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

0 QTRS
↳1 AAECC Innermost (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔)
        ↳9 QDP
        ↳10 QReductionProof (⇔)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔)
        ↳16 QDP
        ↳17 QReductionProof (⇔)
        ↳18 QDP
        ↳19 Rewriting (⇔)
        ↳20 QDP
        ↳21 Rewriting (⇔)
        ↳22 QDP
        ↳23 Rewriting (⇔)
        ↳24 QDP
        ↳25 Rewriting (⇔)
        ↳26 QDP
        ↳27 Rewriting (⇔)
        ↳28 QDP
        ↳29 Rewriting (⇔)
        ↳30 QDP
        ↳31 Rewriting (⇔)
        ↳32 QDP
        ↳33 Rewriting (⇔)
        ↳34 QDP
        ↳35 MNOCProof (⇔)
        ↳36 QDP
        ↳37 NonTerminationProof (⇔)
        ↳38 NO

(0) Obligation:

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

f(s(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
id(s(x)) → s(id(x))
id(0) → 0

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

id(s(x)) → s(id(x))
id(0) → 0

The TRS R 2 is

f(s(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)

The signature Sigma is {f}

(2) Obligation:

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

f(s(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1)
id(s(x0))
id(0)

(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:

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
F(s(s(s(s(s(s(s(s(x)))))))), y, y) → ID(s(s(s(s(s(s(s(s(x)))))))))
ID(s(x)) → ID(x)

The TRS R consists of the following rules:

f(s(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1)
id(s(x0))
id(0)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

(6) Complex Obligation (AND)

(7) Obligation:

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

ID(s(x)) → ID(x)

The TRS R consists of the following rules:

f(s(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1)
id(s(x0))
id(0)

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

(8) 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.

(9) Obligation:

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

ID(s(x)) → ID(x)

R is empty.
The set Q consists of the following terms:

f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1)
id(s(x0))
id(0)

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

(10) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1)
id(s(x0))
id(0)

(11) Obligation:

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

ID(s(x)) → ID(x)

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

(12) 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:

  • ID(s(x)) → ID(x)
    The graph contains the following edges 1 > 1

(13) YES

(14) Obligation:

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(id(s(s(s(s(s(s(s(s(x))))))))), y, y)

The TRS R consists of the following rules:

f(s(s(s(s(s(s(s(s(x)))))))), y, y) → f(id(s(s(s(s(s(s(s(s(x))))))))), y, y)
id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1)
id(s(x0))
id(0)

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:

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(id(s(s(s(s(s(s(s(s(x))))))))), y, y)

The TRS R consists of the following rules:

id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1)
id(s(x0))
id(0)

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

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

f(s(s(s(s(s(s(s(s(x0)))))))), x1, x1)

(18) Obligation:

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(id(s(s(s(s(s(s(s(s(x))))))))), y, y)

The TRS R consists of the following rules:

id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

id(s(x0))
id(0)

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

(19) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(id(s(s(s(s(s(s(s(s(x))))))))), y, y) at position [0] we obtained the following new rules [LPAR04]:

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(id(s(s(s(s(s(s(s(x))))))))), y, y)

(20) Obligation:

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(id(s(s(s(s(s(s(s(x))))))))), y, y)

The TRS R consists of the following rules:

id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

id(s(x0))
id(0)

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

(21) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(id(s(s(s(s(s(s(s(x))))))))), y, y) at position [0,0] we obtained the following new rules [LPAR04]:

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(id(s(s(s(s(s(s(x))))))))), y, y)

(22) Obligation:

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(id(s(s(s(s(s(s(x))))))))), y, y)

The TRS R consists of the following rules:

id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

id(s(x0))
id(0)

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

(23) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(id(s(s(s(s(s(s(x))))))))), y, y) at position [0,0,0] we obtained the following new rules [LPAR04]:

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(id(s(s(s(s(s(x))))))))), y, y)

(24) Obligation:

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(id(s(s(s(s(s(x))))))))), y, y)

The TRS R consists of the following rules:

id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

id(s(x0))
id(0)

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

(25) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(id(s(s(s(s(s(x))))))))), y, y) at position [0,0,0,0] we obtained the following new rules [LPAR04]:

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(id(s(s(s(s(x))))))))), y, y)

(26) Obligation:

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(id(s(s(s(s(x))))))))), y, y)

The TRS R consists of the following rules:

id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

id(s(x0))
id(0)

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

(27) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(id(s(s(s(s(x))))))))), y, y) at position [0,0,0,0,0] we obtained the following new rules [LPAR04]:

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(s(id(s(s(s(x))))))))), y, y)

(28) Obligation:

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(s(id(s(s(s(x))))))))), y, y)

The TRS R consists of the following rules:

id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

id(s(x0))
id(0)

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

(29) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(s(id(s(s(s(x))))))))), y, y) at position [0,0,0,0,0,0] we obtained the following new rules [LPAR04]:

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(s(s(id(s(s(x))))))))), y, y)

(30) Obligation:

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(s(s(id(s(s(x))))))))), y, y)

The TRS R consists of the following rules:

id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

id(s(x0))
id(0)

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

(31) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(s(s(id(s(s(x))))))))), y, y) at position [0,0,0,0,0,0,0] we obtained the following new rules [LPAR04]:

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(s(s(s(id(s(x))))))))), y, y)

(32) Obligation:

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(s(s(s(id(s(x))))))))), y, y)

The TRS R consists of the following rules:

id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

id(s(x0))
id(0)

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

(33) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(s(s(s(id(s(x))))))))), y, y) at position [0,0,0,0,0,0,0,0] we obtained the following new rules [LPAR04]:

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(s(s(s(s(id(x))))))))), y, y)

(34) Obligation:

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(s(s(s(s(id(x))))))))), y, y)

The TRS R consists of the following rules:

id(s(x)) → s(id(x))
id(0) → 0

The set Q consists of the following terms:

id(s(x0))
id(0)

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

(35) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.

(36) Obligation:

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

F(s(s(s(s(s(s(s(s(x)))))))), y, y) → F(s(s(s(s(s(s(s(s(id(x))))))))), y, y)

The TRS R consists of the following rules:

id(s(x)) → s(id(x))
id(0) → 0

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

(37) 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 = F(s(s(s(s(s(s(s(s(x)))))))), y, y) evaluates to t =F(s(s(s(s(s(s(s(s(id(x))))))))), y, y)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [x / id(x)]
  • Semiunifier: [ ]



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from F(s(s(s(s(s(s(s(s(x)))))))), y, y) to F(s(s(s(s(s(s(s(s(id(x))))))))), y, y).



(38) NO