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, p(1, a(y, t))))
a(p(x, y), z) → p(a(x, z), a(y, z))
a(a(x, y), z) → a(x, a(y, z))
lambda(x) → x
a(x, y) → x
a(x, y) → y
p(x, y) → x
p(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, p(1, a(y, t))))
a(p(x, y), z) → p(a(x, z), a(y, z))
a(a(x, y), z) → a(x, a(y, z))
lambda(x) → x
a(x, y) → x
a(x, y) → y
p(x, y) → x
p(x, y) → y
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
A(a(x, y), z) → A(y, z)
A(p(x, y), z) → P(a(x, z), a(y, z))
A(p(x, y), z) → A(x, z)
A(lambda(x), y) → A(y, t)
A(lambda(x), y) → LAMBDA(a(x, p(1, a(y, t))))
A(a(x, y), z) → A(x, a(y, z))
A(p(x, y), z) → A(y, z)
A(lambda(x), y) → P(1, a(y, t))
A(lambda(x), y) → A(x, p(1, a(y, t)))
The TRS R consists of the following rules:
a(lambda(x), y) → lambda(a(x, p(1, a(y, t))))
a(p(x, y), z) → p(a(x, z), a(y, z))
a(a(x, y), z) → a(x, a(y, z))
lambda(x) → x
a(x, y) → x
a(x, y) → y
p(x, y) → x
p(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:
A(a(x, y), z) → A(y, z)
A(p(x, y), z) → P(a(x, z), a(y, z))
A(p(x, y), z) → A(x, z)
A(lambda(x), y) → A(y, t)
A(lambda(x), y) → LAMBDA(a(x, p(1, a(y, t))))
A(a(x, y), z) → A(x, a(y, z))
A(p(x, y), z) → A(y, z)
A(lambda(x), y) → P(1, a(y, t))
A(lambda(x), y) → A(x, p(1, a(y, t)))
The TRS R consists of the following rules:
a(lambda(x), y) → lambda(a(x, p(1, a(y, t))))
a(p(x, y), z) → p(a(x, z), a(y, z))
a(a(x, y), z) → a(x, a(y, z))
lambda(x) → x
a(x, y) → x
a(x, y) → y
p(x, y) → x
p(x, y) → y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
A(a(x, y), z) → A(y, z)
A(p(x, y), z) → A(x, z)
A(lambda(x), y) → A(y, t)
A(a(x, y), z) → A(x, a(y, z))
A(p(x, y), z) → A(y, z)
A(lambda(x), y) → A(x, p(1, a(y, t)))
The TRS R consists of the following rules:
a(lambda(x), y) → lambda(a(x, p(1, a(y, t))))
a(p(x, y), z) → p(a(x, z), a(y, z))
a(a(x, y), z) → a(x, a(y, z))
lambda(x) → x
a(x, y) → x
a(x, y) → y
p(x, y) → x
p(x, y) → y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.