Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
rev1(nil) -> nil
rev1(rev1(x)) -> x
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
++2(nil, y) -> y
++2(x, nil) -> x
++2(.2(x, y), z) -> .2(x, ++2(y, z))
++2(x, ++2(y, z)) -> ++2(++2(x, y), z)
make1(x) -> .2(x, nil)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
rev1(nil) -> nil
rev1(rev1(x)) -> x
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
++2(nil, y) -> y
++2(x, nil) -> x
++2(.2(x, y), z) -> .2(x, ++2(y, z))
++2(x, ++2(y, z)) -> ++2(++2(x, y), z)
make1(x) -> .2(x, nil)
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
REV1(++2(x, y)) -> REV1(x)
REV1(++2(x, y)) -> REV1(y)
++12(x, ++2(y, z)) -> ++12(++2(x, y), z)
REV1(++2(x, y)) -> ++12(rev1(y), rev1(x))
++12(x, ++2(y, z)) -> ++12(x, y)
++12(.2(x, y), z) -> ++12(y, z)
The TRS R consists of the following rules:
rev1(nil) -> nil
rev1(rev1(x)) -> x
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
++2(nil, y) -> y
++2(x, nil) -> x
++2(.2(x, y), z) -> .2(x, ++2(y, z))
++2(x, ++2(y, z)) -> ++2(++2(x, y), z)
make1(x) -> .2(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:
REV1(++2(x, y)) -> REV1(x)
REV1(++2(x, y)) -> REV1(y)
++12(x, ++2(y, z)) -> ++12(++2(x, y), z)
REV1(++2(x, y)) -> ++12(rev1(y), rev1(x))
++12(x, ++2(y, z)) -> ++12(x, y)
++12(.2(x, y), z) -> ++12(y, z)
The TRS R consists of the following rules:
rev1(nil) -> nil
rev1(rev1(x)) -> x
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
++2(nil, y) -> y
++2(x, nil) -> x
++2(.2(x, y), z) -> .2(x, ++2(y, z))
++2(x, ++2(y, z)) -> ++2(++2(x, y), z)
make1(x) -> .2(x, nil)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
++12(x, ++2(y, z)) -> ++12(++2(x, y), z)
++12(x, ++2(y, z)) -> ++12(x, y)
++12(.2(x, y), z) -> ++12(y, z)
The TRS R consists of the following rules:
rev1(nil) -> nil
rev1(rev1(x)) -> x
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
++2(nil, y) -> y
++2(x, nil) -> x
++2(.2(x, y), z) -> .2(x, ++2(y, z))
++2(x, ++2(y, z)) -> ++2(++2(x, y), z)
make1(x) -> .2(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.
++12(x, ++2(y, z)) -> ++12(++2(x, y), z)
++12(x, ++2(y, z)) -> ++12(x, y)
Used argument filtering: ++12(x1, x2) = x2
++2(x1, x2) = ++2(x1, x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
++12(.2(x, y), z) -> ++12(y, z)
The TRS R consists of the following rules:
rev1(nil) -> nil
rev1(rev1(x)) -> x
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
++2(nil, y) -> y
++2(x, nil) -> x
++2(.2(x, y), z) -> .2(x, ++2(y, z))
++2(x, ++2(y, z)) -> ++2(++2(x, y), z)
make1(x) -> .2(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.
++12(.2(x, y), z) -> ++12(y, z)
Used argument filtering: ++12(x1, x2) = x1
.2(x1, x2) = .1(x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
rev1(nil) -> nil
rev1(rev1(x)) -> x
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
++2(nil, y) -> y
++2(x, nil) -> x
++2(.2(x, y), z) -> .2(x, ++2(y, z))
++2(x, ++2(y, z)) -> ++2(++2(x, y), z)
make1(x) -> .2(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
Q DP problem:
The TRS P consists of the following rules:
REV1(++2(x, y)) -> REV1(y)
REV1(++2(x, y)) -> REV1(x)
The TRS R consists of the following rules:
rev1(nil) -> nil
rev1(rev1(x)) -> x
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
++2(nil, y) -> y
++2(x, nil) -> x
++2(.2(x, y), z) -> .2(x, ++2(y, z))
++2(x, ++2(y, z)) -> ++2(++2(x, y), z)
make1(x) -> .2(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.
REV1(++2(x, y)) -> REV1(y)
REV1(++2(x, y)) -> REV1(x)
Used argument filtering: REV1(x1) = x1
++2(x1, x2) = ++2(x1, x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
rev1(nil) -> nil
rev1(rev1(x)) -> x
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
++2(nil, y) -> y
++2(x, nil) -> x
++2(.2(x, y), z) -> .2(x, ++2(y, z))
++2(x, ++2(y, z)) -> ++2(++2(x, y), z)
make1(x) -> .2(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.