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:

*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, f(z))) → *(g(x, z), +(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:

*¹(*(x, y), z) → *¹(x, *(y, z))
*¹(*(x, y), z) → *¹(y, z)
*¹(+(x, y), z) → *¹(x, z)
*¹(+(x, y), z) → *¹(y, z)
*¹(x, +(y, f(z))) → *¹(g(x, z), +(y, y))

The TRS R consists of the following rules:

*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, f(z))) → *(g(x, z), +(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.

(4) Complex Obligation (AND)

(5) Obligation:

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

*¹(x, +(y, f(z))) → *¹(g(x, z), +(y, y))

The TRS R consists of the following rules:

*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, f(z))) → *(g(x, z), +(y, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(6) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule *¹(x, +(y, f(z))) → *¹(g(x, z), +(y, y)) we obtained the following new rules [LPAR04]:

*¹(g(y_0, y_1), +(y_2, f(x2))) → *¹(g(g(y_0, y_1), x2), +(y_2, y_2))

(7) Obligation:

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

*¹(g(y_0, y_1), +(y_2, f(x2))) → *¹(g(g(y_0, y_1), x2), +(y_2, y_2))

The TRS R consists of the following rules:

*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, f(z))) → *(g(x, z), +(y, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(8) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule *¹(g(y_0, y_1), +(y_2, f(x2))) → *¹(g(g(y_0, y_1), x2), +(y_2, y_2)) we obtained the following new rules [LPAR04]:

*¹(g(g(y_0, y_1), y_2), +(y_3, f(x3))) → *¹(g(g(g(y_0, y_1), y_2), x3), +(y_3, y_3))

(9) Obligation:

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

*¹(g(g(y_0, y_1), y_2), +(y_3, f(x3))) → *¹(g(g(g(y_0, y_1), y_2), x3), +(y_3, y_3))

The TRS R consists of the following rules:

*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, f(z))) → *(g(x, z), +(y, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(10) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule *¹(g(g(y_0, y_1), y_2), +(y_3, f(x3))) → *¹(g(g(g(y_0, y_1), y_2), x3), +(y_3, y_3)) we obtained the following new rules [LPAR04]:

*¹(g(g(x0, x1), x2), +(f(y_4), f(x4))) → *¹(g(g(g(x0, x1), x2), x4), +(f(y_4), f(y_4)))

(11) Obligation:

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

*¹(g(g(x0, x1), x2), +(f(y_4), f(x4))) → *¹(g(g(g(x0, x1), x2), x4), +(f(y_4), f(y_4)))

The TRS R consists of the following rules:

*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, f(z))) → *(g(x, z), +(y, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(12) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = *¹(g(g(x0, x1), x2), +(f(y_4), f(x4))) evaluates to t =*¹(g(g(g(x0, x1), x2), x4), +(f(y_4), f(y_4)))

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

• Matcher: [x0 / g(x0, x1), x1 / x2, x2 / x4, x4 / y_4]
• Semiunifier: [ ]

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Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from *^1(g(g(x0, x1), x2), +(f(y_4), f(x4))) to *^1(g(g(g(x0, x1), x2), x4), +(f(y_4), f(y_4))).

(13) FALSE

(14) Obligation:

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

*¹(*(x, y), z) → *¹(y, z)
*¹(*(x, y), z) → *¹(x, *(y, z))
*¹(+(x, y), z) → *¹(x, z)
*¹(+(x, y), z) → *¹(y, z)

The TRS R consists of the following rules:

*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, f(z))) → *(g(x, z), +(y, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(15) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• *¹(*(x, y), z) → *¹(y, z)
  The graph contains the following edges 1 > 1, 2 >= 2

• *¹(*(x, y), z) → *¹(x, *(y, z))
  The graph contains the following edges 1 > 1

• *¹(+(x, y), z) → *¹(x, z)
  The graph contains the following edges 1 > 1, 2 >= 2

• *¹(+(x, y), z) → *¹(y, z)
  The graph contains the following edges 1 > 1, 2 >= 2

(16) TRUE