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:
f(s(a), s(b), x) → f(x, x, x)
g(f(s(x), s(y), z)) → g(f(x, y, z))
cons(x, y) → x
cons(x, y) → y
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(s(a), s(b), x) → f(x, x, x)
g(f(s(x), s(y), z)) → g(f(x, y, z))
cons(x, y) → x
cons(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:
F(s(a), s(b), x) → F(x, x, x)
G(f(s(x), s(y), z)) → F(x, y, z)
G(f(s(x), s(y), z)) → G(f(x, y, z))
The TRS R consists of the following rules:
f(s(a), s(b), x) → f(x, x, x)
g(f(s(x), s(y), z)) → g(f(x, y, z))
cons(x, y) → x
cons(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:
F(s(a), s(b), x) → F(x, x, x)
G(f(s(x), s(y), z)) → F(x, y, z)
G(f(s(x), s(y), z)) → G(f(x, y, z))
The TRS R consists of the following rules:
f(s(a), s(b), x) → f(x, x, x)
g(f(s(x), s(y), z)) → g(f(x, y, z))
cons(x, y) → x
cons(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:
F(s(a), s(b), x) → F(x, x, x)
G(f(s(x), s(y), z)) → F(x, y, z)
G(f(s(x), s(y), z)) → G(f(x, y, z))
The TRS R consists of the following rules:
f(s(a), s(b), x) → f(x, x, x)
g(f(s(x), s(y), z)) → g(f(x, y, z))
cons(x, y) → x
cons(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 2 SCCs with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F(s(a), s(b), x) → F(x, x, x)
The TRS R consists of the following rules:
f(s(a), s(b), x) → f(x, x, x)
g(f(s(x), s(y), z)) → g(f(x, y, z))
cons(x, y) → x
cons(x, y) → y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
G(f(s(x), s(y), z)) → G(f(x, y, z))
The TRS R consists of the following rules:
f(s(a), s(b), x) → f(x, x, x)
g(f(s(x), s(y), z)) → g(f(x, y, z))
cons(x, y) → x
cons(x, y) → y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.