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:
app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
APP'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(minus, x), y)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y))
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(plus, x)
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(app'2(plus, x), y)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(quot, app'2(app'2(minus, x), y))
APP'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> APP'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
APP'2(app'2(plus, app'2(s, x)), y) -> APP'2(app'2(plus, x), y)
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
APP'2(app'2(plus, app'2(s, x)), y) -> APP'2(s, app'2(app'2(plus, x), y))
APP'2(app'2(app, app'2(app'2(cons, x), l)), k) -> APP'2(app'2(cons, x), app'2(app'2(app, l), k))
APP'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> APP'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k))))
APP'2(app'2(plus, app'2(s, x)), y) -> APP'2(plus, x)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(minus, x)
APP'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> APP'2(app'2(plus, y), z)
APP'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> APP'2(plus, y)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(app'2(cons, app'2(app'2(plus, x), y)), l)
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(cons, app'2(app'2(plus, x), y))
APP'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> APP'2(minus, x)
APP'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))
APP'2(app'2(app, app'2(app'2(cons, x), l)), k) -> APP'2(app, l)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(minus, x), y)
APP'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> APP'2(app'2(minus, x), app'2(app'2(plus, y), z))
APP'2(app'2(app, app'2(app'2(cons, x), l)), k) -> APP'2(app'2(app, l), k)
The TRS R consists of the following rules:
app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
APP'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(minus, x), y)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y))
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(plus, x)
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(app'2(plus, x), y)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(quot, app'2(app'2(minus, x), y))
APP'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> APP'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
APP'2(app'2(plus, app'2(s, x)), y) -> APP'2(app'2(plus, x), y)
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
APP'2(app'2(plus, app'2(s, x)), y) -> APP'2(s, app'2(app'2(plus, x), y))
APP'2(app'2(app, app'2(app'2(cons, x), l)), k) -> APP'2(app'2(cons, x), app'2(app'2(app, l), k))
APP'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> APP'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k))))
APP'2(app'2(plus, app'2(s, x)), y) -> APP'2(plus, x)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(minus, x)
APP'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> APP'2(app'2(plus, y), z)
APP'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> APP'2(plus, y)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(app'2(cons, app'2(app'2(plus, x), y)), l)
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(cons, app'2(app'2(plus, x), y))
APP'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> APP'2(minus, x)
APP'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))
APP'2(app'2(app, app'2(app'2(cons, x), l)), k) -> APP'2(app, l)
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(minus, x), y)
APP'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> APP'2(app'2(minus, x), app'2(app'2(plus, y), z))
APP'2(app'2(app, app'2(app'2(cons, x), l)), k) -> APP'2(app'2(app, l), k)
The TRS R consists of the following rules:
app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 6 SCCs with 17 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP'2(app'2(app, app'2(app'2(cons, x), l)), k) -> APP'2(app'2(app, l), k)
The TRS R consists of the following rules:
app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
Q is empty.
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.
APP'2(app'2(app, app'2(app'2(cons, x), l)), k) -> APP'2(app'2(app, l), k)
Used argument filtering: APP'2(x1, x2) = x1
app'2(x1, x2) = app'1(x2)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP'2(app'2(plus, app'2(s, x)), y) -> APP'2(app'2(plus, x), y)
The TRS R consists of the following rules:
app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
Q is empty.
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.
APP'2(app'2(plus, app'2(s, x)), y) -> APP'2(app'2(plus, x), y)
Used argument filtering: APP'2(x1, x2) = x1
app'2(x1, x2) = app'1(x2)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> APP'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
The TRS R consists of the following rules:
app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> APP'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
The TRS R consists of the following rules:
app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(minus, x), y)
APP'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> APP'2(app'2(minus, x), app'2(app'2(plus, y), z))
The TRS R consists of the following rules:
app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
Q is empty.
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.
APP'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(minus, x), y)
APP'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> APP'2(app'2(minus, x), app'2(app'2(plus, y), z))
Used argument filtering: APP'2(x1, x2) = x1
app'2(x1, x2) = app'2(x1, x2)
minus = minus
s = s
plus = plus
0 = 0
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> APP'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y))
The TRS R consists of the following rules:
app'2(app'2(minus, x), 0) -> x
app'2(app'2(minus, app'2(s, x)), app'2(s, y)) -> app'2(app'2(minus, x), y)
app'2(app'2(minus, app'2(app'2(minus, x), y)), z) -> app'2(app'2(minus, x), app'2(app'2(plus, y), z))
app'2(app'2(quot, 0), app'2(s, y)) -> 0
app'2(app'2(quot, app'2(s, x)), app'2(s, y)) -> app'2(s, app'2(app'2(quot, app'2(app'2(minus, x), y)), app'2(s, y)))
app'2(app'2(plus, 0), y) -> y
app'2(app'2(plus, app'2(s, x)), y) -> app'2(s, app'2(app'2(plus, x), y))
app'2(app'2(app, nil), k) -> k
app'2(app'2(app, l), nil) -> l
app'2(app'2(app, app'2(app'2(cons, x), l)), k) -> app'2(app'2(cons, x), app'2(app'2(app, l), k))
app'2(sum, app'2(app'2(cons, x), nil)) -> app'2(app'2(cons, x), nil)
app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), l))) -> app'2(sum, app'2(app'2(cons, app'2(app'2(plus, x), y)), l))
app'2(sum, app'2(app'2(app, l), app'2(app'2(cons, x), app'2(app'2(cons, y), k)))) -> app'2(sum, app'2(app'2(app, l), app'2(sum, app'2(app'2(cons, x), app'2(app'2(cons, y), k)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.