NO Termination w.r.t. Q proof of EEG_IJCAR_12_emmes-nonloop-ex6_2.ari

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

0 QTRS
↳1 AAECC Innermost (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 3 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 MRRProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 DependencyGraphProof (⇔, 0 ms)
        ↳22 TRUE
    ↳23 QDP
        ↳24 UsableRulesProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 QReductionProof (⇔, 0 ms)
        ↳27 QDP
        ↳28 QDPSizeChangeProof (⇔, 0 ms)
        ↳29 YES
    ↳30 QDP
        ↳31 MNOCProof (⇔, 0 ms)
        ↳32 QDP
        ↳33 NonLoopProof (⇒, 34 ms)
        ↳34 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(xs, ys) → appendAkk(reverse(xs), ys)
appendAkk(nil, ys) → ys
appendAkk(cons(x, xs), ys) → appendAkk(xs, cons(x, ys))
reverse(nil) → nil
reverse(cons(x, xs)) → append(reverse(xs), cons(x, nil))

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(xs, ys) → appendAkk(reverse(xs), ys)
appendAkk(nil, ys) → ys
appendAkk(cons(x, xs), ys) → appendAkk(xs, cons(x, ys))
reverse(nil) → nil
reverse(cons(x, xs)) → append(reverse(xs), cons(x, nil))

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(xs, ys) → appendAkk(reverse(xs), ys)
appendAkk(nil, ys) → ys
appendAkk(cons(x, xs), ys) → appendAkk(xs, cons(x, ys))
reverse(nil) → nil
reverse(cons(x, xs)) → append(reverse(xs), cons(x, nil))

The set Q consists of the following terms:

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

(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(xs, ys) → APPENDAKK(reverse(xs), ys)
APPEND(xs, ys) → REVERSE(xs)
APPENDAKK(cons(x, xs), ys) → APPENDAKK(xs, cons(x, ys))
REVERSE(cons(x, xs)) → APPEND(reverse(xs), cons(x, nil))
REVERSE(cons(x, xs)) → REVERSE(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(xs, ys) → appendAkk(reverse(xs), ys)
appendAkk(nil, ys) → ys
appendAkk(cons(x, xs), ys) → appendAkk(xs, cons(x, ys))
reverse(nil) → nil
reverse(cons(x, xs)) → append(reverse(xs), cons(x, nil))

The set Q consists of the following terms:

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

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

APPENDAKK(cons(x, xs), ys) → APPENDAKK(xs, cons(x, 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(xs, ys) → appendAkk(reverse(xs), ys)
appendAkk(nil, ys) → ys
appendAkk(cons(x, xs), ys) → appendAkk(xs, cons(x, ys))
reverse(nil) → nil
reverse(cons(x, xs)) → append(reverse(xs), cons(x, nil))

The set Q consists of the following terms:

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

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:

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

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

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

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(x0, x1)
appendAkk(nil, x0)
appendAkk(cons(x0, x1), x2)
reverse(nil)
reverse(cons(x0, x1))

(11) Obligation:

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

APPENDAKK(cons(x, xs), ys) → APPENDAKK(xs, cons(x, 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:

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

(13) YES

(14) Obligation:

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

APPEND(xs, ys) → REVERSE(xs)
REVERSE(cons(x, xs)) → APPEND(reverse(xs), cons(x, nil))
REVERSE(cons(x, xs)) → REVERSE(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(xs, ys) → appendAkk(reverse(xs), ys)
appendAkk(nil, ys) → ys
appendAkk(cons(x, xs), ys) → appendAkk(xs, cons(x, ys))
reverse(nil) → nil
reverse(cons(x, xs)) → append(reverse(xs), cons(x, nil))

The set Q consists of the following terms:

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

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:

APPEND(xs, ys) → REVERSE(xs)
REVERSE(cons(x, xs)) → APPEND(reverse(xs), cons(x, nil))
REVERSE(cons(x, xs)) → REVERSE(xs)

The TRS R consists of the following rules:

reverse(nil) → nil
reverse(cons(x, xs)) → append(reverse(xs), cons(x, nil))
append(xs, ys) → appendAkk(reverse(xs), ys)
appendAkk(nil, ys) → ys
appendAkk(cons(x, xs), ys) → appendAkk(xs, cons(x, ys))

The set Q consists of the following terms:

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

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))

(18) Obligation:

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

APPEND(xs, ys) → REVERSE(xs)
REVERSE(cons(x, xs)) → APPEND(reverse(xs), cons(x, nil))
REVERSE(cons(x, xs)) → REVERSE(xs)

The TRS R consists of the following rules:

reverse(nil) → nil
reverse(cons(x, xs)) → append(reverse(xs), cons(x, nil))
append(xs, ys) → appendAkk(reverse(xs), ys)
appendAkk(nil, ys) → ys
appendAkk(cons(x, xs), ys) → appendAkk(xs, cons(x, ys))

The set Q consists of the following terms:

append(x0, x1)
appendAkk(nil, x0)
appendAkk(cons(x0, x1), x2)
reverse(nil)
reverse(cons(x0, x1))

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

(19) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

REVERSE(cons(x, xs)) → APPEND(reverse(xs), cons(x, nil))
REVERSE(cons(x, xs)) → REVERSE(xs)


Used ordering: Polynomial interpretation [POLO]:

POL(APPEND(x1, x2)) = 2·x1 + x2   
POL(REVERSE(x1)) = 2·x1   
POL(append(x1, x2)) = x1 + x2   
POL(appendAkk(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = 2 + x1 + x2   
POL(nil) = 0   
POL(reverse(x1)) = x1   

(20) Obligation:

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

APPEND(xs, ys) → REVERSE(xs)

The TRS R consists of the following rules:

reverse(nil) → nil
reverse(cons(x, xs)) → append(reverse(xs), cons(x, nil))
append(xs, ys) → appendAkk(reverse(xs), ys)
appendAkk(nil, ys) → ys
appendAkk(cons(x, xs), ys) → appendAkk(xs, cons(x, ys))

The set Q consists of the following terms:

append(x0, x1)
appendAkk(nil, x0)
appendAkk(cons(x0, x1), x2)
reverse(nil)
reverse(cons(x0, x1))

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

(21) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(22) TRUE

(23) 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(xs, ys) → appendAkk(reverse(xs), ys)
appendAkk(nil, ys) → ys
appendAkk(cons(x, xs), ys) → appendAkk(xs, cons(x, ys))
reverse(nil) → nil
reverse(cons(x, xs)) → append(reverse(xs), cons(x, nil))

The set Q consists of the following terms:

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

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

(24) 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.

(25) 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(x0, x1)
appendAkk(nil, x0)
appendAkk(cons(x0, x1), x2)
reverse(nil)
reverse(cons(x0, x1))

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

(26) 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(x0, x1)
appendAkk(nil, x0)
appendAkk(cons(x0, x1), x2)
reverse(nil)
reverse(cons(x0, x1))

(27) 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.

(28) 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

(29) YES

(30) 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(xs, ys) → appendAkk(reverse(xs), ys)
appendAkk(nil, ys) → ys
appendAkk(cons(x, xs), ys) → appendAkk(xs, cons(x, ys))
reverse(nil) → nil
reverse(cons(x, xs)) → append(reverse(xs), cons(x, nil))

The set Q consists of the following terms:

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

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

(31) MNOCProof (EQUIVALENT transformation)

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

(32) 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(xs, ys) → appendAkk(reverse(xs), ys)
appendAkk(nil, ys) → ys
appendAkk(cons(x, xs), ys) → appendAkk(xs, cons(x, ys))
reverse(nil) → nil
reverse(cons(x, xs)) → append(reverse(xs), cons(x, nil))

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

(33) 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(xs, ys) -> appendAkk(reverse(xs), ys)), ([1,0],reverse(cons(x, xs)) -> append(reverse(xs), cons(x, nil))), ([1,0],append(xs, ys) -> appendAkk(reverse(xs), ys)), ([1,0,0,0],reverse(nil) -> nil), ([1,0,0],reverse(nil) -> nil), ([1,0],appendAkk(nil, ys) -> ys), ([1],appendAkk(cons(x, xs), ys) -> appendAkk(xs, cons(x, ys))), ([1],appendAkk(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

(34) NO