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) -> 1
f1(s1(x)) -> g2(x, s1(x))
g2(0, y) -> y
g2(s1(x), y) -> g2(x, +2(y, s1(x)))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
g2(s1(x), y) -> g2(x, s1(+2(y, x)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(0) -> 1
f1(s1(x)) -> g2(x, s1(x))
g2(0, y) -> y
g2(s1(x), y) -> g2(x, +2(y, s1(x)))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
g2(s1(x), y) -> g2(x, s1(+2(y, x)))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
+12(x, s1(y)) -> +12(x, y)
G2(s1(x), y) -> +12(y, s1(x))
G2(s1(x), y) -> G2(x, +2(y, s1(x)))
G2(s1(x), y) -> G2(x, s1(+2(y, x)))
F1(s1(x)) -> G2(x, s1(x))
G2(s1(x), y) -> +12(y, x)
The TRS R consists of the following rules:
f1(0) -> 1
f1(s1(x)) -> g2(x, s1(x))
g2(0, y) -> y
g2(s1(x), y) -> g2(x, +2(y, s1(x)))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
g2(s1(x), y) -> g2(x, s1(+2(y, x)))
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:
+12(x, s1(y)) -> +12(x, y)
G2(s1(x), y) -> +12(y, s1(x))
G2(s1(x), y) -> G2(x, +2(y, s1(x)))
G2(s1(x), y) -> G2(x, s1(+2(y, x)))
F1(s1(x)) -> G2(x, s1(x))
G2(s1(x), y) -> +12(y, x)
The TRS R consists of the following rules:
f1(0) -> 1
f1(s1(x)) -> g2(x, s1(x))
g2(0, y) -> y
g2(s1(x), y) -> g2(x, +2(y, s1(x)))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
g2(s1(x), y) -> g2(x, s1(+2(y, x)))
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:
+12(x, s1(y)) -> +12(x, y)
The TRS R consists of the following rules:
f1(0) -> 1
f1(s1(x)) -> g2(x, s1(x))
g2(0, y) -> y
g2(s1(x), y) -> g2(x, +2(y, s1(x)))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
g2(s1(x), y) -> g2(x, s1(+2(y, x)))
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, s1(y)) -> +12(x, y)
Used argument filtering: +12(x1, x2) = x2
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) -> 1
f1(s1(x)) -> g2(x, s1(x))
g2(0, y) -> y
g2(s1(x), y) -> g2(x, +2(y, s1(x)))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
g2(s1(x), y) -> g2(x, s1(+2(y, x)))
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(s1(x), y) -> G2(x, +2(y, s1(x)))
G2(s1(x), y) -> G2(x, s1(+2(y, x)))
The TRS R consists of the following rules:
f1(0) -> 1
f1(s1(x)) -> g2(x, s1(x))
g2(0, y) -> y
g2(s1(x), y) -> g2(x, +2(y, s1(x)))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
g2(s1(x), y) -> g2(x, s1(+2(y, x)))
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(s1(x), y) -> G2(x, +2(y, s1(x)))
G2(s1(x), y) -> G2(x, s1(+2(y, x)))
Used argument filtering: G2(x1, x2) = x1
s1(x1) = s1(x1)
+2(x1, x2) = +2(x1, x2)
0 = 0
Used ordering: Quasi Precedence:
+_2 > s_1
↳ 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) -> 1
f1(s1(x)) -> g2(x, s1(x))
g2(0, y) -> y
g2(s1(x), y) -> g2(x, +2(y, s1(x)))
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
g2(s1(x), y) -> g2(x, s1(+2(y, x)))
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.