Left Termination of the query pattern flat_in_2(a, g) w.r.t. the given Prolog program could not be shown:

0 Prolog
  ↳1 PrologToPiTRSProof (⇐)
    ↳2 PiTRS
    ↳3 DependencyPairsProof (⇔)
    ↳4 PiDP
    ↳5 DependencyGraphProof (⇔)
    ↳6 PiDP
    ↳7 UsableRulesProof (⇔)
    ↳8 PiDP
    ↳9 PiDPToQDPProof (⇐)
    ↳10 QDP
    ↳11 UsableRulesReductionPairsProof (⇔)
    ↳12 QDP
    ↳13 UsableRulesProof (⇔)
    ↳14 QDP
    ↳15 QReductionProof (⇔)
    ↳16 QDP
    ↳17 NonTerminationProof (⇔)
    ↳18 FALSE
  ↳19 PrologToPiTRSProof (⇐)
    ↳20 PiTRS
    ↳21 DependencyPairsProof (⇔)
    ↳22 PiDP
    ↳23 DependencyGraphProof (⇔)
    ↳24 PiDP
    ↳25 UsableRulesProof (⇔)
    ↳26 PiDP
    ↳27 PiDPToQDPProof (⇐)
    ↳28 QDP
    ↳29 UsableRulesReductionPairsProof (⇔)
    ↳30 QDP
    ↳31 UsableRulesProof (⇔)
    ↳32 QDP
    ↳33 QReductionProof (⇔)
    ↳34 QDP
    ↳35 NonTerminationProof (⇔)
    ↳36 FALSE

(0) Obligation:

Clauses:

right(tree(X, XS1, XS2), XS2).
flat(niltree, nil).
flat(tree(X, niltree, XS), cons(X, YS)) :- ','(right(tree(X, niltree, XS), ZS), flat(ZS, YS)).
flat(tree(X, tree(Y, YS1, YS2), XS), ZS) :- flat(tree(Y, YS1, tree(X, YS2, XS)), ZS).

Queries:

flat(a,g).

(1) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
flat_in: (f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

flat_in_ag(niltree, nil) → flat_out_ag(niltree, nil)
flat_in_ag(tree(X, niltree, XS), cons(X, YS)) → U1_ag(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
right_in_aa(tree(X, XS1, XS2), XS2) → right_out_aa(tree(X, XS1, XS2), XS2)
U1_ag(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_ag(X, XS, YS, flat_in_ag(ZS, YS))
flat_in_ag(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_ag(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
U3_ag(X, Y, YS1, YS2, XS, ZS, flat_out_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS)) → flat_out_ag(tree(X, tree(Y, YS1, YS2), XS), ZS)
U2_ag(X, XS, YS, flat_out_ag(ZS, YS)) → flat_out_ag(tree(X, niltree, XS), cons(X, YS))

The argument filtering Pi contains the following mapping:
flat_in_ag(x1, x2)  =  flat_in_ag(x2)
nil  =  nil
flat_out_ag(x1, x2)  =  flat_out_ag
cons(x1, x2)  =  cons(x1, x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x3, x4)
right_in_aa(x1, x2)  =  right_in_aa
right_out_aa(x1, x2)  =  right_out_aa
tree(x1, x2, x3)  =  tree(x2, x3)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x4)
U3_ag(x1, x2, x3, x4, x5, x6, x7)  =  U3_ag(x7)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(2) Obligation:

Pi-finite rewrite system:
The TRS R consists of the following rules:

flat_in_ag(niltree, nil) → flat_out_ag(niltree, nil)
flat_in_ag(tree(X, niltree, XS), cons(X, YS)) → U1_ag(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
right_in_aa(tree(X, XS1, XS2), XS2) → right_out_aa(tree(X, XS1, XS2), XS2)
U1_ag(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_ag(X, XS, YS, flat_in_ag(ZS, YS))
flat_in_ag(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_ag(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
U3_ag(X, Y, YS1, YS2, XS, ZS, flat_out_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS)) → flat_out_ag(tree(X, tree(Y, YS1, YS2), XS), ZS)
U2_ag(X, XS, YS, flat_out_ag(ZS, YS)) → flat_out_ag(tree(X, niltree, XS), cons(X, YS))

The argument filtering Pi contains the following mapping:
flat_in_ag(x1, x2)  =  flat_in_ag(x2)
nil  =  nil
flat_out_ag(x1, x2)  =  flat_out_ag
cons(x1, x2)  =  cons(x1, x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x3, x4)
right_in_aa(x1, x2)  =  right_in_aa
right_out_aa(x1, x2)  =  right_out_aa
tree(x1, x2, x3)  =  tree(x2, x3)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x4)
U3_ag(x1, x2, x3, x4, x5, x6, x7)  =  U3_ag(x7)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

FLAT_IN_AG(tree(X, niltree, XS), cons(X, YS)) → U1_AG(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
FLAT_IN_AG(tree(X, niltree, XS), cons(X, YS)) → RIGHT_IN_AA(tree(X, niltree, XS), ZS)
U1_AG(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_AG(X, XS, YS, flat_in_ag(ZS, YS))
U1_AG(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → FLAT_IN_AG(ZS, YS)
FLAT_IN_AG(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_AG(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
FLAT_IN_AG(tree(X, tree(Y, YS1, YS2), XS), ZS) → FLAT_IN_AG(tree(Y, YS1, tree(X, YS2, XS)), ZS)

The TRS R consists of the following rules:

flat_in_ag(niltree, nil) → flat_out_ag(niltree, nil)
flat_in_ag(tree(X, niltree, XS), cons(X, YS)) → U1_ag(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
right_in_aa(tree(X, XS1, XS2), XS2) → right_out_aa(tree(X, XS1, XS2), XS2)
U1_ag(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_ag(X, XS, YS, flat_in_ag(ZS, YS))
flat_in_ag(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_ag(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
U3_ag(X, Y, YS1, YS2, XS, ZS, flat_out_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS)) → flat_out_ag(tree(X, tree(Y, YS1, YS2), XS), ZS)
U2_ag(X, XS, YS, flat_out_ag(ZS, YS)) → flat_out_ag(tree(X, niltree, XS), cons(X, YS))

The argument filtering Pi contains the following mapping:
flat_in_ag(x1, x2)  =  flat_in_ag(x2)
nil  =  nil
flat_out_ag(x1, x2)  =  flat_out_ag
cons(x1, x2)  =  cons(x1, x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x3, x4)
right_in_aa(x1, x2)  =  right_in_aa
right_out_aa(x1, x2)  =  right_out_aa
tree(x1, x2, x3)  =  tree(x2, x3)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x4)
U3_ag(x1, x2, x3, x4, x5, x6, x7)  =  U3_ag(x7)
FLAT_IN_AG(x1, x2)  =  FLAT_IN_AG(x2)
U1_AG(x1, x2, x3, x4)  =  U1_AG(x3, x4)
RIGHT_IN_AA(x1, x2)  =  RIGHT_IN_AA
U2_AG(x1, x2, x3, x4)  =  U2_AG(x4)
U3_AG(x1, x2, x3, x4, x5, x6, x7)  =  U3_AG(x7)

We have to consider all (P,R,Pi)-chains

(4) Obligation:

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

FLAT_IN_AG(tree(X, niltree, XS), cons(X, YS)) → U1_AG(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
FLAT_IN_AG(tree(X, niltree, XS), cons(X, YS)) → RIGHT_IN_AA(tree(X, niltree, XS), ZS)
U1_AG(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_AG(X, XS, YS, flat_in_ag(ZS, YS))
U1_AG(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → FLAT_IN_AG(ZS, YS)
FLAT_IN_AG(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_AG(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
FLAT_IN_AG(tree(X, tree(Y, YS1, YS2), XS), ZS) → FLAT_IN_AG(tree(Y, YS1, tree(X, YS2, XS)), ZS)

The TRS R consists of the following rules:

flat_in_ag(niltree, nil) → flat_out_ag(niltree, nil)
flat_in_ag(tree(X, niltree, XS), cons(X, YS)) → U1_ag(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
right_in_aa(tree(X, XS1, XS2), XS2) → right_out_aa(tree(X, XS1, XS2), XS2)
U1_ag(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_ag(X, XS, YS, flat_in_ag(ZS, YS))
flat_in_ag(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_ag(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
U3_ag(X, Y, YS1, YS2, XS, ZS, flat_out_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS)) → flat_out_ag(tree(X, tree(Y, YS1, YS2), XS), ZS)
U2_ag(X, XS, YS, flat_out_ag(ZS, YS)) → flat_out_ag(tree(X, niltree, XS), cons(X, YS))

The argument filtering Pi contains the following mapping:
flat_in_ag(x1, x2)  =  flat_in_ag(x2)
nil  =  nil
flat_out_ag(x1, x2)  =  flat_out_ag
cons(x1, x2)  =  cons(x1, x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x3, x4)
right_in_aa(x1, x2)  =  right_in_aa
right_out_aa(x1, x2)  =  right_out_aa
tree(x1, x2, x3)  =  tree(x2, x3)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x4)
U3_ag(x1, x2, x3, x4, x5, x6, x7)  =  U3_ag(x7)
FLAT_IN_AG(x1, x2)  =  FLAT_IN_AG(x2)
U1_AG(x1, x2, x3, x4)  =  U1_AG(x3, x4)
RIGHT_IN_AA(x1, x2)  =  RIGHT_IN_AA
U2_AG(x1, x2, x3, x4)  =  U2_AG(x4)
U3_AG(x1, x2, x3, x4, x5, x6, x7)  =  U3_AG(x7)

We have to consider all (P,R,Pi)-chains

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.

(6) Obligation:

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

U1_AG(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → FLAT_IN_AG(ZS, YS)
FLAT_IN_AG(tree(X, niltree, XS), cons(X, YS)) → U1_AG(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
FLAT_IN_AG(tree(X, tree(Y, YS1, YS2), XS), ZS) → FLAT_IN_AG(tree(Y, YS1, tree(X, YS2, XS)), ZS)

The TRS R consists of the following rules:

flat_in_ag(niltree, nil) → flat_out_ag(niltree, nil)
flat_in_ag(tree(X, niltree, XS), cons(X, YS)) → U1_ag(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
right_in_aa(tree(X, XS1, XS2), XS2) → right_out_aa(tree(X, XS1, XS2), XS2)
U1_ag(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_ag(X, XS, YS, flat_in_ag(ZS, YS))
flat_in_ag(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_ag(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
U3_ag(X, Y, YS1, YS2, XS, ZS, flat_out_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS)) → flat_out_ag(tree(X, tree(Y, YS1, YS2), XS), ZS)
U2_ag(X, XS, YS, flat_out_ag(ZS, YS)) → flat_out_ag(tree(X, niltree, XS), cons(X, YS))

The argument filtering Pi contains the following mapping:
flat_in_ag(x1, x2)  =  flat_in_ag(x2)
nil  =  nil
flat_out_ag(x1, x2)  =  flat_out_ag
cons(x1, x2)  =  cons(x1, x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x3, x4)
right_in_aa(x1, x2)  =  right_in_aa
right_out_aa(x1, x2)  =  right_out_aa
tree(x1, x2, x3)  =  tree(x2, x3)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x4)
U3_ag(x1, x2, x3, x4, x5, x6, x7)  =  U3_ag(x7)
FLAT_IN_AG(x1, x2)  =  FLAT_IN_AG(x2)
U1_AG(x1, x2, x3, x4)  =  U1_AG(x3, x4)

We have to consider all (P,R,Pi)-chains

(7) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(8) Obligation:

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

U1_AG(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → FLAT_IN_AG(ZS, YS)
FLAT_IN_AG(tree(X, niltree, XS), cons(X, YS)) → U1_AG(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
FLAT_IN_AG(tree(X, tree(Y, YS1, YS2), XS), ZS) → FLAT_IN_AG(tree(Y, YS1, tree(X, YS2, XS)), ZS)

The TRS R consists of the following rules:

right_in_aa(tree(X, XS1, XS2), XS2) → right_out_aa(tree(X, XS1, XS2), XS2)

The argument filtering Pi contains the following mapping:
cons(x1, x2)  =  cons(x1, x2)
right_in_aa(x1, x2)  =  right_in_aa
right_out_aa(x1, x2)  =  right_out_aa
tree(x1, x2, x3)  =  tree(x2, x3)
FLAT_IN_AG(x1, x2)  =  FLAT_IN_AG(x2)
U1_AG(x1, x2, x3, x4)  =  U1_AG(x3, x4)

We have to consider all (P,R,Pi)-chains

(9) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(10) Obligation:

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

U1_AG(YS, right_out_aa) → FLAT_IN_AG(YS)
FLAT_IN_AG(cons(X, YS)) → U1_AG(YS, right_in_aa)
FLAT_IN_AG(ZS) → FLAT_IN_AG(ZS)

The TRS R consists of the following rules:

right_in_aa → right_out_aa

The set Q consists of the following terms:

right_in_aa

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

(11) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

U1_AG(YS, right_out_aa) → FLAT_IN_AG(YS)
FLAT_IN_AG(cons(X, YS)) → U1_AG(YS, right_in_aa)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(FLAT_IN_AG(x1)) = 2 + x1   
POL(U1_AG(x1, x2)) = 1 + x1 + x2   
POL(cons(x1, x2)) = 2 + x1 + 2·x2   
POL(right_in_aa) = 2   
POL(right_out_aa) = 2   

(12) Obligation:

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

FLAT_IN_AG(ZS) → FLAT_IN_AG(ZS)

The TRS R consists of the following rules:

right_in_aa → right_out_aa

The set Q consists of the following terms:

right_in_aa

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

(13) 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.

(14) Obligation:

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

FLAT_IN_AG(ZS) → FLAT_IN_AG(ZS)

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

right_in_aa

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

(15) 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].

right_in_aa

(16) Obligation:

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

FLAT_IN_AG(ZS) → FLAT_IN_AG(ZS)

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

(17) 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 = FLAT_IN_AG(ZS) evaluates to t =FLAT_IN_AG(ZS)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Semiunifier: [ ]
  • Matcher: [ ]



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from FLAT_IN_AG(ZS) to FLAT_IN_AG(ZS).



(18) FALSE

(19) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
flat_in: (f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

flat_in_ag(niltree, nil) → flat_out_ag(niltree, nil)
flat_in_ag(tree(X, niltree, XS), cons(X, YS)) → U1_ag(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
right_in_aa(tree(X, XS1, XS2), XS2) → right_out_aa(tree(X, XS1, XS2), XS2)
U1_ag(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_ag(X, XS, YS, flat_in_ag(ZS, YS))
flat_in_ag(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_ag(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
U3_ag(X, Y, YS1, YS2, XS, ZS, flat_out_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS)) → flat_out_ag(tree(X, tree(Y, YS1, YS2), XS), ZS)
U2_ag(X, XS, YS, flat_out_ag(ZS, YS)) → flat_out_ag(tree(X, niltree, XS), cons(X, YS))

The argument filtering Pi contains the following mapping:
flat_in_ag(x1, x2)  =  flat_in_ag(x2)
nil  =  nil
flat_out_ag(x1, x2)  =  flat_out_ag(x2)
cons(x1, x2)  =  cons(x1, x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x1, x3, x4)
right_in_aa(x1, x2)  =  right_in_aa
right_out_aa(x1, x2)  =  right_out_aa
tree(x1, x2, x3)  =  tree(x2, x3)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x1, x3, x4)
U3_ag(x1, x2, x3, x4, x5, x6, x7)  =  U3_ag(x6, x7)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(20) Obligation:

Pi-finite rewrite system:
The TRS R consists of the following rules:

flat_in_ag(niltree, nil) → flat_out_ag(niltree, nil)
flat_in_ag(tree(X, niltree, XS), cons(X, YS)) → U1_ag(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
right_in_aa(tree(X, XS1, XS2), XS2) → right_out_aa(tree(X, XS1, XS2), XS2)
U1_ag(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_ag(X, XS, YS, flat_in_ag(ZS, YS))
flat_in_ag(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_ag(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
U3_ag(X, Y, YS1, YS2, XS, ZS, flat_out_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS)) → flat_out_ag(tree(X, tree(Y, YS1, YS2), XS), ZS)
U2_ag(X, XS, YS, flat_out_ag(ZS, YS)) → flat_out_ag(tree(X, niltree, XS), cons(X, YS))

The argument filtering Pi contains the following mapping:
flat_in_ag(x1, x2)  =  flat_in_ag(x2)
nil  =  nil
flat_out_ag(x1, x2)  =  flat_out_ag(x2)
cons(x1, x2)  =  cons(x1, x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x1, x3, x4)
right_in_aa(x1, x2)  =  right_in_aa
right_out_aa(x1, x2)  =  right_out_aa
tree(x1, x2, x3)  =  tree(x2, x3)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x1, x3, x4)
U3_ag(x1, x2, x3, x4, x5, x6, x7)  =  U3_ag(x6, x7)

(21) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

FLAT_IN_AG(tree(X, niltree, XS), cons(X, YS)) → U1_AG(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
FLAT_IN_AG(tree(X, niltree, XS), cons(X, YS)) → RIGHT_IN_AA(tree(X, niltree, XS), ZS)
U1_AG(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_AG(X, XS, YS, flat_in_ag(ZS, YS))
U1_AG(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → FLAT_IN_AG(ZS, YS)
FLAT_IN_AG(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_AG(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
FLAT_IN_AG(tree(X, tree(Y, YS1, YS2), XS), ZS) → FLAT_IN_AG(tree(Y, YS1, tree(X, YS2, XS)), ZS)

The TRS R consists of the following rules:

flat_in_ag(niltree, nil) → flat_out_ag(niltree, nil)
flat_in_ag(tree(X, niltree, XS), cons(X, YS)) → U1_ag(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
right_in_aa(tree(X, XS1, XS2), XS2) → right_out_aa(tree(X, XS1, XS2), XS2)
U1_ag(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_ag(X, XS, YS, flat_in_ag(ZS, YS))
flat_in_ag(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_ag(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
U3_ag(X, Y, YS1, YS2, XS, ZS, flat_out_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS)) → flat_out_ag(tree(X, tree(Y, YS1, YS2), XS), ZS)
U2_ag(X, XS, YS, flat_out_ag(ZS, YS)) → flat_out_ag(tree(X, niltree, XS), cons(X, YS))

The argument filtering Pi contains the following mapping:
flat_in_ag(x1, x2)  =  flat_in_ag(x2)
nil  =  nil
flat_out_ag(x1, x2)  =  flat_out_ag(x2)
cons(x1, x2)  =  cons(x1, x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x1, x3, x4)
right_in_aa(x1, x2)  =  right_in_aa
right_out_aa(x1, x2)  =  right_out_aa
tree(x1, x2, x3)  =  tree(x2, x3)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x1, x3, x4)
U3_ag(x1, x2, x3, x4, x5, x6, x7)  =  U3_ag(x6, x7)
FLAT_IN_AG(x1, x2)  =  FLAT_IN_AG(x2)
U1_AG(x1, x2, x3, x4)  =  U1_AG(x1, x3, x4)
RIGHT_IN_AA(x1, x2)  =  RIGHT_IN_AA
U2_AG(x1, x2, x3, x4)  =  U2_AG(x1, x3, x4)
U3_AG(x1, x2, x3, x4, x5, x6, x7)  =  U3_AG(x6, x7)

We have to consider all (P,R,Pi)-chains

(22) Obligation:

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

FLAT_IN_AG(tree(X, niltree, XS), cons(X, YS)) → U1_AG(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
FLAT_IN_AG(tree(X, niltree, XS), cons(X, YS)) → RIGHT_IN_AA(tree(X, niltree, XS), ZS)
U1_AG(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_AG(X, XS, YS, flat_in_ag(ZS, YS))
U1_AG(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → FLAT_IN_AG(ZS, YS)
FLAT_IN_AG(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_AG(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
FLAT_IN_AG(tree(X, tree(Y, YS1, YS2), XS), ZS) → FLAT_IN_AG(tree(Y, YS1, tree(X, YS2, XS)), ZS)

The TRS R consists of the following rules:

flat_in_ag(niltree, nil) → flat_out_ag(niltree, nil)
flat_in_ag(tree(X, niltree, XS), cons(X, YS)) → U1_ag(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
right_in_aa(tree(X, XS1, XS2), XS2) → right_out_aa(tree(X, XS1, XS2), XS2)
U1_ag(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_ag(X, XS, YS, flat_in_ag(ZS, YS))
flat_in_ag(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_ag(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
U3_ag(X, Y, YS1, YS2, XS, ZS, flat_out_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS)) → flat_out_ag(tree(X, tree(Y, YS1, YS2), XS), ZS)
U2_ag(X, XS, YS, flat_out_ag(ZS, YS)) → flat_out_ag(tree(X, niltree, XS), cons(X, YS))

The argument filtering Pi contains the following mapping:
flat_in_ag(x1, x2)  =  flat_in_ag(x2)
nil  =  nil
flat_out_ag(x1, x2)  =  flat_out_ag(x2)
cons(x1, x2)  =  cons(x1, x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x1, x3, x4)
right_in_aa(x1, x2)  =  right_in_aa
right_out_aa(x1, x2)  =  right_out_aa
tree(x1, x2, x3)  =  tree(x2, x3)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x1, x3, x4)
U3_ag(x1, x2, x3, x4, x5, x6, x7)  =  U3_ag(x6, x7)
FLAT_IN_AG(x1, x2)  =  FLAT_IN_AG(x2)
U1_AG(x1, x2, x3, x4)  =  U1_AG(x1, x3, x4)
RIGHT_IN_AA(x1, x2)  =  RIGHT_IN_AA
U2_AG(x1, x2, x3, x4)  =  U2_AG(x1, x3, x4)
U3_AG(x1, x2, x3, x4, x5, x6, x7)  =  U3_AG(x6, x7)

We have to consider all (P,R,Pi)-chains

(23) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.

(24) Obligation:

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

U1_AG(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → FLAT_IN_AG(ZS, YS)
FLAT_IN_AG(tree(X, niltree, XS), cons(X, YS)) → U1_AG(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
FLAT_IN_AG(tree(X, tree(Y, YS1, YS2), XS), ZS) → FLAT_IN_AG(tree(Y, YS1, tree(X, YS2, XS)), ZS)

The TRS R consists of the following rules:

flat_in_ag(niltree, nil) → flat_out_ag(niltree, nil)
flat_in_ag(tree(X, niltree, XS), cons(X, YS)) → U1_ag(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
right_in_aa(tree(X, XS1, XS2), XS2) → right_out_aa(tree(X, XS1, XS2), XS2)
U1_ag(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → U2_ag(X, XS, YS, flat_in_ag(ZS, YS))
flat_in_ag(tree(X, tree(Y, YS1, YS2), XS), ZS) → U3_ag(X, Y, YS1, YS2, XS, ZS, flat_in_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS))
U3_ag(X, Y, YS1, YS2, XS, ZS, flat_out_ag(tree(Y, YS1, tree(X, YS2, XS)), ZS)) → flat_out_ag(tree(X, tree(Y, YS1, YS2), XS), ZS)
U2_ag(X, XS, YS, flat_out_ag(ZS, YS)) → flat_out_ag(tree(X, niltree, XS), cons(X, YS))

The argument filtering Pi contains the following mapping:
flat_in_ag(x1, x2)  =  flat_in_ag(x2)
nil  =  nil
flat_out_ag(x1, x2)  =  flat_out_ag(x2)
cons(x1, x2)  =  cons(x1, x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x1, x3, x4)
right_in_aa(x1, x2)  =  right_in_aa
right_out_aa(x1, x2)  =  right_out_aa
tree(x1, x2, x3)  =  tree(x2, x3)
U2_ag(x1, x2, x3, x4)  =  U2_ag(x1, x3, x4)
U3_ag(x1, x2, x3, x4, x5, x6, x7)  =  U3_ag(x6, x7)
FLAT_IN_AG(x1, x2)  =  FLAT_IN_AG(x2)
U1_AG(x1, x2, x3, x4)  =  U1_AG(x1, x3, x4)

We have to consider all (P,R,Pi)-chains

(25) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(26) Obligation:

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

U1_AG(X, XS, YS, right_out_aa(tree(X, niltree, XS), ZS)) → FLAT_IN_AG(ZS, YS)
FLAT_IN_AG(tree(X, niltree, XS), cons(X, YS)) → U1_AG(X, XS, YS, right_in_aa(tree(X, niltree, XS), ZS))
FLAT_IN_AG(tree(X, tree(Y, YS1, YS2), XS), ZS) → FLAT_IN_AG(tree(Y, YS1, tree(X, YS2, XS)), ZS)

The TRS R consists of the following rules:

right_in_aa(tree(X, XS1, XS2), XS2) → right_out_aa(tree(X, XS1, XS2), XS2)

The argument filtering Pi contains the following mapping:
cons(x1, x2)  =  cons(x1, x2)
right_in_aa(x1, x2)  =  right_in_aa
right_out_aa(x1, x2)  =  right_out_aa
tree(x1, x2, x3)  =  tree(x2, x3)
FLAT_IN_AG(x1, x2)  =  FLAT_IN_AG(x2)
U1_AG(x1, x2, x3, x4)  =  U1_AG(x1, x3, x4)

We have to consider all (P,R,Pi)-chains

(27) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(28) Obligation:

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

U1_AG(X, YS, right_out_aa) → FLAT_IN_AG(YS)
FLAT_IN_AG(cons(X, YS)) → U1_AG(X, YS, right_in_aa)
FLAT_IN_AG(ZS) → FLAT_IN_AG(ZS)

The TRS R consists of the following rules:

right_in_aa → right_out_aa

The set Q consists of the following terms:

right_in_aa

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

(29) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

U1_AG(X, YS, right_out_aa) → FLAT_IN_AG(YS)
FLAT_IN_AG(cons(X, YS)) → U1_AG(X, YS, right_in_aa)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(FLAT_IN_AG(x1)) = 2 + x1   
POL(U1_AG(x1, x2, x3)) = 1 + x1 + x2 + x3   
POL(cons(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(right_in_aa) = 2   
POL(right_out_aa) = 2   

(30) Obligation:

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

FLAT_IN_AG(ZS) → FLAT_IN_AG(ZS)

The TRS R consists of the following rules:

right_in_aa → right_out_aa

The set Q consists of the following terms:

right_in_aa

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

(31) 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.

(32) Obligation:

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

FLAT_IN_AG(ZS) → FLAT_IN_AG(ZS)

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

right_in_aa

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

(33) 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].

right_in_aa

(34) Obligation:

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

FLAT_IN_AG(ZS) → FLAT_IN_AG(ZS)

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

(35) 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 = FLAT_IN_AG(ZS) evaluates to t =FLAT_IN_AG(ZS)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Semiunifier: [ ]
  • Matcher: [ ]



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from FLAT_IN_AG(ZS) to FLAT_IN_AG(ZS).



(36) FALSE