NO Termination w.r.t. Q proof of EEG_IJCAR_12_emmes-nonloop-ex6_1.ari

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

0 QTRS
↳1 AAECC Innermost (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 12 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QReductionProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔, 0 ms)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QReductionProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 QDPSizeChangeProof (⇔, 0 ms)
        ↳20 YES
    ↳21 QDP
        ↳22 MNOCProof (⇔, 0 ms)
        ↳23 QDP
        ↳24 NonLoopProof (⇒, 0 ms)
        ↳25 NO

(0) Obligation:

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

f(true, xs) → f(isList(xs), append(cons(a, nil), xs))
isList(nil) → true
isList(cons(x, xs)) → isList(xs)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

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

isList(nil) → true
isList(cons(x, xs)) → isList(xs)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))

The TRS R 2 is

f(true, xs) → f(isList(xs), append(cons(a, nil), xs))

The signature Sigma is {f}

(2) Obligation:

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

f(true, xs) → f(isList(xs), append(cons(a, nil), xs))
isList(nil) → true
isList(cons(x, xs)) → isList(xs)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))

The set Q consists of the following terms:

f(true, x0)
isList(nil)
isList(cons(x0, x1))
append(nil, x0)
append(cons(x0, 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(true, xs) → F(isList(xs), append(cons(a, nil), xs))
F(true, xs) → ISLIST(xs)
F(true, xs) → APPEND(cons(a, nil), xs)
ISLIST(cons(x, xs)) → ISLIST(xs)
APPEND(cons(x, xs), ys) → APPEND(xs, ys)

The TRS R consists of the following rules:

f(true, xs) → f(isList(xs), append(cons(a, nil), xs))
isList(nil) → true
isList(cons(x, xs)) → isList(xs)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))

The set Q consists of the following terms:

f(true, x0)
isList(nil)
isList(cons(x0, x1))
append(nil, x0)
append(cons(x0, 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 2 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

APPEND(cons(x, xs), ys) → APPEND(xs, ys)

The TRS R consists of the following rules:

f(true, xs) → f(isList(xs), append(cons(a, nil), xs))
isList(nil) → true
isList(cons(x, xs)) → isList(xs)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))

The set Q consists of the following terms:

f(true, x0)
isList(nil)
isList(cons(x0, x1))
append(nil, x0)
append(cons(x0, 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:

APPEND(cons(x, xs), ys) → APPEND(xs, ys)

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

f(true, x0)
isList(nil)
isList(cons(x0, x1))
append(nil, x0)
append(cons(x0, 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(true, x0)
isList(nil)
isList(cons(x0, x1))
append(nil, x0)
append(cons(x0, x1), x2)

(11) Obligation:

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

APPEND(cons(x, xs), ys) → APPEND(xs, ys)

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:

  • APPEND(cons(x, xs), ys) → APPEND(xs, ys)
    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:

ISLIST(cons(x, xs)) → ISLIST(xs)

The TRS R consists of the following rules:

f(true, xs) → f(isList(xs), append(cons(a, nil), xs))
isList(nil) → true
isList(cons(x, xs)) → isList(xs)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))

The set Q consists of the following terms:

f(true, x0)
isList(nil)
isList(cons(x0, x1))
append(nil, x0)
append(cons(x0, 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:

ISLIST(cons(x, xs)) → ISLIST(xs)

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

f(true, x0)
isList(nil)
isList(cons(x0, x1))
append(nil, x0)
append(cons(x0, 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(true, x0)
isList(nil)
isList(cons(x0, x1))
append(nil, x0)
append(cons(x0, x1), x2)

(18) Obligation:

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

ISLIST(cons(x, xs)) → ISLIST(xs)

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:

  • ISLIST(cons(x, xs)) → ISLIST(xs)
    The graph contains the following edges 1 > 1

(20) YES

(21) Obligation:

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

F(true, xs) → F(isList(xs), append(cons(a, nil), xs))

The TRS R consists of the following rules:

f(true, xs) → f(isList(xs), append(cons(a, nil), xs))
isList(nil) → true
isList(cons(x, xs)) → isList(xs)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))

The set Q consists of the following terms:

f(true, x0)
isList(nil)
isList(cons(x0, x1))
append(nil, x0)
append(cons(x0, x1), x2)

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

(22) MNOCProof (EQUIVALENT transformation)

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

(23) Obligation:

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

F(true, xs) → F(isList(xs), append(cons(a, nil), xs))

The TRS R consists of the following rules:

f(true, xs) → f(isList(xs), append(cons(a, nil), xs))
isList(nil) → true
isList(cons(x, xs)) → isList(xs)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))

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

(24) NonLoopProof (COMPLETE transformation)

By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP.
We apply the theorem with m = 1, b = 1,
σ' = [ ], and μ' = [ ] on the rule
F(true, cons(a, zr1))[zr1 / cons(a, zr1)]n[zr1 / nil] → F(true, cons(a, cons(a, zr1)))[zr1 / cons(a, zr1)]n[zr1 / nil]
This rule is correct for the QDP as the following derivation shows:

F(true, cons(a, zr1))[zr1 / cons(a, zr1)]n[zr1 / nil] → F(true, cons(a, cons(a, zr1)))[zr1 / cons(a, zr1)]n[zr1 / nil]
    by Equivalence by Domain Renaming of the lhs with [zl0 / zr1]
    intermediate steps: Equiv DR (rhs) - Equiv DR (lhs) - Instantiation - Equiv DR (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs)
    F(true, cons(x1, zl1))[zl1 / cons(x1, zl1)]n[zl1 / nil] → F(true, cons(a, cons(x1, zr1)))[zr1 / cons(x1, zr1)]n[zr1 / nil]
        by Rewrite t with the rewrite sequence : [([1],append(cons(x, xs), ys) -> cons(x, append(xs, ys))), ([1,1],append(nil, ys) -> ys)]
        intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs)
        F(true, cons(x1, zl1))[zr1 / cons(x1, zr1), zl1 / cons(x1, zl1)]n[zr1 / nil, zl1 / nil] → F(true, append(cons(a, nil), cons(x1, zr1)))[zr1 / cons(x1, zr1), zl1 / cons(x1, zl1)]n[zr1 / nil, zl1 / nil]
            by Narrowing at position: [0]
            intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs) - Instantiation - Equiv DR (rhs) - Equiv DR (lhs)
            F(true, cons(y0, zs1))[zs1 / cons(y0, zs1)]n[zs1 / y1] → F(isList(y1), append(cons(a, nil), cons(y0, zs1)))[zs1 / cons(y0, zs1)]n[zs1 / y1]
                by Narrowing at position: [0]
                intermediate steps: Instantiate mu - Instantiate Sigma - Instantiation - Instantiation
                F(true, xs)[ ]n[ ] → F(isList(xs), append(cons(a, nil), xs))[ ]n[ ]
                    by Rule from TRS P

                intermediate steps: Equiv IPS (rhs) - Equiv IPS (rhs) - Instantiation - Equiv DR (lhs) - Instantiation - Equiv DR (lhs)
                isList(cons(x, xs))[xs / cons(x, xs)]n[ ] → isList(xs)[ ]n[ ]
                    by PatternCreation I with delta: [ ], theta: [ ], sigma: [xs / cons(x, xs)]
                    isList(cons(x, xs))[ ]n[ ] → isList(xs)[ ]n[ ]
                        by Rule from TRS R

            intermediate steps: Equiv IPS (rhs) - Equiv IPS (lhs) - Equiv IPS (rhs) - Equiv IPS (lhs)
            isList(nil)[ ]n[ ] → true[ ]n[ ]
                by Rule from TRS R

(25) NO