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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
    ↳5 QDP
        ↳6 Instantiation (⇔)
        ↳7 QDP
        ↳8 Instantiation (⇔)
        ↳9 QDP
        ↳10 ForwardInstantiation (⇔)
        ↳11 QDP
        ↳12 NonTerminationProof (⇔)
        ↳13 NO
    ↳14 QDP
        ↳15 QDPSizeChangeProof (⇔)
        ↳16 YES

(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: [ ]



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) NO

(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) YES