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(nil, k) -> k
app2(l, nil) -> l
app2(cons2(x, l), k) -> cons2(x, app2(l, k))
sum1(cons2(x, nil)) -> cons2(x, nil)
sum1(cons2(x, cons2(y, l))) -> sum1(cons2(plus2(x, y), l))
sum1(app2(l, cons2(x, cons2(y, k)))) -> sum1(app2(l, sum1(cons2(x, cons2(y, k)))))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
sum1(plus2(cons2(0, x), cons2(y, l))) -> pred1(sum1(cons2(s1(x), cons2(y, l))))
pred1(cons2(s1(x), nil)) -> cons2(x, nil)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
app2(nil, k) -> k
app2(l, nil) -> l
app2(cons2(x, l), k) -> cons2(x, app2(l, k))
sum1(cons2(x, nil)) -> cons2(x, nil)
sum1(cons2(x, cons2(y, l))) -> sum1(cons2(plus2(x, y), l))
sum1(app2(l, cons2(x, cons2(y, k)))) -> sum1(app2(l, sum1(cons2(x, cons2(y, k)))))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
sum1(plus2(cons2(0, x), cons2(y, l))) -> pred1(sum1(cons2(s1(x), cons2(y, l))))
pred1(cons2(s1(x), nil)) -> cons2(x, nil)
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
PLUS2(s1(x), y) -> PLUS2(x, y)
SUM1(app2(l, cons2(x, cons2(y, k)))) -> APP2(l, sum1(cons2(x, cons2(y, k))))
SUM1(cons2(x, cons2(y, l))) -> PLUS2(x, y)
SUM1(plus2(cons2(0, x), cons2(y, l))) -> SUM1(cons2(s1(x), cons2(y, l)))
SUM1(cons2(x, cons2(y, l))) -> SUM1(cons2(plus2(x, y), l))
SUM1(plus2(cons2(0, x), cons2(y, l))) -> PRED1(sum1(cons2(s1(x), cons2(y, l))))
SUM1(app2(l, cons2(x, cons2(y, k)))) -> SUM1(cons2(x, cons2(y, k)))
SUM1(app2(l, cons2(x, cons2(y, k)))) -> SUM1(app2(l, sum1(cons2(x, cons2(y, k)))))
APP2(cons2(x, l), k) -> APP2(l, k)
The TRS R consists of the following rules:
app2(nil, k) -> k
app2(l, nil) -> l
app2(cons2(x, l), k) -> cons2(x, app2(l, k))
sum1(cons2(x, nil)) -> cons2(x, nil)
sum1(cons2(x, cons2(y, l))) -> sum1(cons2(plus2(x, y), l))
sum1(app2(l, cons2(x, cons2(y, k)))) -> sum1(app2(l, sum1(cons2(x, cons2(y, k)))))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
sum1(plus2(cons2(0, x), cons2(y, l))) -> pred1(sum1(cons2(s1(x), cons2(y, l))))
pred1(cons2(s1(x), nil)) -> cons2(x, nil)
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:
PLUS2(s1(x), y) -> PLUS2(x, y)
SUM1(app2(l, cons2(x, cons2(y, k)))) -> APP2(l, sum1(cons2(x, cons2(y, k))))
SUM1(cons2(x, cons2(y, l))) -> PLUS2(x, y)
SUM1(plus2(cons2(0, x), cons2(y, l))) -> SUM1(cons2(s1(x), cons2(y, l)))
SUM1(cons2(x, cons2(y, l))) -> SUM1(cons2(plus2(x, y), l))
SUM1(plus2(cons2(0, x), cons2(y, l))) -> PRED1(sum1(cons2(s1(x), cons2(y, l))))
SUM1(app2(l, cons2(x, cons2(y, k)))) -> SUM1(cons2(x, cons2(y, k)))
SUM1(app2(l, cons2(x, cons2(y, k)))) -> SUM1(app2(l, sum1(cons2(x, cons2(y, k)))))
APP2(cons2(x, l), k) -> APP2(l, k)
The TRS R consists of the following rules:
app2(nil, k) -> k
app2(l, nil) -> l
app2(cons2(x, l), k) -> cons2(x, app2(l, k))
sum1(cons2(x, nil)) -> cons2(x, nil)
sum1(cons2(x, cons2(y, l))) -> sum1(cons2(plus2(x, y), l))
sum1(app2(l, cons2(x, cons2(y, k)))) -> sum1(app2(l, sum1(cons2(x, cons2(y, k)))))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
sum1(plus2(cons2(0, x), cons2(y, l))) -> pred1(sum1(cons2(s1(x), cons2(y, l))))
pred1(cons2(s1(x), nil)) -> cons2(x, nil)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 4 SCCs with 5 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PLUS2(s1(x), y) -> PLUS2(x, y)
The TRS R consists of the following rules:
app2(nil, k) -> k
app2(l, nil) -> l
app2(cons2(x, l), k) -> cons2(x, app2(l, k))
sum1(cons2(x, nil)) -> cons2(x, nil)
sum1(cons2(x, cons2(y, l))) -> sum1(cons2(plus2(x, y), l))
sum1(app2(l, cons2(x, cons2(y, k)))) -> sum1(app2(l, sum1(cons2(x, cons2(y, k)))))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
sum1(plus2(cons2(0, x), cons2(y, l))) -> pred1(sum1(cons2(s1(x), cons2(y, l))))
pred1(cons2(s1(x), nil)) -> cons2(x, nil)
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.
PLUS2(s1(x), y) -> PLUS2(x, y)
Used argument filtering: PLUS2(x1, x2) = x1
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
app2(nil, k) -> k
app2(l, nil) -> l
app2(cons2(x, l), k) -> cons2(x, app2(l, k))
sum1(cons2(x, nil)) -> cons2(x, nil)
sum1(cons2(x, cons2(y, l))) -> sum1(cons2(plus2(x, y), l))
sum1(app2(l, cons2(x, cons2(y, k)))) -> sum1(app2(l, sum1(cons2(x, cons2(y, k)))))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
sum1(plus2(cons2(0, x), cons2(y, l))) -> pred1(sum1(cons2(s1(x), cons2(y, l))))
pred1(cons2(s1(x), nil)) -> cons2(x, nil)
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
Q DP problem:
The TRS P consists of the following rules:
APP2(cons2(x, l), k) -> APP2(l, k)
The TRS R consists of the following rules:
app2(nil, k) -> k
app2(l, nil) -> l
app2(cons2(x, l), k) -> cons2(x, app2(l, k))
sum1(cons2(x, nil)) -> cons2(x, nil)
sum1(cons2(x, cons2(y, l))) -> sum1(cons2(plus2(x, y), l))
sum1(app2(l, cons2(x, cons2(y, k)))) -> sum1(app2(l, sum1(cons2(x, cons2(y, k)))))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
sum1(plus2(cons2(0, x), cons2(y, l))) -> pred1(sum1(cons2(s1(x), cons2(y, l))))
pred1(cons2(s1(x), nil)) -> cons2(x, nil)
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.
APP2(cons2(x, l), k) -> APP2(l, k)
Used argument filtering: APP2(x1, x2) = x1
cons2(x1, x2) = cons1(x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
app2(nil, k) -> k
app2(l, nil) -> l
app2(cons2(x, l), k) -> cons2(x, app2(l, k))
sum1(cons2(x, nil)) -> cons2(x, nil)
sum1(cons2(x, cons2(y, l))) -> sum1(cons2(plus2(x, y), l))
sum1(app2(l, cons2(x, cons2(y, k)))) -> sum1(app2(l, sum1(cons2(x, cons2(y, k)))))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
sum1(plus2(cons2(0, x), cons2(y, l))) -> pred1(sum1(cons2(s1(x), cons2(y, l))))
pred1(cons2(s1(x), nil)) -> cons2(x, nil)
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
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SUM1(cons2(x, cons2(y, l))) -> SUM1(cons2(plus2(x, y), l))
The TRS R consists of the following rules:
app2(nil, k) -> k
app2(l, nil) -> l
app2(cons2(x, l), k) -> cons2(x, app2(l, k))
sum1(cons2(x, nil)) -> cons2(x, nil)
sum1(cons2(x, cons2(y, l))) -> sum1(cons2(plus2(x, y), l))
sum1(app2(l, cons2(x, cons2(y, k)))) -> sum1(app2(l, sum1(cons2(x, cons2(y, k)))))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
sum1(plus2(cons2(0, x), cons2(y, l))) -> pred1(sum1(cons2(s1(x), cons2(y, l))))
pred1(cons2(s1(x), nil)) -> cons2(x, nil)
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.
SUM1(cons2(x, cons2(y, l))) -> SUM1(cons2(plus2(x, y), l))
Used argument filtering: SUM1(x1) = x1
cons2(x1, x2) = cons1(x2)
plus2(x1, x2) = x2
0 = 0
s1(x1) = s
Used ordering: Quasi Precedence:
cons_1 > s
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
app2(nil, k) -> k
app2(l, nil) -> l
app2(cons2(x, l), k) -> cons2(x, app2(l, k))
sum1(cons2(x, nil)) -> cons2(x, nil)
sum1(cons2(x, cons2(y, l))) -> sum1(cons2(plus2(x, y), l))
sum1(app2(l, cons2(x, cons2(y, k)))) -> sum1(app2(l, sum1(cons2(x, cons2(y, k)))))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
sum1(plus2(cons2(0, x), cons2(y, l))) -> pred1(sum1(cons2(s1(x), cons2(y, l))))
pred1(cons2(s1(x), nil)) -> cons2(x, nil)
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
Q DP problem:
The TRS P consists of the following rules:
SUM1(app2(l, cons2(x, cons2(y, k)))) -> SUM1(app2(l, sum1(cons2(x, cons2(y, k)))))
The TRS R consists of the following rules:
app2(nil, k) -> k
app2(l, nil) -> l
app2(cons2(x, l), k) -> cons2(x, app2(l, k))
sum1(cons2(x, nil)) -> cons2(x, nil)
sum1(cons2(x, cons2(y, l))) -> sum1(cons2(plus2(x, y), l))
sum1(app2(l, cons2(x, cons2(y, k)))) -> sum1(app2(l, sum1(cons2(x, cons2(y, k)))))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
sum1(plus2(cons2(0, x), cons2(y, l))) -> pred1(sum1(cons2(s1(x), cons2(y, l))))
pred1(cons2(s1(x), nil)) -> cons2(x, nil)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.