Termination w.r.t. Q of the following Term Rewriting System could be disproven:

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 [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

G(f(s(x), s(y), z)) → G(f(x, y, z))
G(f(s(x), s(y), z)) → F(x, y, z)
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
      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

G(f(s(x), s(y), z)) → G(f(x, y, z))
G(f(s(x), s(y), z)) → F(x, y, z)
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.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ NonTerminationProof
          ↳ 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.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

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


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:



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 [y' / s(b), x' / s(a)]

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 [y / s(b), x' / s(a)]

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.





↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ NonTerminationProof

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.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

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


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:



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 [y' / s(b), x' / s(a)]

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 [y / s(b), x' / s(a)]

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.