Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
a2(lambda1(x), y) -> lambda1(a2(x, p2(1, a2(y, t))))
a2(p2(x, y), z) -> p2(a2(x, z), a2(y, z))
a2(a2(x, y), z) -> a2(x, a2(y, z))
lambda1(x) -> x
a2(x, y) -> x
a2(x, y) -> y
p2(x, y) -> x
p2(x, y) -> y
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a2(lambda1(x), y) -> lambda1(a2(x, p2(1, a2(y, t))))
a2(p2(x, y), z) -> p2(a2(x, z), a2(y, z))
a2(a2(x, y), z) -> a2(x, a2(y, z))
lambda1(x) -> x
a2(x, y) -> x
a2(x, y) -> y
p2(x, y) -> x
p2(x, y) -> y
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
A2(p2(x, y), z) -> A2(y, z)
A2(p2(x, y), z) -> A2(x, z)
A2(lambda1(x), y) -> A2(x, p2(1, a2(y, t)))
A2(lambda1(x), y) -> A2(y, t)
A2(lambda1(x), y) -> LAMBDA1(a2(x, p2(1, a2(y, t))))
A2(a2(x, y), z) -> A2(x, a2(y, z))
A2(a2(x, y), z) -> A2(y, z)
A2(p2(x, y), z) -> P2(a2(x, z), a2(y, z))
A2(lambda1(x), y) -> P2(1, a2(y, t))
The TRS R consists of the following rules:
a2(lambda1(x), y) -> lambda1(a2(x, p2(1, a2(y, t))))
a2(p2(x, y), z) -> p2(a2(x, z), a2(y, z))
a2(a2(x, y), z) -> a2(x, a2(y, z))
lambda1(x) -> x
a2(x, y) -> x
a2(x, y) -> y
p2(x, y) -> x
p2(x, y) -> 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:
A2(p2(x, y), z) -> A2(y, z)
A2(p2(x, y), z) -> A2(x, z)
A2(lambda1(x), y) -> A2(x, p2(1, a2(y, t)))
A2(lambda1(x), y) -> A2(y, t)
A2(lambda1(x), y) -> LAMBDA1(a2(x, p2(1, a2(y, t))))
A2(a2(x, y), z) -> A2(x, a2(y, z))
A2(a2(x, y), z) -> A2(y, z)
A2(p2(x, y), z) -> P2(a2(x, z), a2(y, z))
A2(lambda1(x), y) -> P2(1, a2(y, t))
The TRS R consists of the following rules:
a2(lambda1(x), y) -> lambda1(a2(x, p2(1, a2(y, t))))
a2(p2(x, y), z) -> p2(a2(x, z), a2(y, z))
a2(a2(x, y), z) -> a2(x, a2(y, z))
lambda1(x) -> x
a2(x, y) -> x
a2(x, y) -> y
p2(x, y) -> x
p2(x, y) -> y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
A2(p2(x, y), z) -> A2(y, z)
A2(p2(x, y), z) -> A2(x, z)
A2(lambda1(x), y) -> A2(x, p2(1, a2(y, t)))
A2(lambda1(x), y) -> A2(y, t)
A2(a2(x, y), z) -> A2(x, a2(y, z))
A2(a2(x, y), z) -> A2(y, z)
The TRS R consists of the following rules:
a2(lambda1(x), y) -> lambda1(a2(x, p2(1, a2(y, t))))
a2(p2(x, y), z) -> p2(a2(x, z), a2(y, z))
a2(a2(x, y), z) -> a2(x, a2(y, z))
lambda1(x) -> x
a2(x, y) -> x
a2(x, y) -> y
p2(x, y) -> x
p2(x, y) -> y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.