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 UsableRulesProof (⇔)
        ↳7 QDP
        ↳8 QDPSizeChangeProof (⇔)
        ↳9 YES
    ↳10 QDP
        ↳11 NonTerminationProof (⇔)
        ↳12 NO

(0) Obligation:

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

f(0, 1, g(x, y), z) → f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) → 0
g(0, 1) → 1
h(g(x, y)) → h(x)

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(0, 1, g(x, y), z) → F(g(x, y), g(x, y), g(x, y), h(x))
F(0, 1, g(x, y), z) → H(x)
H(g(x, y)) → H(x)

The TRS R consists of the following rules:

f(0, 1, g(x, y), z) → f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) → 0
g(0, 1) → 1
h(g(x, y)) → h(x)

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:

H(g(x, y)) → H(x)

The TRS R consists of the following rules:

f(0, 1, g(x, y), z) → f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) → 0
g(0, 1) → 1
h(g(x, y)) → h(x)

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

(6) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(7) Obligation:

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

H(g(x, y)) → H(x)

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

(8) 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:

  • H(g(x, y)) → H(x)
    The graph contains the following edges 1 > 1

(9) YES

(10) Obligation:

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

F(0, 1, g(x, y), z) → F(g(x, y), g(x, y), g(x, y), h(x))

The TRS R consists of the following rules:

f(0, 1, g(x, y), z) → f(g(x, y), g(x, y), g(x, y), h(x))
g(0, 1) → 0
g(0, 1) → 1
h(g(x, y)) → h(x)

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

(11) NonTerminationProof (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(g(0, 1), g(0, 1), g(x, y), z) evaluates to t =F(g(x, y), g(x, y), g(x, y), h(x))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [ ]
  • Semiunifier: [x / 0, y / 1, z / h(0)]



Rewriting sequence

F(g(0, 1), g(0, 1), g(0, 1), h(0)) → F(g(0, 1), 1, g(0, 1), h(0))
with rule g(0, 1) → 1 at position [1] and matcher [ ]

F(g(0, 1), 1, g(0, 1), h(0)) → F(0, 1, g(0, 1), h(0))
with rule g(0, 1) → 0 at position [0] and matcher [ ]

F(0, 1, g(0, 1), h(0)) → F(g(0, 1), g(0, 1), g(0, 1), h(0))
with rule F(0, 1, g(x, y), z) → F(g(x, y), g(x, y), g(x, y), h(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.



(12) NO