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, YS) -> YS
app2(cons2(X, XS), YS) -> cons2(X, app2(XS, YS))
from1(X) -> cons2(X, from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS))
prefix1(L) -> cons2(nil, zWadr2(L, prefix1(L)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
app2(nil, YS) -> YS
app2(cons2(X, XS), YS) -> cons2(X, app2(XS, YS))
from1(X) -> cons2(X, from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS))
prefix1(L) -> cons2(nil, zWadr2(L, prefix1(L)))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
PREFIX1(L) -> PREFIX1(L)
ZWADR2(cons2(X, XS), cons2(Y, YS)) -> ZWADR2(XS, YS)
PREFIX1(L) -> ZWADR2(L, prefix1(L))
ZWADR2(cons2(X, XS), cons2(Y, YS)) -> APP2(Y, cons2(X, nil))
FROM1(X) -> FROM1(s1(X))
APP2(cons2(X, XS), YS) -> APP2(XS, YS)
The TRS R consists of the following rules:
app2(nil, YS) -> YS
app2(cons2(X, XS), YS) -> cons2(X, app2(XS, YS))
from1(X) -> cons2(X, from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS))
prefix1(L) -> cons2(nil, zWadr2(L, prefix1(L)))
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:
PREFIX1(L) -> PREFIX1(L)
ZWADR2(cons2(X, XS), cons2(Y, YS)) -> ZWADR2(XS, YS)
PREFIX1(L) -> ZWADR2(L, prefix1(L))
ZWADR2(cons2(X, XS), cons2(Y, YS)) -> APP2(Y, cons2(X, nil))
FROM1(X) -> FROM1(s1(X))
APP2(cons2(X, XS), YS) -> APP2(XS, YS)
The TRS R consists of the following rules:
app2(nil, YS) -> YS
app2(cons2(X, XS), YS) -> cons2(X, app2(XS, YS))
from1(X) -> cons2(X, from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS))
prefix1(L) -> cons2(nil, zWadr2(L, prefix1(L)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 4 SCCs with 2 less nodes.
↳ 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:
app2(nil, YS) -> YS
app2(cons2(X, XS), YS) -> cons2(X, app2(XS, YS))
from1(X) -> cons2(X, from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS))
prefix1(L) -> cons2(nil, zWadr2(L, prefix1(L)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP2(cons2(X, XS), YS) -> APP2(XS, YS)
The TRS R consists of the following rules:
app2(nil, YS) -> YS
app2(cons2(X, XS), YS) -> cons2(X, app2(XS, YS))
from1(X) -> cons2(X, from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS))
prefix1(L) -> cons2(nil, zWadr2(L, prefix1(L)))
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, XS), YS) -> APP2(XS, YS)
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, YS) -> YS
app2(cons2(X, XS), YS) -> cons2(X, app2(XS, YS))
from1(X) -> cons2(X, from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS))
prefix1(L) -> cons2(nil, zWadr2(L, prefix1(L)))
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:
ZWADR2(cons2(X, XS), cons2(Y, YS)) -> ZWADR2(XS, YS)
The TRS R consists of the following rules:
app2(nil, YS) -> YS
app2(cons2(X, XS), YS) -> cons2(X, app2(XS, YS))
from1(X) -> cons2(X, from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS))
prefix1(L) -> cons2(nil, zWadr2(L, prefix1(L)))
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.
ZWADR2(cons2(X, XS), cons2(Y, YS)) -> ZWADR2(XS, YS)
Used argument filtering: ZWADR2(x1, x2) = x2
cons2(x1, x2) = cons1(x2)
Used ordering: Quasi Precedence:
trivial
↳ 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, YS) -> YS
app2(cons2(X, XS), YS) -> cons2(X, app2(XS, YS))
from1(X) -> cons2(X, from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS))
prefix1(L) -> cons2(nil, zWadr2(L, prefix1(L)))
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:
PREFIX1(L) -> PREFIX1(L)
The TRS R consists of the following rules:
app2(nil, YS) -> YS
app2(cons2(X, XS), YS) -> cons2(X, app2(XS, YS))
from1(X) -> cons2(X, from1(s1(X)))
zWadr2(nil, YS) -> nil
zWadr2(XS, nil) -> nil
zWadr2(cons2(X, XS), cons2(Y, YS)) -> cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS))
prefix1(L) -> cons2(nil, zWadr2(L, prefix1(L)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.