Termination w.r.t. Q of the given QTRS could be disproven:

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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)) → G(f(x, y, z))
G(f(s(x), s(y), z)) → 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.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

(4) Complex Obligation (AND)

(5) Obligation:

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.

(6) NonTerminationLoopProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = F(cons(s(a), y), cons(x', s(b)), x) evaluates to t =F(x, x, x)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

• Matcher: [ ]
• Semiunifier: [y / s(b), x' / s(a), x / cons(s(a), s(b))]

---

Rewriting sequence

F(cons(s(a), s(b)), cons(s(a), s(b)), cons(s(a), s(b))) → F(cons(s(a), s(b)), s(b), cons(s(a), s(b)))
with rule cons(x', y') → y' at position [1] and matcher [x' / s(a), y' / s(b)]

F(cons(s(a), s(b)), s(b), cons(s(a), s(b))) → F(s(a), s(b), cons(s(a), s(b)))
with rule cons(x', y) → x' at position [0] and matcher [x' / s(a), y / s(b)]

F(s(a), s(b), cons(s(a), s(b))) → F(cons(s(a), s(b)), cons(s(a), s(b)), cons(s(a), s(b)))
with rule F(s(a), s(b), x) → F(x, x, x)

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.

(7) NO

(8) Obligation:

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.

(9) NonTerminationLoopProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = G(f(cons(s(a), y), cons(x', s(b)), s(x))) evaluates to t =G(f(x, x, s(x)))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

• Matcher: [ ]
• Semiunifier: [y / s(b), x' / s(a), x / cons(s(a), s(b))]

---

Rewriting sequence

G(f(cons(s(a), s(b)), cons(s(a), s(b)), s(cons(s(a), s(b))))) → G(f(cons(s(a), s(b)), s(b), s(cons(s(a), s(b)))))
with rule cons(x', y') → y' at position [0,1] and matcher [x' / s(a), y' / s(b)]

G(f(cons(s(a), s(b)), s(b), s(cons(s(a), s(b))))) → G(f(s(a), s(b), s(cons(s(a), s(b)))))
with rule cons(x', y) → x' at position [0,0] and matcher [x' / s(a), y / s(b)]

G(f(s(a), s(b), s(cons(s(a), s(b))))) → G(f(s(cons(s(a), s(b))), s(cons(s(a), s(b))), s(cons(s(a), s(b)))))
with rule f(s(a), s(b), x') → f(x', x', x') at position [0] and matcher [x' / s(cons(s(a), s(b)))]

G(f(s(cons(s(a), s(b))), s(cons(s(a), s(b))), s(cons(s(a), s(b))))) → G(f(cons(s(a), s(b)), cons(s(a), s(b)), s(cons(s(a), s(b)))))
with rule G(f(s(x), s(y), z)) → G(f(x, y, z))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.

(10) NO