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:
a(lambda(x), y) → lambda(a(x, 1))
a(lambda(x), y) → lambda(a(x, a(y, t)))
a(a(x, y), z) → a(x, a(y, z))
lambda(x) → x
a(x, y) → x
a(x, y) → y
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(lambda(x), y) → lambda(a(x, 1))
a(lambda(x), y) → lambda(a(x, a(y, t)))
a(a(x, y), z) → a(x, a(y, z))
lambda(x) → x
a(x, y) → x
a(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:
A(lambda(x), y) → A(y, t)
A(a(x, y), z) → A(x, a(y, z))
A(lambda(x), y) → A(x, a(y, t))
A(a(x, y), z) → A(y, z)
A(lambda(x), y) → A(x, 1)
A(lambda(x), y) → LAMBDA(a(x, a(y, t)))
A(lambda(x), y) → LAMBDA(a(x, 1))
The TRS R consists of the following rules:
a(lambda(x), y) → lambda(a(x, 1))
a(lambda(x), y) → lambda(a(x, a(y, t)))
a(a(x, y), z) → a(x, a(y, z))
lambda(x) → x
a(x, y) → x
a(x, y) → y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
Q DP problem:
The TRS P consists of the following rules:
A(lambda(x), y) → A(y, t)
A(a(x, y), z) → A(x, a(y, z))
A(lambda(x), y) → A(x, a(y, t))
A(a(x, y), z) → A(y, z)
A(lambda(x), y) → A(x, 1)
A(lambda(x), y) → LAMBDA(a(x, a(y, t)))
A(lambda(x), y) → LAMBDA(a(x, 1))
The TRS R consists of the following rules:
a(lambda(x), y) → lambda(a(x, 1))
a(lambda(x), y) → lambda(a(x, a(y, t)))
a(a(x, y), z) → a(x, a(y, z))
lambda(x) → x
a(x, y) → x
a(x, y) → y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
A(lambda(x), y) → A(y, t)
A(lambda(x), y) → A(x, a(y, t))
A(a(x, y), z) → A(x, a(y, z))
A(a(x, y), z) → A(y, z)
A(lambda(x), y) → A(x, 1)
A(lambda(x), y) → LAMBDA(a(x, a(y, t)))
A(lambda(x), y) → LAMBDA(a(x, 1))
The TRS R consists of the following rules:
a(lambda(x), y) → lambda(a(x, 1))
a(lambda(x), y) → lambda(a(x, a(y, t)))
a(a(x, y), z) → a(x, a(y, z))
lambda(x) → x
a(x, y) → x
a(x, y) → y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
A(lambda(x), y) → A(y, t)
A(a(x, y), z) → A(x, a(y, z))
A(lambda(x), y) → A(x, a(y, t))
A(a(x, y), z) → A(y, z)
A(lambda(x), y) → A(x, 1)
The TRS R consists of the following rules:
a(lambda(x), y) → lambda(a(x, 1))
a(lambda(x), y) → lambda(a(x, a(y, t)))
a(a(x, y), z) → a(x, a(y, z))
lambda(x) → x
a(x, y) → x
a(x, y) → y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.