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:
norm1(nil) -> 0
norm1(g2(x, y)) -> s1(norm1(x))
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
rem2(nil, y) -> nil
rem2(g2(x, y), 0) -> g2(x, y)
rem2(g2(x, y), s1(z)) -> rem2(x, z)
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
norm1(nil) -> 0
norm1(g2(x, y)) -> s1(norm1(x))
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
rem2(nil, y) -> nil
rem2(g2(x, y), 0) -> g2(x, y)
rem2(g2(x, y), s1(z)) -> rem2(x, 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:
norm1(nil) -> 0
norm1(g2(x, y)) -> s1(norm1(x))
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
rem2(nil, y) -> nil
rem2(g2(x, y), 0) -> g2(x, y)
rem2(g2(x, y), s1(z)) -> rem2(x, z)
The set Q consists of the following terms:
norm1(nil)
norm1(g2(x0, x1))
f2(x0, nil)
f2(x0, g2(x1, x2))
rem2(nil, x0)
rem2(g2(x0, x1), 0)
rem2(g2(x0, x1), s1(x2))
Q DP problem:
The TRS P consists of the following rules:
NORM1(g2(x, y)) -> NORM1(x)
F2(x, g2(y, z)) -> F2(x, y)
REM2(g2(x, y), s1(z)) -> REM2(x, z)
The TRS R consists of the following rules:
norm1(nil) -> 0
norm1(g2(x, y)) -> s1(norm1(x))
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
rem2(nil, y) -> nil
rem2(g2(x, y), 0) -> g2(x, y)
rem2(g2(x, y), s1(z)) -> rem2(x, z)
The set Q consists of the following terms:
norm1(nil)
norm1(g2(x0, x1))
f2(x0, nil)
f2(x0, g2(x1, x2))
rem2(nil, x0)
rem2(g2(x0, x1), 0)
rem2(g2(x0, x1), s1(x2))
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:
NORM1(g2(x, y)) -> NORM1(x)
F2(x, g2(y, z)) -> F2(x, y)
REM2(g2(x, y), s1(z)) -> REM2(x, z)
The TRS R consists of the following rules:
norm1(nil) -> 0
norm1(g2(x, y)) -> s1(norm1(x))
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
rem2(nil, y) -> nil
rem2(g2(x, y), 0) -> g2(x, y)
rem2(g2(x, y), s1(z)) -> rem2(x, z)
The set Q consists of the following terms:
norm1(nil)
norm1(g2(x0, x1))
f2(x0, nil)
f2(x0, g2(x1, x2))
rem2(nil, x0)
rem2(g2(x0, x1), 0)
rem2(g2(x0, x1), s1(x2))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
REM2(g2(x, y), s1(z)) -> REM2(x, z)
The TRS R consists of the following rules:
norm1(nil) -> 0
norm1(g2(x, y)) -> s1(norm1(x))
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
rem2(nil, y) -> nil
rem2(g2(x, y), 0) -> g2(x, y)
rem2(g2(x, y), s1(z)) -> rem2(x, z)
The set Q consists of the following terms:
norm1(nil)
norm1(g2(x0, x1))
f2(x0, nil)
f2(x0, g2(x1, x2))
rem2(nil, x0)
rem2(g2(x0, x1), 0)
rem2(g2(x0, x1), s1(x2))
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.
REM2(g2(x, y), s1(z)) -> REM2(x, z)
Used argument filtering: REM2(x1, x2) = x2
s1(x1) = s1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
norm1(nil) -> 0
norm1(g2(x, y)) -> s1(norm1(x))
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
rem2(nil, y) -> nil
rem2(g2(x, y), 0) -> g2(x, y)
rem2(g2(x, y), s1(z)) -> rem2(x, z)
The set Q consists of the following terms:
norm1(nil)
norm1(g2(x0, x1))
f2(x0, nil)
f2(x0, g2(x1, x2))
rem2(nil, x0)
rem2(g2(x0, x1), 0)
rem2(g2(x0, x1), s1(x2))
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
Q DP problem:
The TRS P consists of the following rules:
F2(x, g2(y, z)) -> F2(x, y)
The TRS R consists of the following rules:
norm1(nil) -> 0
norm1(g2(x, y)) -> s1(norm1(x))
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
rem2(nil, y) -> nil
rem2(g2(x, y), 0) -> g2(x, y)
rem2(g2(x, y), s1(z)) -> rem2(x, z)
The set Q consists of the following terms:
norm1(nil)
norm1(g2(x0, x1))
f2(x0, nil)
f2(x0, g2(x1, x2))
rem2(nil, x0)
rem2(g2(x0, x1), 0)
rem2(g2(x0, x1), s1(x2))
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.
F2(x, g2(y, z)) -> F2(x, y)
Used argument filtering: F2(x1, x2) = x2
g2(x1, x2) = g1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
norm1(nil) -> 0
norm1(g2(x, y)) -> s1(norm1(x))
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
rem2(nil, y) -> nil
rem2(g2(x, y), 0) -> g2(x, y)
rem2(g2(x, y), s1(z)) -> rem2(x, z)
The set Q consists of the following terms:
norm1(nil)
norm1(g2(x0, x1))
f2(x0, nil)
f2(x0, g2(x1, x2))
rem2(nil, x0)
rem2(g2(x0, x1), 0)
rem2(g2(x0, x1), s1(x2))
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
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
NORM1(g2(x, y)) -> NORM1(x)
The TRS R consists of the following rules:
norm1(nil) -> 0
norm1(g2(x, y)) -> s1(norm1(x))
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
rem2(nil, y) -> nil
rem2(g2(x, y), 0) -> g2(x, y)
rem2(g2(x, y), s1(z)) -> rem2(x, z)
The set Q consists of the following terms:
norm1(nil)
norm1(g2(x0, x1))
f2(x0, nil)
f2(x0, g2(x1, x2))
rem2(nil, x0)
rem2(g2(x0, x1), 0)
rem2(g2(x0, x1), s1(x2))
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.
NORM1(g2(x, y)) -> NORM1(x)
Used argument filtering: NORM1(x1) = x1
g2(x1, x2) = g1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
norm1(nil) -> 0
norm1(g2(x, y)) -> s1(norm1(x))
f2(x, nil) -> g2(nil, x)
f2(x, g2(y, z)) -> g2(f2(x, y), z)
rem2(nil, y) -> nil
rem2(g2(x, y), 0) -> g2(x, y)
rem2(g2(x, y), s1(z)) -> rem2(x, z)
The set Q consists of the following terms:
norm1(nil)
norm1(g2(x0, x1))
f2(x0, nil)
f2(x0, g2(x1, x2))
rem2(nil, x0)
rem2(g2(x0, x1), 0)
rem2(g2(x0, x1), s1(x2))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.