/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.pl /export/starexec/sandbox2/output/output_files


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YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.pl
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Left Termination of the query pattern

hidden_flatten(g,a,a)

w.r.t. the given Prolog program could successfully be proven:

(0) Prolog
(1) CutEliminatorProof [SOUND, 0 ms]
(2) Prolog
(3) PrologToPiTRSProof [SOUND, 0 ms]
(4) PiTRS
(5) DependencyPairsProof [EQUIVALENT, 2 ms]
(6) PiDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) PiDP
(9) PiDPToQDPProof [SOUND, 12 ms]
(10) QDP
(11) UsableRulesReductionPairsProof [EQUIVALENT, 17 ms]
(12) QDP
(13) DependencyGraphProof [EQUIVALENT, 0 ms]
(14) TRUE


----------------------------------------

(0)
Obligation:
Clauses:

hidden_flatten([], L, L).
hidden_flatten(.(.(H, T), L), S, F) :- ','(!, ','(hidden_flatten(L, S, Lf), hidden_flatten(.(H, T), Lf, F))).
hidden_flatten(.(H, T), S, .(H, L)) :- hidden_flatten(T, S, L).


Query: hidden_flatten(g,a,a)
----------------------------------------

(1) CutEliminatorProof (SOUND)
Eliminated all cuts by simply ignoring them[PROLOG].
----------------------------------------

(2)
Obligation:
Clauses:

hidden_flatten([], L, L).
hidden_flatten(.(.(H, T), L), S, F) :- ','(hidden_flatten(L, S, Lf), hidden_flatten(.(H, T), Lf, F)).
hidden_flatten(.(H, T), S, .(H, L)) :- hidden_flatten(T, S, L).


Query: hidden_flatten(g,a,a)
----------------------------------------

(3) PrologToPiTRSProof (SOUND)
We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:

hidden_flatten_in_3: (b,f,f)

Transforming Prolog into the following Term Rewriting System:

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

   hidden_flatten_in_gaa([], L, L) -> hidden_flatten_out_gaa([], L, L)
   hidden_flatten_in_gaa(.(.(H, T), L), S, F) -> U1_gaa(H, T, L, S, F, hidden_flatten_in_gaa(L, S, Lf))
   hidden_flatten_in_gaa(.(H, T), S, .(H, L)) -> U3_gaa(H, T, S, L, hidden_flatten_in_gaa(T, S, L))
   U3_gaa(H, T, S, L, hidden_flatten_out_gaa(T, S, L)) -> hidden_flatten_out_gaa(.(H, T), S, .(H, L))
   U1_gaa(H, T, L, S, F, hidden_flatten_out_gaa(L, S, Lf)) -> U2_gaa(H, T, L, S, F, hidden_flatten_in_gaa(.(H, T), Lf, F))
   U2_gaa(H, T, L, S, F, hidden_flatten_out_gaa(.(H, T), Lf, F)) -> hidden_flatten_out_gaa(.(.(H, T), L), S, F)

The argument filtering Pi contains the following mapping:
hidden_flatten_in_gaa(x1, x2, x3)  =  hidden_flatten_in_gaa(x1)

[]  =  []

hidden_flatten_out_gaa(x1, x2, x3)  =  hidden_flatten_out_gaa

.(x1, x2)  =  .(x1, x2)

U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x1, x2, x6)

U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x5)

U2_gaa(x1, x2, x3, x4, x5, x6)  =  U2_gaa(x6)





Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



----------------------------------------

(4)
Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:

   hidden_flatten_in_gaa([], L, L) -> hidden_flatten_out_gaa([], L, L)
   hidden_flatten_in_gaa(.(.(H, T), L), S, F) -> U1_gaa(H, T, L, S, F, hidden_flatten_in_gaa(L, S, Lf))
   hidden_flatten_in_gaa(.(H, T), S, .(H, L)) -> U3_gaa(H, T, S, L, hidden_flatten_in_gaa(T, S, L))
   U3_gaa(H, T, S, L, hidden_flatten_out_gaa(T, S, L)) -> hidden_flatten_out_gaa(.(H, T), S, .(H, L))
   U1_gaa(H, T, L, S, F, hidden_flatten_out_gaa(L, S, Lf)) -> U2_gaa(H, T, L, S, F, hidden_flatten_in_gaa(.(H, T), Lf, F))
   U2_gaa(H, T, L, S, F, hidden_flatten_out_gaa(.(H, T), Lf, F)) -> hidden_flatten_out_gaa(.(.(H, T), L), S, F)

The argument filtering Pi contains the following mapping:
hidden_flatten_in_gaa(x1, x2, x3)  =  hidden_flatten_in_gaa(x1)

[]  =  []

hidden_flatten_out_gaa(x1, x2, x3)  =  hidden_flatten_out_gaa

.(x1, x2)  =  .(x1, x2)

U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x1, x2, x6)

U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x5)

U2_gaa(x1, x2, x3, x4, x5, x6)  =  U2_gaa(x6)



----------------------------------------

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

   HIDDEN_FLATTEN_IN_GAA(.(.(H, T), L), S, F) -> U1_GAA(H, T, L, S, F, hidden_flatten_in_gaa(L, S, Lf))
   HIDDEN_FLATTEN_IN_GAA(.(.(H, T), L), S, F) -> HIDDEN_FLATTEN_IN_GAA(L, S, Lf)
   HIDDEN_FLATTEN_IN_GAA(.(H, T), S, .(H, L)) -> U3_GAA(H, T, S, L, hidden_flatten_in_gaa(T, S, L))
   HIDDEN_FLATTEN_IN_GAA(.(H, T), S, .(H, L)) -> HIDDEN_FLATTEN_IN_GAA(T, S, L)
   U1_GAA(H, T, L, S, F, hidden_flatten_out_gaa(L, S, Lf)) -> U2_GAA(H, T, L, S, F, hidden_flatten_in_gaa(.(H, T), Lf, F))
   U1_GAA(H, T, L, S, F, hidden_flatten_out_gaa(L, S, Lf)) -> HIDDEN_FLATTEN_IN_GAA(.(H, T), Lf, F)

The TRS R consists of the following rules:

   hidden_flatten_in_gaa([], L, L) -> hidden_flatten_out_gaa([], L, L)
   hidden_flatten_in_gaa(.(.(H, T), L), S, F) -> U1_gaa(H, T, L, S, F, hidden_flatten_in_gaa(L, S, Lf))
   hidden_flatten_in_gaa(.(H, T), S, .(H, L)) -> U3_gaa(H, T, S, L, hidden_flatten_in_gaa(T, S, L))
   U3_gaa(H, T, S, L, hidden_flatten_out_gaa(T, S, L)) -> hidden_flatten_out_gaa(.(H, T), S, .(H, L))
   U1_gaa(H, T, L, S, F, hidden_flatten_out_gaa(L, S, Lf)) -> U2_gaa(H, T, L, S, F, hidden_flatten_in_gaa(.(H, T), Lf, F))
   U2_gaa(H, T, L, S, F, hidden_flatten_out_gaa(.(H, T), Lf, F)) -> hidden_flatten_out_gaa(.(.(H, T), L), S, F)

The argument filtering Pi contains the following mapping:
hidden_flatten_in_gaa(x1, x2, x3)  =  hidden_flatten_in_gaa(x1)

[]  =  []

hidden_flatten_out_gaa(x1, x2, x3)  =  hidden_flatten_out_gaa

.(x1, x2)  =  .(x1, x2)

U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x1, x2, x6)

U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x5)

U2_gaa(x1, x2, x3, x4, x5, x6)  =  U2_gaa(x6)

HIDDEN_FLATTEN_IN_GAA(x1, x2, x3)  =  HIDDEN_FLATTEN_IN_GAA(x1)

U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x1, x2, x6)

U3_GAA(x1, x2, x3, x4, x5)  =  U3_GAA(x5)

U2_GAA(x1, x2, x3, x4, x5, x6)  =  U2_GAA(x6)


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

(6)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

   HIDDEN_FLATTEN_IN_GAA(.(.(H, T), L), S, F) -> U1_GAA(H, T, L, S, F, hidden_flatten_in_gaa(L, S, Lf))
   HIDDEN_FLATTEN_IN_GAA(.(.(H, T), L), S, F) -> HIDDEN_FLATTEN_IN_GAA(L, S, Lf)
   HIDDEN_FLATTEN_IN_GAA(.(H, T), S, .(H, L)) -> U3_GAA(H, T, S, L, hidden_flatten_in_gaa(T, S, L))
   HIDDEN_FLATTEN_IN_GAA(.(H, T), S, .(H, L)) -> HIDDEN_FLATTEN_IN_GAA(T, S, L)
   U1_GAA(H, T, L, S, F, hidden_flatten_out_gaa(L, S, Lf)) -> U2_GAA(H, T, L, S, F, hidden_flatten_in_gaa(.(H, T), Lf, F))
   U1_GAA(H, T, L, S, F, hidden_flatten_out_gaa(L, S, Lf)) -> HIDDEN_FLATTEN_IN_GAA(.(H, T), Lf, F)

The TRS R consists of the following rules:

   hidden_flatten_in_gaa([], L, L) -> hidden_flatten_out_gaa([], L, L)
   hidden_flatten_in_gaa(.(.(H, T), L), S, F) -> U1_gaa(H, T, L, S, F, hidden_flatten_in_gaa(L, S, Lf))
   hidden_flatten_in_gaa(.(H, T), S, .(H, L)) -> U3_gaa(H, T, S, L, hidden_flatten_in_gaa(T, S, L))
   U3_gaa(H, T, S, L, hidden_flatten_out_gaa(T, S, L)) -> hidden_flatten_out_gaa(.(H, T), S, .(H, L))
   U1_gaa(H, T, L, S, F, hidden_flatten_out_gaa(L, S, Lf)) -> U2_gaa(H, T, L, S, F, hidden_flatten_in_gaa(.(H, T), Lf, F))
   U2_gaa(H, T, L, S, F, hidden_flatten_out_gaa(.(H, T), Lf, F)) -> hidden_flatten_out_gaa(.(.(H, T), L), S, F)

The argument filtering Pi contains the following mapping:
hidden_flatten_in_gaa(x1, x2, x3)  =  hidden_flatten_in_gaa(x1)

[]  =  []

hidden_flatten_out_gaa(x1, x2, x3)  =  hidden_flatten_out_gaa

.(x1, x2)  =  .(x1, x2)

U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x1, x2, x6)

U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x5)

U2_gaa(x1, x2, x3, x4, x5, x6)  =  U2_gaa(x6)

HIDDEN_FLATTEN_IN_GAA(x1, x2, x3)  =  HIDDEN_FLATTEN_IN_GAA(x1)

U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x1, x2, x6)

U3_GAA(x1, x2, x3, x4, x5)  =  U3_GAA(x5)

U2_GAA(x1, x2, x3, x4, x5, x6)  =  U2_GAA(x6)


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

(7) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 2 less nodes.
----------------------------------------

(8)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

   U1_GAA(H, T, L, S, F, hidden_flatten_out_gaa(L, S, Lf)) -> HIDDEN_FLATTEN_IN_GAA(.(H, T), Lf, F)
   HIDDEN_FLATTEN_IN_GAA(.(.(H, T), L), S, F) -> U1_GAA(H, T, L, S, F, hidden_flatten_in_gaa(L, S, Lf))
   HIDDEN_FLATTEN_IN_GAA(.(.(H, T), L), S, F) -> HIDDEN_FLATTEN_IN_GAA(L, S, Lf)
   HIDDEN_FLATTEN_IN_GAA(.(H, T), S, .(H, L)) -> HIDDEN_FLATTEN_IN_GAA(T, S, L)

The TRS R consists of the following rules:

   hidden_flatten_in_gaa([], L, L) -> hidden_flatten_out_gaa([], L, L)
   hidden_flatten_in_gaa(.(.(H, T), L), S, F) -> U1_gaa(H, T, L, S, F, hidden_flatten_in_gaa(L, S, Lf))
   hidden_flatten_in_gaa(.(H, T), S, .(H, L)) -> U3_gaa(H, T, S, L, hidden_flatten_in_gaa(T, S, L))
   U3_gaa(H, T, S, L, hidden_flatten_out_gaa(T, S, L)) -> hidden_flatten_out_gaa(.(H, T), S, .(H, L))
   U1_gaa(H, T, L, S, F, hidden_flatten_out_gaa(L, S, Lf)) -> U2_gaa(H, T, L, S, F, hidden_flatten_in_gaa(.(H, T), Lf, F))
   U2_gaa(H, T, L, S, F, hidden_flatten_out_gaa(.(H, T), Lf, F)) -> hidden_flatten_out_gaa(.(.(H, T), L), S, F)

The argument filtering Pi contains the following mapping:
hidden_flatten_in_gaa(x1, x2, x3)  =  hidden_flatten_in_gaa(x1)

[]  =  []

hidden_flatten_out_gaa(x1, x2, x3)  =  hidden_flatten_out_gaa

.(x1, x2)  =  .(x1, x2)

U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x1, x2, x6)

U3_gaa(x1, x2, x3, x4, x5)  =  U3_gaa(x5)

U2_gaa(x1, x2, x3, x4, x5, x6)  =  U2_gaa(x6)

HIDDEN_FLATTEN_IN_GAA(x1, x2, x3)  =  HIDDEN_FLATTEN_IN_GAA(x1)

U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x1, x2, x6)


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

(9) PiDPToQDPProof (SOUND)
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_GAA(H, T, hidden_flatten_out_gaa) -> HIDDEN_FLATTEN_IN_GAA(.(H, T))
   HIDDEN_FLATTEN_IN_GAA(.(.(H, T), L)) -> U1_GAA(H, T, hidden_flatten_in_gaa(L))
   HIDDEN_FLATTEN_IN_GAA(.(.(H, T), L)) -> HIDDEN_FLATTEN_IN_GAA(L)
   HIDDEN_FLATTEN_IN_GAA(.(H, T)) -> HIDDEN_FLATTEN_IN_GAA(T)

The TRS R consists of the following rules:

   hidden_flatten_in_gaa([]) -> hidden_flatten_out_gaa
   hidden_flatten_in_gaa(.(.(H, T), L)) -> U1_gaa(H, T, hidden_flatten_in_gaa(L))
   hidden_flatten_in_gaa(.(H, T)) -> U3_gaa(hidden_flatten_in_gaa(T))
   U3_gaa(hidden_flatten_out_gaa) -> hidden_flatten_out_gaa
   U1_gaa(H, T, hidden_flatten_out_gaa) -> U2_gaa(hidden_flatten_in_gaa(.(H, T)))
   U2_gaa(hidden_flatten_out_gaa) -> hidden_flatten_out_gaa

The set Q consists of the following terms:

   hidden_flatten_in_gaa(x0)
   U3_gaa(x0)
   U1_gaa(x0, x1, x2)
   U2_gaa(x0)

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

(11) UsableRulesReductionPairsProof (EQUIVALENT)
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:

   HIDDEN_FLATTEN_IN_GAA(.(.(H, T), L)) -> U1_GAA(H, T, hidden_flatten_in_gaa(L))
   HIDDEN_FLATTEN_IN_GAA(.(.(H, T), L)) -> HIDDEN_FLATTEN_IN_GAA(L)
   HIDDEN_FLATTEN_IN_GAA(.(H, T)) -> HIDDEN_FLATTEN_IN_GAA(T)
The following rules are removed from R:

   hidden_flatten_in_gaa([]) -> hidden_flatten_out_gaa
   hidden_flatten_in_gaa(.(.(H, T), L)) -> U1_gaa(H, T, hidden_flatten_in_gaa(L))
   hidden_flatten_in_gaa(.(H, T)) -> U3_gaa(hidden_flatten_in_gaa(T))
   U3_gaa(hidden_flatten_out_gaa) -> hidden_flatten_out_gaa
   U2_gaa(hidden_flatten_out_gaa) -> hidden_flatten_out_gaa
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(.(x_1, x_2)) = 2 + 2*x_1 + 2*x_2
   POL(HIDDEN_FLATTEN_IN_GAA(x_1)) = x_1
   POL(U1_GAA(x_1, x_2, x_3)) = 2*x_1 + 2*x_2 + x_3
   POL(U1_gaa(x_1, x_2, x_3)) = 2*x_1 + 2*x_2 + 2*x_3
   POL(U2_gaa(x_1)) = 2 + x_1
   POL(U3_gaa(x_1)) = 2*x_1
   POL([]) = 2
   POL(hidden_flatten_in_gaa(x_1)) = x_1
   POL(hidden_flatten_out_gaa) = 2


----------------------------------------

(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   U1_GAA(H, T, hidden_flatten_out_gaa) -> HIDDEN_FLATTEN_IN_GAA(.(H, T))

The TRS R consists of the following rules:

   U1_gaa(H, T, hidden_flatten_out_gaa) -> U2_gaa(hidden_flatten_in_gaa(.(H, T)))

The set Q consists of the following terms:

   hidden_flatten_in_gaa(x0)
   U3_gaa(x0)
   U1_gaa(x0, x1, x2)
   U2_gaa(x0)

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

(13) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
----------------------------------------

(14)
TRUE