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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳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 QDPSizeChangeProof (⇔)
        ↳20 YES
    ↳21 QDP
        ↳22 UsableRulesProof (⇔)
        ↳23 QDP
        ↳24 QReductionProof (⇔)
        ↳25 QDP
        ↳26 MNOCProof (⇔)
        ↳27 QDP
        ↳28 NonTerminationProof (⇔)
        ↳29 NO

(0) Obligation:

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

f(X) → cons(X, f(g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

f(X) → cons(X, f(g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)

The set Q consists of the following terms:

f(x0)
g(0)
g(s(x0))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))

(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(X) → F(g(X))
F(X) → G(X)
G(s(X)) → G(X)
SEL(s(X), cons(Y, Z)) → SEL(X, Z)

The TRS R consists of the following rules:

f(X) → cons(X, f(g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)

The set Q consists of the following terms:

f(x0)
g(0)
g(s(x0))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 1 less node.

(6) Complex Obligation (AND)

(7) Obligation:

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

SEL(s(X), cons(Y, Z)) → SEL(X, Z)

The TRS R consists of the following rules:

f(X) → cons(X, f(g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)

The set Q consists of the following terms:

f(x0)
g(0)
g(s(x0))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))

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:

SEL(s(X), cons(Y, Z)) → SEL(X, Z)

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

f(x0)
g(0)
g(s(x0))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))

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(x0)
g(0)
g(s(x0))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))

(11) Obligation:

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

SEL(s(X), cons(Y, Z)) → SEL(X, Z)

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:

  • SEL(s(X), cons(Y, Z)) → SEL(X, Z)
    The graph contains the following edges 1 > 1, 2 > 2

(13) YES

(14) Obligation:

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

G(s(X)) → G(X)

The TRS R consists of the following rules:

f(X) → cons(X, f(g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)

The set Q consists of the following terms:

f(x0)
g(0)
g(s(x0))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))

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:

G(s(X)) → G(X)

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

f(x0)
g(0)
g(s(x0))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))

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(x0)
g(0)
g(s(x0))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))

(18) Obligation:

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

G(s(X)) → G(X)

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

(19) 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:

  • G(s(X)) → G(X)
    The graph contains the following edges 1 > 1

(20) YES

(21) Obligation:

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

F(X) → F(g(X))

The TRS R consists of the following rules:

f(X) → cons(X, f(g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)

The set Q consists of the following terms:

f(x0)
g(0)
g(s(x0))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))

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

(22) 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.

(23) Obligation:

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

F(X) → F(g(X))

The TRS R consists of the following rules:

g(0) → s(0)
g(s(X)) → s(s(g(X)))

The set Q consists of the following terms:

f(x0)
g(0)
g(s(x0))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))

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

(24) 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(x0)
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))

(25) Obligation:

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

F(X) → F(g(X))

The TRS R consists of the following rules:

g(0) → s(0)
g(s(X)) → s(s(g(X)))

The set Q consists of the following terms:

g(0)
g(s(x0))

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

(26) MNOCProof (EQUIVALENT transformation)

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

(27) Obligation:

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

F(X) → F(g(X))

The TRS R consists of the following rules:

g(0) → s(0)
g(s(X)) → s(s(g(X)))

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

(28) 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(X) evaluates to t =F(g(X))

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



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from F(X) to F(g(X)).



(29) NO