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:
fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, fst2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(len1(Z))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, fst2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(len1(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:
fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, fst2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(len1(Z))
The set Q consists of the following terms:
fst2(0, x0)
fst2(s1(x0), cons2(x1, x2))
from1(x0)
add2(0, x0)
add2(s1(x0), x1)
len1(nil)
len1(cons2(x0, x1))
Q DP problem:
The TRS P consists of the following rules:
ADD2(s1(X), Y) -> ADD2(X, Y)
LEN1(cons2(X, Z)) -> LEN1(Z)
FROM1(X) -> FROM1(s1(X))
FST2(s1(X), cons2(Y, Z)) -> FST2(X, Z)
The TRS R consists of the following rules:
fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, fst2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(len1(Z))
The set Q consists of the following terms:
fst2(0, x0)
fst2(s1(x0), cons2(x1, x2))
from1(x0)
add2(0, x0)
add2(s1(x0), x1)
len1(nil)
len1(cons2(x0, x1))
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:
ADD2(s1(X), Y) -> ADD2(X, Y)
LEN1(cons2(X, Z)) -> LEN1(Z)
FROM1(X) -> FROM1(s1(X))
FST2(s1(X), cons2(Y, Z)) -> FST2(X, Z)
The TRS R consists of the following rules:
fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, fst2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(len1(Z))
The set Q consists of the following terms:
fst2(0, x0)
fst2(s1(x0), cons2(x1, x2))
from1(x0)
add2(0, x0)
add2(s1(x0), x1)
len1(nil)
len1(cons2(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 4 SCCs.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
LEN1(cons2(X, Z)) -> LEN1(Z)
The TRS R consists of the following rules:
fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, fst2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(len1(Z))
The set Q consists of the following terms:
fst2(0, x0)
fst2(s1(x0), cons2(x1, x2))
from1(x0)
add2(0, x0)
add2(s1(x0), x1)
len1(nil)
len1(cons2(x0, x1))
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.
LEN1(cons2(X, Z)) -> LEN1(Z)
Used argument filtering: LEN1(x1) = x1
cons2(x1, x2) = cons1(x2)
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ 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:
fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, fst2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(len1(Z))
The set Q consists of the following terms:
fst2(0, x0)
fst2(s1(x0), cons2(x1, x2))
from1(x0)
add2(0, x0)
add2(s1(x0), x1)
len1(nil)
len1(cons2(x0, x1))
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
Q DP problem:
The TRS P consists of the following rules:
ADD2(s1(X), Y) -> ADD2(X, Y)
The TRS R consists of the following rules:
fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, fst2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(len1(Z))
The set Q consists of the following terms:
fst2(0, x0)
fst2(s1(x0), cons2(x1, x2))
from1(x0)
add2(0, x0)
add2(s1(x0), x1)
len1(nil)
len1(cons2(x0, x1))
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.
ADD2(s1(X), Y) -> ADD2(X, Y)
Used argument filtering: ADD2(x1, x2) = x1
s1(x1) = s1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ 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:
fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, fst2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(len1(Z))
The set Q consists of the following terms:
fst2(0, x0)
fst2(s1(x0), cons2(x1, x2))
from1(x0)
add2(0, x0)
add2(s1(x0), x1)
len1(nil)
len1(cons2(x0, x1))
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
Q DP problem:
The TRS P consists of the following rules:
FROM1(X) -> FROM1(s1(X))
The TRS R consists of the following rules:
fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, fst2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(len1(Z))
The set Q consists of the following terms:
fst2(0, x0)
fst2(s1(x0), cons2(x1, x2))
from1(x0)
add2(0, x0)
add2(s1(x0), x1)
len1(nil)
len1(cons2(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
FST2(s1(X), cons2(Y, Z)) -> FST2(X, Z)
The TRS R consists of the following rules:
fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, fst2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(len1(Z))
The set Q consists of the following terms:
fst2(0, x0)
fst2(s1(x0), cons2(x1, x2))
from1(x0)
add2(0, x0)
add2(s1(x0), x1)
len1(nil)
len1(cons2(x0, x1))
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.
FST2(s1(X), cons2(Y, Z)) -> FST2(X, Z)
Used argument filtering: FST2(x1, x2) = x2
cons2(x1, x2) = cons1(x2)
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, fst2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(len1(Z))
The set Q consists of the following terms:
fst2(0, x0)
fst2(s1(x0), cons2(x1, x2))
from1(x0)
add2(0, x0)
add2(s1(x0), x1)
len1(nil)
len1(cons2(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.