s, x)), app2(s, y)) -> APP2(app2(sub, x), y)
Used argument filtering: APP2(x1, x2) = x2
app2(x1, x2) = app1(x2)
s = s
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), xs))
The set Q consists of the following terms:
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(sub, x0), 0)
app2(app2(sub, app2(s, x0)), app2(s, x1))
app2(app2(gtr, 0), x0)
app2(app2(gtr, app2(s, x0)), 0)
app2(app2(gtr, app2(s, x0)), app2(s, x1))
app2(app2(d, x0), 0)
app2(app2(d, app2(s, x0)), app2(s, x1))
app2(len, nil)
app2(len, app2(app2(cons, x0), x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(d, app2(s, x)), app2(s, y)) -> APP2(app2(d, app2(s, x)), app2(app2(sub, y), x))
The TRS R consists of the following rules:
app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), xs))
The set Q consists of the following terms:
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(sub, x0), 0)
app2(app2(sub, app2(s, x0)), app2(s, x1))
app2(app2(gtr, 0), x0)
app2(app2(gtr, app2(s, x0)), 0)
app2(app2(gtr, app2(s, x0)), app2(s, x1))
app2(app2(d, x0), 0)
app2(app2(d, app2(s, x0)), app2(s, x1))
app2(len, nil)
app2(len, app2(app2(cons, x0), x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(p, x)
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(app2(filter, p), xs)
The TRS R consists of the following rules:
app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), xs))
The set Q consists of the following terms:
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(sub, x0), 0)
app2(app2(sub, app2(s, x0)), app2(s, x1))
app2(app2(gtr, 0), x0)
app2(app2(gtr, app2(s, x0)), 0)
app2(app2(gtr, app2(s, x0)), app2(s, x1))
app2(app2(d, x0), 0)
app2(app2(d, app2(s, x0)), app2(s, x1))
app2(len, nil)
app2(len, app2(app2(cons, x0), x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(p, x)
APP2(app2(filter, p), app2(app2(cons, x), xs)) -> APP2(app2(filter, p), xs)
Used argument filtering: APP2(x1, x2) = x2
app2(x1, x2) = app2(x1, x2)
cons = cons
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
app2(app2(app2(if, true), xs), ys) -> xs
app2(app2(app2(if, false), xs), ys) -> ys
app2(app2(sub, x), 0) -> x
app2(app2(sub, app2(s, x)), app2(s, y)) -> app2(app2(sub, x), y)
app2(app2(gtr, 0), y) -> false
app2(app2(gtr, app2(s, x)), 0) -> true
app2(app2(gtr, app2(s, x)), app2(s, y)) -> app2(app2(gtr, x), y)
app2(app2(d, x), 0) -> true
app2(app2(d, app2(s, x)), app2(s, y)) -> app2(app2(app2(if, app2(app2(gtr, x), y)), false), app2(app2(d, app2(s, x)), app2(app2(sub, y), x)))
app2(len, nil) -> 0
app2(len, app2(app2(cons, x), xs)) -> app2(s, app2(len, xs))
app2(app2(filter, p), nil) -> nil
app2(app2(filter, p), app2(app2(cons, x), xs)) -> app2(app2(app2(if, app2(p, x)), app2(app2(cons, x), app2(app2(filter, p), xs))), app2(app2(filter, p), xs))
The set Q consists of the following terms:
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(sub, x0), 0)
app2(app2(sub, app2(s, x0)), app2(s, x1))
app2(app2(gtr, 0), x0)
app2(app2(gtr, app2(s, x0)), 0)
app2(app2(gtr, app2(s, x0)), app2(s, x1))
app2(app2(d, x0), 0)
app2(app2(d, app2(s, x0)), app2(s, x1))
app2(len, nil)
app2(len, app2(app2(cons, x0), x1))
app2(app2(filter, x0), nil)
app2(app2(filter, x0), app2(app2(cons, x1), x2))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.