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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
↳5 UsableRulesProof (⇔)
↳6 QDP
↳7 Narrowing (⇔)
↳8 QDP
↳9 DependencyGraphProof (⇔)
↳10 QDP
↳11 UsableRulesProof (⇔)
↳12 QDP
↳13 NonTerminationProof (⇔)
↳14 NO

(0) Obligation:

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

*(X, +(Y, 1)) → +(*(X, +(Y, *(1, 0))), X)
*(X, 1) → X
*(X, 0) → X
*(X, 0) → 0

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, 1)) → *1(X, +(Y, *(1, 0)))
*1(X, +(Y, 1)) → *1(1, 0)

The TRS R consists of the following rules:

*(X, +(Y, 1)) → +(*(X, +(Y, *(1, 0))), X)
*(X, 1) → X
*(X, 0) → X
*(X, 0) → 0

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 1 SCC with 1 less node.

(4) Obligation:

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

*1(X, +(Y, 1)) → *1(X, +(Y, *(1, 0)))

The TRS R consists of the following rules:

*(X, +(Y, 1)) → +(*(X, +(Y, *(1, 0))), X)
*(X, 1) → X
*(X, 0) → X
*(X, 0) → 0

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

(5) 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.

(6) Obligation:

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

*1(X, +(Y, 1)) → *1(X, +(Y, *(1, 0)))

The TRS R consists of the following rules:

*(X, 0) → X
*(X, 0) → 0

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

(7) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule *1(X, +(Y, 1)) → *1(X, +(Y, *(1, 0))) at position [1,1] we obtained the following new rules [LPAR04]:

*1(y0, +(y1, 1)) → *1(y0, +(y1, 1))
*1(y0, +(y1, 1)) → *1(y0, +(y1, 0))

(8) Obligation:

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

*1(y0, +(y1, 1)) → *1(y0, +(y1, 1))
*1(y0, +(y1, 1)) → *1(y0, +(y1, 0))

The TRS R consists of the following rules:

*(X, 0) → X
*(X, 0) → 0

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

(9) DependencyGraphProof (EQUIVALENT transformation)

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

(10) Obligation:

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

*1(y0, +(y1, 1)) → *1(y0, +(y1, 1))

The TRS R consists of the following rules:

*(X, 0) → X
*(X, 0) → 0

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

(11) 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.

(12) Obligation:

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

*1(y0, +(y1, 1)) → *1(y0, +(y1, 1))

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

(13) 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(y0, +(y1, 1)) evaluates to t =*1(y0, +(y1, 1))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [ ]
  • Semiunifier: [ ]



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from *^1(y0, +(y1, 1)) to *^1(y0, +(y1, 1)).



(14) NO