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:
f1(s1(X)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, Y)))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(s1(X)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, Y)))
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:
f1(s1(X)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, Y)))
The set Q consists of the following terms:
f1(s1(x0))
g1(cons2(0, x0))
g1(cons2(s1(x0), x1))
h1(cons2(x0, x1))
Q DP problem:
The TRS P consists of the following rules:
H1(cons2(X, Y)) -> G1(cons2(X, Y))
G1(cons2(0, Y)) -> G1(Y)
H1(cons2(X, Y)) -> H1(g1(cons2(X, Y)))
F1(s1(X)) -> F1(X)
The TRS R consists of the following rules:
f1(s1(X)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, Y)))
The set Q consists of the following terms:
f1(s1(x0))
g1(cons2(0, x0))
g1(cons2(s1(x0), x1))
h1(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:
H1(cons2(X, Y)) -> G1(cons2(X, Y))
G1(cons2(0, Y)) -> G1(Y)
H1(cons2(X, Y)) -> H1(g1(cons2(X, Y)))
F1(s1(X)) -> F1(X)
The TRS R consists of the following rules:
f1(s1(X)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, Y)))
The set Q consists of the following terms:
f1(s1(x0))
g1(cons2(0, x0))
g1(cons2(s1(x0), x1))
h1(cons2(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 2 less nodes.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
G1(cons2(0, Y)) -> G1(Y)
The TRS R consists of the following rules:
f1(s1(X)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, Y)))
The set Q consists of the following terms:
f1(s1(x0))
g1(cons2(0, x0))
g1(cons2(s1(x0), x1))
h1(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.
G1(cons2(0, Y)) -> G1(Y)
Used argument filtering: G1(x1) = x1
cons2(x1, x2) = cons1(x2)
0 = 0
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f1(s1(X)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, Y)))
The set Q consists of the following terms:
f1(s1(x0))
g1(cons2(0, x0))
g1(cons2(s1(x0), x1))
h1(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
Q DP problem:
The TRS P consists of the following rules:
F1(s1(X)) -> F1(X)
The TRS R consists of the following rules:
f1(s1(X)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, Y)))
The set Q consists of the following terms:
f1(s1(x0))
g1(cons2(0, x0))
g1(cons2(s1(x0), x1))
h1(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.
F1(s1(X)) -> F1(X)
Used argument filtering: F1(x1) = x1
s1(x1) = s1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f1(s1(X)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, Y)))
The set Q consists of the following terms:
f1(s1(x0))
g1(cons2(0, x0))
g1(cons2(s1(x0), x1))
h1(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.