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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 QDPOrderProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 QDP
↳7 NonTerminationProof (⇔)
↳8 FALSE

(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: Matrix interpretation [MATRO]:

POL(+1(x1, x2)) =
/0\
\0/
 +
/10\
\00/
·x1 +
/10\
\00/
·x2

POL(+(x1, x2)) =
/1\
\0/
 +
/10\
\00/
·x1 +
/10\
\00/
·x2

POL(*1(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/10\
\00/
·x2

POL(*(x1, x2)) =
/0\
\0/
 +
/00\
\11/
·x1 +
/10\
\00/
·x2

The following usable rules [FROCOS05] were oriented:

*(x, +(y, z)) → +(*(x, y), *(x, z))
+(+(x, *(y, z)), *(y, u)) → +(x, *(y, +(z, u)))
+(x, +(y, z)) → +(+(x, y), 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:
  • Semiunifier: [ ]
  • Matcher: [ ]



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) FALSE