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:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(xs))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(xs))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
SUM1(++2(xs, :2(x, :2(y, ys)))) -> ++12(xs, sum1(:2(x, :2(y, ys))))
AVG1(xs) -> HD1(sum1(xs))
QUOT2(s1(x), s1(y)) -> -12(x, y)
LENGTH1(:2(x, xs)) -> LENGTH1(xs)
-12(s1(x), s1(y)) -> -12(x, y)
+12(s1(x), y) -> +12(x, y)
AVG1(xs) -> QUOT2(hd1(sum1(xs)), length1(xs))
AVG1(xs) -> SUM1(xs)
++12(:2(x, xs), ys) -> ++12(xs, ys)
AVG1(xs) -> LENGTH1(xs)
SUM1(:2(x, :2(y, xs))) -> +12(x, y)
SUM1(:2(x, :2(y, xs))) -> SUM1(:2(+2(x, y), xs))
QUOT2(s1(x), s1(y)) -> QUOT2(-2(x, y), s1(y))
SUM1(++2(xs, :2(x, :2(y, ys)))) -> SUM1(:2(x, :2(y, ys)))
SUM1(++2(xs, :2(x, :2(y, ys)))) -> SUM1(++2(xs, sum1(:2(x, :2(y, ys)))))
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(xs))
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:
SUM1(++2(xs, :2(x, :2(y, ys)))) -> ++12(xs, sum1(:2(x, :2(y, ys))))
AVG1(xs) -> HD1(sum1(xs))
QUOT2(s1(x), s1(y)) -> -12(x, y)
LENGTH1(:2(x, xs)) -> LENGTH1(xs)
-12(s1(x), s1(y)) -> -12(x, y)
+12(s1(x), y) -> +12(x, y)
AVG1(xs) -> QUOT2(hd1(sum1(xs)), length1(xs))
AVG1(xs) -> SUM1(xs)
++12(:2(x, xs), ys) -> ++12(xs, ys)
AVG1(xs) -> LENGTH1(xs)
SUM1(:2(x, :2(y, xs))) -> +12(x, y)
SUM1(:2(x, :2(y, xs))) -> SUM1(:2(+2(x, y), xs))
QUOT2(s1(x), s1(y)) -> QUOT2(-2(x, y), s1(y))
SUM1(++2(xs, :2(x, :2(y, ys)))) -> SUM1(:2(x, :2(y, ys)))
SUM1(++2(xs, :2(x, :2(y, ys)))) -> SUM1(++2(xs, sum1(:2(x, :2(y, ys)))))
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(xs))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 7 SCCs with 8 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
LENGTH1(:2(x, xs)) -> LENGTH1(xs)
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(xs))
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.
LENGTH1(:2(x, xs)) -> LENGTH1(xs)
Used argument filtering: LENGTH1(x1) = x1
:2(x1, x2) = :1(x2)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(xs))
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
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
-12(s1(x), s1(y)) -> -12(x, y)
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(xs))
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.
-12(s1(x), s1(y)) -> -12(x, y)
Used argument filtering: -12(x1, x2) = x2
s1(x1) = s1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(xs))
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
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
QUOT2(s1(x), s1(y)) -> QUOT2(-2(x, y), s1(y))
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(xs))
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.
QUOT2(s1(x), s1(y)) -> QUOT2(-2(x, y), s1(y))
Used argument filtering: QUOT2(x1, x2) = x1
s1(x1) = s1(x1)
-2(x1, x2) = x1
0 = 0
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(xs))
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
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
++12(:2(x, xs), ys) -> ++12(xs, ys)
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(xs))
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.
++12(:2(x, xs), ys) -> ++12(xs, ys)
Used argument filtering: ++12(x1, x2) = x1
:2(x1, x2) = :1(x2)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(xs))
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
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
+12(s1(x), y) -> +12(x, y)
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(xs))
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.
+12(s1(x), y) -> +12(x, y)
Used argument filtering: +12(x1, x2) = x1
s1(x1) = s1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(xs))
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
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SUM1(:2(x, :2(y, xs))) -> SUM1(:2(+2(x, y), xs))
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(xs))
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(:2(x, :2(y, xs))) -> SUM1(:2(+2(x, y), xs))
Used argument filtering: SUM1(x1) = x1
:2(x1, x2) = :1(x2)
+2(x1, x2) = x2
0 = 0
s1(x1) = s
Used ordering: Precedence:
:1 > s
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(xs))
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
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SUM1(++2(xs, :2(x, :2(y, ys)))) -> SUM1(++2(xs, sum1(:2(x, :2(y, ys)))))
The TRS R consists of the following rules:
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
++2(nil, ys) -> ys
++2(:2(x, xs), ys) -> :2(x, ++2(xs, ys))
sum1(:2(x, nil)) -> :2(x, nil)
sum1(:2(x, :2(y, xs))) -> sum1(:2(+2(x, y), xs))
sum1(++2(xs, :2(x, :2(y, ys)))) -> sum1(++2(xs, sum1(:2(x, :2(y, ys)))))
-2(x, 0) -> x
-2(0, s1(y)) -> 0
-2(s1(x), s1(y)) -> -2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(-2(x, y), s1(y)))
length1(nil) -> 0
length1(:2(x, xs)) -> s1(length1(xs))
hd1(:2(x, xs)) -> x
avg1(xs) -> quot2(hd1(sum1(xs)), length1(xs))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.