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:
+2(a, b) -> +2(b, a)
+2(a, +2(b, z)) -> +2(b, +2(a, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
f2(a, y) -> a
f2(b, y) -> b
f2(+2(x, y), z) -> +2(f2(x, z), f2(y, z))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
+2(a, b) -> +2(b, a)
+2(a, +2(b, z)) -> +2(b, +2(a, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
f2(a, y) -> a
f2(b, y) -> b
f2(+2(x, y), z) -> +2(f2(x, z), f2(y, z))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
+12(a, b) -> +12(b, a)
F2(+2(x, y), z) -> F2(y, z)
+12(+2(x, y), z) -> +12(y, z)
+12(a, +2(b, z)) -> +12(a, z)
+12(a, +2(b, z)) -> +12(b, +2(a, z))
F2(+2(x, y), z) -> F2(x, z)
F2(+2(x, y), z) -> +12(f2(x, z), f2(y, z))
+12(+2(x, y), z) -> +12(x, +2(y, z))
The TRS R consists of the following rules:
+2(a, b) -> +2(b, a)
+2(a, +2(b, z)) -> +2(b, +2(a, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
f2(a, y) -> a
f2(b, y) -> b
f2(+2(x, y), z) -> +2(f2(x, z), f2(y, z))
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(a, b) -> +12(b, a)
F2(+2(x, y), z) -> F2(y, z)
+12(+2(x, y), z) -> +12(y, z)
+12(a, +2(b, z)) -> +12(a, z)
+12(a, +2(b, z)) -> +12(b, +2(a, z))
F2(+2(x, y), z) -> F2(x, z)
F2(+2(x, y), z) -> +12(f2(x, z), f2(y, z))
+12(+2(x, y), z) -> +12(x, +2(y, z))
The TRS R consists of the following rules:
+2(a, b) -> +2(b, a)
+2(a, +2(b, z)) -> +2(b, +2(a, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
f2(a, y) -> a
f2(b, y) -> b
f2(+2(x, y), z) -> +2(f2(x, z), f2(y, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
+12(a, +2(b, z)) -> +12(a, z)
The TRS R consists of the following rules:
+2(a, b) -> +2(b, a)
+2(a, +2(b, z)) -> +2(b, +2(a, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
f2(a, y) -> a
f2(b, y) -> b
f2(+2(x, y), z) -> +2(f2(x, z), f2(y, z))
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(a, +2(b, z)) -> +12(a, z)
Used argument filtering: +12(x1, x2) = x2
+2(x1, x2) = +1(x2)
b = b
Used ordering: Quasi Precedence:
trivial
↳ 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:
+2(a, b) -> +2(b, a)
+2(a, +2(b, z)) -> +2(b, +2(a, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
f2(a, y) -> a
f2(b, y) -> b
f2(+2(x, y), z) -> +2(f2(x, z), f2(y, z))
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
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
+12(+2(x, y), z) -> +12(y, z)
+12(+2(x, y), z) -> +12(x, +2(y, z))
The TRS R consists of the following rules:
+2(a, b) -> +2(b, a)
+2(a, +2(b, z)) -> +2(b, +2(a, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
f2(a, y) -> a
f2(b, y) -> b
f2(+2(x, y), z) -> +2(f2(x, z), f2(y, z))
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)
+12(+2(x, y), z) -> +12(x, +2(y, z))
Used argument filtering: +12(x1, x2) = x1
+2(x1, x2) = +2(x1, x2)
Used ordering: Quasi Precedence:
trivial
↳ 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:
+2(a, b) -> +2(b, a)
+2(a, +2(b, z)) -> +2(b, +2(a, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
f2(a, y) -> a
f2(b, y) -> b
f2(+2(x, y), z) -> +2(f2(x, z), f2(y, z))
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
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
F2(+2(x, y), z) -> F2(y, z)
F2(+2(x, y), z) -> F2(x, z)
The TRS R consists of the following rules:
+2(a, b) -> +2(b, a)
+2(a, +2(b, z)) -> +2(b, +2(a, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
f2(a, y) -> a
f2(b, y) -> b
f2(+2(x, y), z) -> +2(f2(x, z), f2(y, z))
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.
F2(+2(x, y), z) -> F2(y, z)
F2(+2(x, y), z) -> F2(x, z)
Used argument filtering: F2(x1, x2) = x1
+2(x1, x2) = +2(x1, x2)
Used ordering: Quasi Precedence:
trivial
↳ 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:
+2(a, b) -> +2(b, a)
+2(a, +2(b, z)) -> +2(b, +2(a, z))
+2(+2(x, y), z) -> +2(x, +2(y, z))
f2(a, y) -> a
f2(b, y) -> b
f2(+2(x, y), z) -> +2(f2(x, z), f2(y, z))
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.