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:

ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))

The set Q consists of the following terms:

ap(f, x0)
ap(ap(ap(g, x0), x1), ap(s, x2))

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

AP(ap(ap(g, x), y), ap(s, z)) → AP(ap(ap(g, x), y), ap(ap(x, y), 0))
AP(ap(ap(g, x), y), ap(s, z)) → AP(ap(x, y), 0)
AP(ap(ap(g, x), y), ap(s, z)) → AP(x, y)

The TRS R consists of the following rules:

ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))

The set Q consists of the following terms:

ap(f, x0)
ap(ap(ap(g, x0), x1), ap(s, x2))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Obligation:

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

AP(ap(ap(g, x), y), ap(s, z)) → AP(x, y)
AP(ap(ap(g, x), y), ap(s, z)) → AP(ap(ap(g, x), y), ap(ap(x, y), 0))

The TRS R consists of the following rules:

ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))

The set Q consists of the following terms:

ap(f, x0)
ap(ap(ap(g, x0), x1), ap(s, x2))

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

AP(ap(ap(g, x), y), ap(s, z)) → AP(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(AP(x₁, x₂)) = x₁   
POL(ap(x₁, x₂)) = 1 + x₁ + x₂   
POL(f) = 0   
POL(g) = 0   
POL(s) = 0   

The following usable rules [FROCOS05] were oriented: none

(8) Obligation:

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

AP(ap(ap(g, x), y), ap(s, z)) → AP(ap(ap(g, x), y), ap(ap(x, y), 0))

The TRS R consists of the following rules:

ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))

The set Q consists of the following terms:

ap(f, x0)
ap(ap(ap(g, x0), x1), ap(s, x2))

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

(9) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule AP(ap(ap(g, x), y), ap(s, z)) → AP(ap(ap(g, x), y), ap(ap(x, y), 0)) at position [1] we obtained the following new rules [LPAR04]:

AP(ap(ap(g, f), x0), ap(s, y2)) → AP(ap(ap(g, f), x0), ap(x0, 0))
AP(ap(ap(g, ap(ap(g, x0), x1)), ap(s, x2)), ap(s, y2)) → AP(ap(ap(g, ap(ap(g, x0), x1)), ap(s, x2)), ap(ap(ap(ap(g, x0), x1), ap(ap(x0, x1), 0)), 0))

(10) Obligation:

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

AP(ap(ap(g, f), x0), ap(s, y2)) → AP(ap(ap(g, f), x0), ap(x0, 0))
AP(ap(ap(g, ap(ap(g, x0), x1)), ap(s, x2)), ap(s, y2)) → AP(ap(ap(g, ap(ap(g, x0), x1)), ap(s, x2)), ap(ap(ap(ap(g, x0), x1), ap(ap(x0, x1), 0)), 0))

The TRS R consists of the following rules:

ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))

The set Q consists of the following terms:

ap(f, x0)
ap(ap(ap(g, x0), x1), ap(s, x2))

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

(11) DependencyGraphProof (EQUIVALENT transformation)

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

(12) Complex Obligation (AND)

(13) Obligation:

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

AP(ap(ap(g, ap(ap(g, x0), x1)), ap(s, x2)), ap(s, y2)) → AP(ap(ap(g, ap(ap(g, x0), x1)), ap(s, x2)), ap(ap(ap(ap(g, x0), x1), ap(ap(x0, x1), 0)), 0))

The TRS R consists of the following rules:

ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))

The set Q consists of the following terms:

ap(f, x0)
ap(ap(ap(g, x0), x1), ap(s, x2))

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

(14) Obligation:

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

AP(ap(ap(g, f), x0), ap(s, y2)) → AP(ap(ap(g, f), x0), ap(x0, 0))

The TRS R consists of the following rules:

ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))

The set Q consists of the following terms:

ap(f, x0)
ap(ap(ap(g, x0), x1), ap(s, x2))

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

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

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

AP(ap(ap(g, f), x0), ap(s, y2)) → AP(ap(ap(g, f), x0), ap(x0, 0))

The TRS R consists of the following rules:

ap(f, x) → x

The set Q consists of the following terms:

ap(f, x0)
ap(ap(ap(g, x0), x1), ap(s, x2))

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

(17) 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 = AP(ap(ap(g, f), x0), ap(s, y2)) evaluates to t =AP(ap(ap(g, f), x0), ap(x0, 0))

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

• Semiunifier: [y2 / 0, x0 / s]
• Matcher: [ ]

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Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from AP(ap(ap(g, f), s), ap(s, 0)) to AP(ap(ap(g, f), s), ap(s, 0)).

(18) FALSE