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

0 QTRS
↳1 DependencyPairsProof (⇔, 27 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 QDPSizeChangeProof (⇔, 0 ms)
        ↳7 YES
    ↳8 QDP
        ↳9 MNOCProof (⇔, 0 ms)
        ↳10 QDP
        ↳11 UsableRulesProof (⇔, 0 ms)
        ↳12 QDP
        ↳13 QReductionProof (⇔, 0 ms)
        ↳14 QDP
        ↳15 MRRProof (⇔, 16 ms)
        ↳16 QDP
        ↳17 MRRProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 PisEmptyProof (⇔, 0 ms)
        ↳20 YES
    ↳21 QDP
        ↳22 QDPSizeChangeProof (⇔, 0 ms)
        ↳23 YES
    ↳24 QDP
        ↳25 NonLoopProof (⇔, 578 ms)
        ↳26 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) 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:

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))

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 4 SCCs with 3 less nodes.

(4) Complex Obligation (AND)

(5) 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))

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

(6) 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

(7) YES

(8) 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))

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

(9) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.

(10) 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.

(11) 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.

(12) 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.

(13) 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))

(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:

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.

(15) 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)) → REVERSE(xs)


Used ordering: Polynomial interpretation [POLO]:

POL(APPEND(x1, x2)) = x1 + x2   
POL(REVERSE(x1)) = x1   
POL(append(x1, x2)) = x1 + x2   
POL(appendAkk(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = 1 + x1 + x2   
POL(nil) = 0   
POL(reverse(x1)) = x1   

(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))

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.

(17) 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:

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


Used ordering: Polynomial interpretation [POLO]:

POL(APPEND(x1, x2)) = 1 + 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   

(18) Obligation:

Q DP problem:
P is empty.
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) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(20) YES

(21) 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))

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

(22) 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

(23) YES

(24) 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 minimal (P,Q,R)-chains.

(25) NonLoopProof (EQUIVALENT 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:

intermediate steps: Equivalent (Domain Renaming) - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming)
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
        intermediate steps: Equivalent (Remove Unused)
        F(true, cons(x1, zl1))[zl1 / cons(x1, zl1), zr1 / cons(x1, zr1)]n[zl1 / nil, zr1 / nil] → F(true, append(cons(a, nil), cons(x1, zr1)))[zl1 / cons(x1, zl1), zr1 / cons(x1, zr1)]n[zl1 / nil, zr1 / nil]
            by Narrowing at position: [0]
                intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Instantiation - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming)
                F(true, cons(y1, zs1))[zs1 / cons(y1, zs1)]n[zs1 / y0] → F(isList(y0), append(cons(a, nil), cons(y1, zs1)))[zs1 / cons(y1, zs1)]n[zs1 / y0]
                    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 OriginalRule from TRS P

                        intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming)
                        isList(cons(x, xs))[xs / cons(x, xs)]n[ ] → isList(xs)[ ]n[ ]
                            by PatternCreation I
                                isList(cons(x, xs))[ ]n[ ] → isList(xs)[ ]n[ ]
                                    by OriginalRule from TRS R

                intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused)
                isList(nil)[ ]n[ ] → true[ ]n[ ]
                    by OriginalRule from TRS R

(26) NO