(0) Obligation:
Clauses:
flat([], []).
flat(.([], T), R) :- flat(T, R).
flat(.(.(H, T), TT), .(H, R)) :- flat(.(T, TT), R).
Query: flat(g,a)
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph DT10.
(2) Obligation:
Triples:
flatA(.([], .([], X1)), X2) :- flatA(X1, X2).
flatA(.([], .(.(X1, X2), X3)), .(X1, X4)) :- flatA(.(X2, X3), X4).
flatA(.(.(X1, []), X2), .(X1, X3)) :- flatA(X2, X3).
flatA(.(.(X1, .(X2, X3)), X4), .(X1, .(X2, X5))) :- flatA(.(X3, X4), X5).
Clauses:
flatcA([], []).
flatcA(.([], []), []).
flatcA(.([], .([], X1)), X2) :- flatcA(X1, X2).
flatcA(.([], .(.(X1, X2), X3)), .(X1, X4)) :- flatcA(.(X2, X3), X4).
flatcA(.(.(X1, []), X2), .(X1, X3)) :- flatcA(X2, X3).
flatcA(.(.(X1, .(X2, X3)), X4), .(X1, .(X2, X5))) :- flatcA(.(X3, X4), X5).
Afs:
flatA(x1, x2) = flatA(x1)
(3) TriplesToPiDPProof (SOUND transformation)
We use the technique of [DT09]. With regard to the inferred argument filtering the predicates were used in the following modes:
flatA_in: (b,f)
Transforming
TRIPLES into the following
Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
FLATA_IN_GA(.([], .([], X1)), X2) → U1_GA(X1, X2, flatA_in_ga(X1, X2))
FLATA_IN_GA(.([], .([], X1)), X2) → FLATA_IN_GA(X1, X2)
FLATA_IN_GA(.([], .(.(X1, X2), X3)), .(X1, X4)) → U2_GA(X1, X2, X3, X4, flatA_in_ga(.(X2, X3), X4))
FLATA_IN_GA(.([], .(.(X1, X2), X3)), .(X1, X4)) → FLATA_IN_GA(.(X2, X3), X4)
FLATA_IN_GA(.(.(X1, []), X2), .(X1, X3)) → U3_GA(X1, X2, X3, flatA_in_ga(X2, X3))
FLATA_IN_GA(.(.(X1, []), X2), .(X1, X3)) → FLATA_IN_GA(X2, X3)
FLATA_IN_GA(.(.(X1, .(X2, X3)), X4), .(X1, .(X2, X5))) → U4_GA(X1, X2, X3, X4, X5, flatA_in_ga(.(X3, X4), X5))
FLATA_IN_GA(.(.(X1, .(X2, X3)), X4), .(X1, .(X2, X5))) → FLATA_IN_GA(.(X3, X4), X5)
R is empty.
The argument filtering Pi contains the following mapping:
flatA_in_ga(
x1,
x2) =
flatA_in_ga(
x1)
.(
x1,
x2) =
.(
x1,
x2)
[] =
[]
FLATA_IN_GA(
x1,
x2) =
FLATA_IN_GA(
x1)
U1_GA(
x1,
x2,
x3) =
U1_GA(
x1,
x3)
U2_GA(
x1,
x2,
x3,
x4,
x5) =
U2_GA(
x1,
x2,
x3,
x5)
U3_GA(
x1,
x2,
x3,
x4) =
U3_GA(
x1,
x2,
x4)
U4_GA(
x1,
x2,
x3,
x4,
x5,
x6) =
U4_GA(
x1,
x2,
x3,
x4,
x6)
We have to consider all (P,R,Pi)-chains
Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES
(4) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
FLATA_IN_GA(.([], .([], X1)), X2) → U1_GA(X1, X2, flatA_in_ga(X1, X2))
FLATA_IN_GA(.([], .([], X1)), X2) → FLATA_IN_GA(X1, X2)
FLATA_IN_GA(.([], .(.(X1, X2), X3)), .(X1, X4)) → U2_GA(X1, X2, X3, X4, flatA_in_ga(.(X2, X3), X4))
FLATA_IN_GA(.([], .(.(X1, X2), X3)), .(X1, X4)) → FLATA_IN_GA(.(X2, X3), X4)
FLATA_IN_GA(.(.(X1, []), X2), .(X1, X3)) → U3_GA(X1, X2, X3, flatA_in_ga(X2, X3))
FLATA_IN_GA(.(.(X1, []), X2), .(X1, X3)) → FLATA_IN_GA(X2, X3)
FLATA_IN_GA(.(.(X1, .(X2, X3)), X4), .(X1, .(X2, X5))) → U4_GA(X1, X2, X3, X4, X5, flatA_in_ga(.(X3, X4), X5))
FLATA_IN_GA(.(.(X1, .(X2, X3)), X4), .(X1, .(X2, X5))) → FLATA_IN_GA(.(X3, X4), X5)
R is empty.
The argument filtering Pi contains the following mapping:
flatA_in_ga(
x1,
x2) =
flatA_in_ga(
x1)
.(
x1,
x2) =
.(
x1,
x2)
[] =
[]
FLATA_IN_GA(
x1,
x2) =
FLATA_IN_GA(
x1)
U1_GA(
x1,
x2,
x3) =
U1_GA(
x1,
x3)
U2_GA(
x1,
x2,
x3,
x4,
x5) =
U2_GA(
x1,
x2,
x3,
x5)
U3_GA(
x1,
x2,
x3,
x4) =
U3_GA(
x1,
x2,
x4)
U4_GA(
x1,
x2,
x3,
x4,
x5,
x6) =
U4_GA(
x1,
x2,
x3,
x4,
x6)
We have to consider all (P,R,Pi)-chains
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 4 less nodes.
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
FLATA_IN_GA(.([], .(.(X1, X2), X3)), .(X1, X4)) → FLATA_IN_GA(.(X2, X3), X4)
FLATA_IN_GA(.([], .([], X1)), X2) → FLATA_IN_GA(X1, X2)
FLATA_IN_GA(.(.(X1, []), X2), .(X1, X3)) → FLATA_IN_GA(X2, X3)
FLATA_IN_GA(.(.(X1, .(X2, X3)), X4), .(X1, .(X2, X5))) → FLATA_IN_GA(.(X3, X4), X5)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
[] =
[]
FLATA_IN_GA(
x1,
x2) =
FLATA_IN_GA(
x1)
We have to consider all (P,R,Pi)-chains
(7) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FLATA_IN_GA(.([], .(.(X1, X2), X3))) → FLATA_IN_GA(.(X2, X3))
FLATA_IN_GA(.([], .([], X1))) → FLATA_IN_GA(X1)
FLATA_IN_GA(.(.(X1, []), X2)) → FLATA_IN_GA(X2)
FLATA_IN_GA(.(.(X1, .(X2, X3)), X4)) → FLATA_IN_GA(.(X3, X4))
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(9) 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:
FLATA_IN_GA(.([], .(.(X1, X2), X3))) → FLATA_IN_GA(.(X2, X3))
FLATA_IN_GA(.([], .([], X1))) → FLATA_IN_GA(X1)
FLATA_IN_GA(.(.(X1, []), X2)) → FLATA_IN_GA(X2)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [POLO]:
POL(.(x1, x2)) = 2·x1 + 2·x2
POL(FLATA_IN_GA(x1)) = 2·x1
POL([]) = 0
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FLATA_IN_GA(.(.(X1, .(X2, X3)), X4)) → FLATA_IN_GA(.(X3, X4))
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(11) 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:
FLATA_IN_GA(.(.(X1, .(X2, X3)), X4)) → FLATA_IN_GA(.(X3, X4))
Used ordering: Polynomial interpretation [POLO]:
POL(.(x1, x2)) = 1 + 2·x1 + 2·x2
POL(FLATA_IN_GA(x1)) = 2·x1
(12) Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(13) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(14) YES