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))
a2(id, x) -> x
a2(1, id) -> 1
a2(t, id) -> t
a2(1, p2(x, y)) -> x
a2(t, 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))
a2(id, x) -> x
a2(1, id) -> 1
a2(t, id) -> t
a2(1, p2(x, y)) -> x
a2(t, p2(x, y)) -> y
Q is empty.
Using Dependency Pairs [1,13] we result in the following initial DP problem:
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))
a2(id, x) -> x
a2(1, id) -> 1
a2(t, id) -> t
a2(1, p2(x, y)) -> x
a2(t, p2(x, y)) -> y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ 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))
a2(id, x) -> x
a2(1, id) -> 1
a2(t, id) -> t
a2(1, p2(x, y)) -> x
a2(t, p2(x, y)) -> y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.