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(0) -> true
f1(1) -> false
f1(s1(x)) -> f1(x)
if3(true, s1(x), s1(y)) -> s1(x)
if3(false, s1(x), s1(y)) -> s1(y)
g2(x, c1(y)) -> c1(g2(x, y))
g2(x, c1(y)) -> g2(x, if3(f1(x), c1(g2(s1(x), y)), c1(y)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(0) -> true
f1(1) -> false
f1(s1(x)) -> f1(x)
if3(true, s1(x), s1(y)) -> s1(x)
if3(false, s1(x), s1(y)) -> s1(y)
g2(x, c1(y)) -> c1(g2(x, y))
g2(x, c1(y)) -> g2(x, if3(f1(x), c1(g2(s1(x), y)), c1(y)))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
G2(x, c1(y)) -> G2(x, if3(f1(x), c1(g2(s1(x), y)), c1(y)))
G2(x, c1(y)) -> G2(x, y)
G2(x, c1(y)) -> IF3(f1(x), c1(g2(s1(x), y)), c1(y))
G2(x, c1(y)) -> G2(s1(x), y)
G2(x, c1(y)) -> F1(x)
F1(s1(x)) -> F1(x)
The TRS R consists of the following rules:
f1(0) -> true
f1(1) -> false
f1(s1(x)) -> f1(x)
if3(true, s1(x), s1(y)) -> s1(x)
if3(false, s1(x), s1(y)) -> s1(y)
g2(x, c1(y)) -> c1(g2(x, y))
g2(x, c1(y)) -> g2(x, if3(f1(x), c1(g2(s1(x), y)), c1(y)))
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:
G2(x, c1(y)) -> G2(x, if3(f1(x), c1(g2(s1(x), y)), c1(y)))
G2(x, c1(y)) -> G2(x, y)
G2(x, c1(y)) -> IF3(f1(x), c1(g2(s1(x), y)), c1(y))
G2(x, c1(y)) -> G2(s1(x), y)
G2(x, c1(y)) -> F1(x)
F1(s1(x)) -> F1(x)
The TRS R consists of the following rules:
f1(0) -> true
f1(1) -> false
f1(s1(x)) -> f1(x)
if3(true, s1(x), s1(y)) -> s1(x)
if3(false, s1(x), s1(y)) -> s1(y)
g2(x, c1(y)) -> c1(g2(x, y))
g2(x, c1(y)) -> g2(x, if3(f1(x), c1(g2(s1(x), y)), c1(y)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
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(0) -> true
f1(1) -> false
f1(s1(x)) -> f1(x)
if3(true, s1(x), s1(y)) -> s1(x)
if3(false, s1(x), s1(y)) -> s1(y)
g2(x, c1(y)) -> c1(g2(x, y))
g2(x, c1(y)) -> g2(x, if3(f1(x), c1(g2(s1(x), y)), c1(y)))
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.
F1(s1(x)) -> F1(x)
Used argument filtering: F1(x1) = x1
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ 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(0) -> true
f1(1) -> false
f1(s1(x)) -> f1(x)
if3(true, s1(x), s1(y)) -> s1(x)
if3(false, s1(x), s1(y)) -> s1(y)
g2(x, c1(y)) -> c1(g2(x, y))
g2(x, c1(y)) -> g2(x, if3(f1(x), c1(g2(s1(x), y)), c1(y)))
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:
G2(x, c1(y)) -> G2(x, y)
G2(x, c1(y)) -> G2(s1(x), y)
The TRS R consists of the following rules:
f1(0) -> true
f1(1) -> false
f1(s1(x)) -> f1(x)
if3(true, s1(x), s1(y)) -> s1(x)
if3(false, s1(x), s1(y)) -> s1(y)
g2(x, c1(y)) -> c1(g2(x, y))
g2(x, c1(y)) -> g2(x, if3(f1(x), c1(g2(s1(x), y)), c1(y)))
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.
G2(x, c1(y)) -> G2(x, y)
G2(x, c1(y)) -> G2(s1(x), y)
Used argument filtering: G2(x1, x2) = x2
c1(x1) = c1(x1)
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:
f1(0) -> true
f1(1) -> false
f1(s1(x)) -> f1(x)
if3(true, s1(x), s1(y)) -> s1(x)
if3(false, s1(x), s1(y)) -> s1(y)
g2(x, c1(y)) -> c1(g2(x, y))
g2(x, c1(y)) -> g2(x, if3(f1(x), c1(g2(s1(x), y)), c1(y)))
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.