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(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
app2(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))
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(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))
The set Q consists of the following terms:
app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(eq, 0), 0)
app2(app2(eq, 0), app2(s, x0))
app2(app2(eq, app2(s, x0)), 0)
app2(app2(eq, app2(s, x0)), app2(s, x1))
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(minsort, nil)
app2(minsort, app2(app2(cons, x0), x1))
app2(app2(min, x0), nil)
app2(app2(min, x0), app2(app2(cons, x1), x2))
app2(app2(del, x0), nil)
app2(app2(del, x0), app2(app2(cons, x1), x2))
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(eq, x)
APP2(minsort, app2(app2(cons, x), y)) -> APP2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y)))
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(le, x), y)
APP2(minsort, app2(app2(cons, x), y)) -> APP2(cons, app2(app2(min, x), y))
APP2(app2(le, app2(s, x)), app2(s, y)) -> APP2(le, x)
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(if, app2(app2(le, x), y))
APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(app2(del, x), z)
APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(min, y), z)
APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(app2(eq, x), y)
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
APP2(minsort, app2(app2(cons, x), y)) -> APP2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))
APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(app2(if, app2(app2(eq, x), y)), z)
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(min, x), z)
APP2(app2(le, app2(s, x)), app2(s, y)) -> APP2(app2(le, x), y)
APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(app2(cons, y), app2(app2(del, x), z))
APP2(app2(eq, app2(s, x)), app2(s, y)) -> APP2(eq, x)
APP2(minsort, app2(app2(cons, x), y)) -> APP2(min, x)
APP2(minsort, app2(app2(cons, x), y)) -> APP2(app2(min, x), y)
APP2(app2(eq, app2(s, x)), app2(s, y)) -> APP2(app2(eq, x), y)
APP2(minsort, app2(app2(cons, x), y)) -> APP2(del, app2(app2(min, x), y))
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(le, x)
APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(if, app2(app2(eq, x), y))
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z))
APP2(minsort, app2(app2(cons, x), y)) -> APP2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(min, y)
The TRS R consists of the following rules:
app2(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))
The set Q consists of the following terms:
app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(eq, 0), 0)
app2(app2(eq, 0), app2(s, x0))
app2(app2(eq, app2(s, x0)), 0)
app2(app2(eq, app2(s, x0)), app2(s, x1))
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(minsort, nil)
app2(minsort, app2(app2(cons, x0), x1))
app2(app2(min, x0), nil)
app2(app2(min, x0), app2(app2(cons, x1), x2))
app2(app2(del, x0), nil)
app2(app2(del, 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(del, x), app2(app2(cons, y), z)) -> APP2(eq, x)
APP2(minsort, app2(app2(cons, x), y)) -> APP2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y)))
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(le, x), y)
APP2(minsort, app2(app2(cons, x), y)) -> APP2(cons, app2(app2(min, x), y))
APP2(app2(le, app2(s, x)), app2(s, y)) -> APP2(le, x)
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(if, app2(app2(le, x), y))
APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(app2(del, x), z)
APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(min, y), z)
APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(app2(eq, x), y)
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
APP2(minsort, app2(app2(cons, x), y)) -> APP2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))
APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(app2(if, app2(app2(eq, x), y)), z)
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(min, x), z)
APP2(app2(le, app2(s, x)), app2(s, y)) -> APP2(app2(le, x), y)
APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(app2(cons, y), app2(app2(del, x), z))
APP2(app2(eq, app2(s, x)), app2(s, y)) -> APP2(eq, x)
APP2(minsort, app2(app2(cons, x), y)) -> APP2(min, x)
APP2(minsort, app2(app2(cons, x), y)) -> APP2(app2(min, x), y)
APP2(app2(eq, app2(s, x)), app2(s, y)) -> APP2(app2(eq, x), y)
APP2(minsort, app2(app2(cons, x), y)) -> APP2(del, app2(app2(min, x), y))
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(le, x)
APP2(app2(del, x), app2(app2(cons, y), z)) -> APP2(if, app2(app2(eq, x), y))
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z))
APP2(minsort, app2(app2(cons, x), y)) -> APP2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(min, y)
The TRS R consists of the following rules:
app2(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))
The set Q consists of the following terms:
app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(eq, 0), 0)
app2(app2(eq, 0), app2(s, x0))
app2(app2(eq, app2(s, x0)), 0)
app2(app2(eq, app2(s, x0)), app2(s, x1))
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(minsort, nil)
app2(minsort, app2(app2(cons, x0), x1))
app2(app2(min, x0), nil)
app2(app2(min, x0), app2(app2(cons, x1), x2))
app2(app2(del, x0), nil)
app2(app2(del, 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 20 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(app2(eq, app2(s, x)), app2(s, y)) -> APP2(app2(eq, x), y)
The TRS R consists of the following rules:
app2(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))
The set Q consists of the following terms:
app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(eq, 0), 0)
app2(app2(eq, 0), app2(s, x0))
app2(app2(eq, app2(s, x0)), 0)
app2(app2(eq, app2(s, x0)), app2(s, x1))
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(minsort, nil)
app2(minsort, app2(app2(cons, x0), x1))
app2(app2(min, x0), nil)
app2(app2(min, x0), app2(app2(cons, x1), x2))
app2(app2(del, x0), nil)
app2(app2(del, 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(eq, app2(s, x)), app2(s, y)) -> APP2(app2(eq, 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
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
app2(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))
The set Q consists of the following terms:
app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(eq, 0), 0)
app2(app2(eq, 0), app2(s, x0))
app2(app2(eq, app2(s, x0)), 0)
app2(app2(eq, app2(s, x0)), app2(s, x1))
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(minsort, nil)
app2(minsort, app2(app2(cons, x0), x1))
app2(app2(min, x0), nil)
app2(app2(min, x0), app2(app2(cons, x1), x2))
app2(app2(del, x0), nil)
app2(app2(del, 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(del, x), app2(app2(cons, y), z)) -> APP2(app2(del, x), z)
The TRS R consists of the following rules:
app2(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))
The set Q consists of the following terms:
app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(eq, 0), 0)
app2(app2(eq, 0), app2(s, x0))
app2(app2(eq, app2(s, x0)), 0)
app2(app2(eq, app2(s, x0)), app2(s, x1))
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(minsort, nil)
app2(minsort, app2(app2(cons, x0), x1))
app2(app2(min, x0), nil)
app2(app2(min, x0), app2(app2(cons, x1), x2))
app2(app2(del, x0), nil)
app2(app2(del, 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(del, x), app2(app2(cons, y), z)) -> APP2(app2(del, x), z)
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
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
app2(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))
The set Q consists of the following terms:
app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(eq, 0), 0)
app2(app2(eq, 0), app2(s, x0))
app2(app2(eq, app2(s, x0)), 0)
app2(app2(eq, app2(s, x0)), app2(s, x1))
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(minsort, nil)
app2(minsort, app2(app2(cons, x0), x1))
app2(app2(min, x0), nil)
app2(app2(min, x0), app2(app2(cons, x1), x2))
app2(app2(del, x0), nil)
app2(app2(del, 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(le, app2(s, x)), app2(s, y)) -> APP2(app2(le, x), y)
The TRS R consists of the following rules:
app2(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))
The set Q consists of the following terms:
app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(eq, 0), 0)
app2(app2(eq, 0), app2(s, x0))
app2(app2(eq, app2(s, x0)), 0)
app2(app2(eq, app2(s, x0)), app2(s, x1))
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(minsort, nil)
app2(minsort, app2(app2(cons, x0), x1))
app2(app2(min, x0), nil)
app2(app2(min, x0), app2(app2(cons, x1), x2))
app2(app2(del, x0), nil)
app2(app2(del, 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(le, app2(s, x)), app2(s, y)) -> APP2(app2(le, 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(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))
The set Q consists of the following terms:
app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(eq, 0), 0)
app2(app2(eq, 0), app2(s, x0))
app2(app2(eq, app2(s, x0)), 0)
app2(app2(eq, app2(s, x0)), app2(s, x1))
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(minsort, nil)
app2(minsort, app2(app2(cons, x0), x1))
app2(app2(min, x0), nil)
app2(app2(min, x0), app2(app2(cons, x1), x2))
app2(app2(del, x0), nil)
app2(app2(del, 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
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(min, x), z)
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(min, y), z)
The TRS R consists of the following rules:
app2(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))
The set Q consists of the following terms:
app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(eq, 0), 0)
app2(app2(eq, 0), app2(s, x0))
app2(app2(eq, app2(s, x0)), 0)
app2(app2(eq, app2(s, x0)), app2(s, x1))
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(minsort, nil)
app2(minsort, app2(app2(cons, x0), x1))
app2(app2(min, x0), nil)
app2(app2(min, x0), app2(app2(cons, x1), x2))
app2(app2(del, x0), nil)
app2(app2(del, 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(min, x), app2(app2(cons, y), z)) -> APP2(app2(min, x), z)
APP2(app2(min, x), app2(app2(cons, y), z)) -> APP2(app2(min, y), z)
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
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
app2(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))
The set Q consists of the following terms:
app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(eq, 0), 0)
app2(app2(eq, 0), app2(s, x0))
app2(app2(eq, app2(s, x0)), 0)
app2(app2(eq, app2(s, x0)), app2(s, x1))
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(minsort, nil)
app2(minsort, app2(app2(cons, x0), x1))
app2(app2(min, x0), nil)
app2(app2(min, x0), app2(app2(cons, x1), x2))
app2(app2(del, x0), nil)
app2(app2(del, 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(minsort, app2(app2(cons, x), y)) -> APP2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y)))
The TRS R consists of the following rules:
app2(app2(le, 0), y) -> true
app2(app2(le, app2(s, x)), 0) -> false
app2(app2(le, app2(s, x)), app2(s, y)) -> app2(app2(le, x), y)
app2(app2(eq, 0), 0) -> true
app2(app2(eq, 0), app2(s, y)) -> false
app2(app2(eq, app2(s, x)), 0) -> false
app2(app2(eq, app2(s, x)), app2(s, y)) -> app2(app2(eq, x), y)
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(minsort, nil) -> nil
app2(minsort, app2(app2(cons, x), y)) -> app2(app2(cons, app2(app2(min, x), y)), app2(minsort, app2(app2(del, app2(app2(min, x), y)), app2(app2(cons, x), y))))
app2(app2(min, x), nil) -> x
app2(app2(min, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(le, x), y)), app2(app2(min, x), z)), app2(app2(min, y), z))
app2(app2(del, x), nil) -> nil
app2(app2(del, x), app2(app2(cons, y), z)) -> app2(app2(app2(if, app2(app2(eq, x), y)), z), app2(app2(cons, y), app2(app2(del, x), z)))
The set Q consists of the following terms:
app2(app2(le, 0), x0)
app2(app2(le, app2(s, x0)), 0)
app2(app2(le, app2(s, x0)), app2(s, x1))
app2(app2(eq, 0), 0)
app2(app2(eq, 0), app2(s, x0))
app2(app2(eq, app2(s, x0)), 0)
app2(app2(eq, app2(s, x0)), app2(s, x1))
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(minsort, nil)
app2(minsort, app2(app2(cons, x0), x1))
app2(app2(min, x0), nil)
app2(app2(min, x0), app2(app2(cons, x1), x2))
app2(app2(del, x0), nil)
app2(app2(del, x0), app2(app2(cons, x1), x2))
We have to consider all minimal (P,Q,R)-chains.