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

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

The TRS R consists of the following rules:

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

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

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( *¹(x₁, x₂) ) = max{0, 2x₂ - 2}

POL( +¹(x₁, x₂) ) = max{0, 2x₁ + 2x₂ - 2}

POL( +(x₁, x₂) ) = x₁ + x₂ + 2

POL( *(x₁, x₂) ) = x₂

The following usable rules [FROCOS05] were oriented:

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

(4) Obligation:

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

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

The TRS R consists of the following rules:

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

Q is empty.
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:

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

The TRS R consists of the following rules:

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

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

(7) 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 = +¹(x', *(x, +(y', z'))) evaluates to t =+¹(x', *(x, +(y', z')))

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

• Matcher: [ ]
• Semiunifier: [ ]

---

Rewriting sequence

+¹(x', *(x, +(y', z'))) → +¹(x', +(*(x, y'), *(x, z')))
with rule *(x'', +(y'', z'')) → +(*(x'', y''), *(x'', z'')) at position [1] and matcher [x'' / x, y'' / y', z'' / z']

+¹(x', +(*(x, y'), *(x, z'))) → +¹(+(x', *(x, y')), *(x, z'))
with rule +¹(x'', +(y'', z'')) → +¹(+(x'', y''), z'') at position [] and matcher [x'' / x', y'' / *(x, y'), z'' / *(x, z')]

+¹(+(x', *(x, y')), *(x, z')) → +¹(x', *(x, +(y', z')))
with rule +¹(+(x, *(y, z)), *(y, u)) → +¹(x, *(y, +(z, u)))

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.

(8) NO