(0) Obligation:
Clauses:fl([], [], 0).
fl(.(E, X), R, s(Z)) :- ','(append(E, Y, R), fl(X, Y, Z)).
append([], X, X).
append(.(X, Xs), Ys, .(X, Zs)) :- append(Xs, Ys, Zs).
Queries:
fl(g,a,a).
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph.
(2) Obligation:
Triples:p7([], T19, T19, T9, T20) :- fl1(T9, T19, T20).
p7(.(T27, T28), X43, .(T27, T30), T9, T31) :- p7(T28, X43, T30, T9, T31).
fl1(.(T8, T9), T12, s(T13)) :- p7(T8, X13, T12, T9, T13).
Clauses:
flc1([], [], 0).
flc1(.(T8, T9), T12, s(T13)) :- qc7(T8, X13, T12, T9, T13).
qc7([], T19, T19, T9, T20) :- flc1(T9, T19, T20).
qc7(.(T27, T28), X43, .(T27, T30), T9, T31) :- qc7(T28, X43, T30, T9, T31).
Afs:
fl1(x1, x2, x3) = fl1(x1)
(3) TriplesToPiDPProof (SOUND transformation)
We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:fl1_in: (b,f,f)
p7_in: (b,f,f,b,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
FL1_IN_GAA(.(T8, T9), T12, s(T13)) → U3_GAA(T8, T9, T12, T13, p7_in_gaaga(T8, X13, T12, T9, T13))
FL1_IN_GAA(.(T8, T9), T12, s(T13)) → P7_IN_GAAGA(T8, X13, T12, T9, T13)
P7_IN_GAAGA([], T19, T19, T9, T20) → U1_GAAGA(T19, T9, T20, fl1_in_gaa(T9, T19, T20))
P7_IN_GAAGA([], T19, T19, T9, T20) → FL1_IN_GAA(T9, T19, T20)
P7_IN_GAAGA(.(T27, T28), X43, .(T27, T30), T9, T31) → U2_GAAGA(T27, T28, X43, T30, T9, T31, p7_in_gaaga(T28, X43, T30, T9, T31))
P7_IN_GAAGA(.(T27, T28), X43, .(T27, T30), T9, T31) → P7_IN_GAAGA(T28, X43, T30, T9, T31)
R is empty.
The argument filtering Pi contains the following mapping:
fl1_in_gaa(x1, x2, x3) = fl1_in_gaa(x1)
.(x1, x2) = .(x1, x2)
p7_in_gaaga(x1, x2, x3, x4, x5) = p7_in_gaaga(x1, x4)
[] = []
s(x1) = s(x1)
FL1_IN_GAA(x1, x2, x3) = FL1_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4, x5) = U3_GAA(x1, x2, x5)
P7_IN_GAAGA(x1, x2, x3, x4, x5) = P7_IN_GAAGA(x1, x4)
U1_GAAGA(x1, x2, x3, x4) = U1_GAAGA(x2, x4)
U2_GAAGA(x1, x2, x3, x4, x5, x6, x7) = U2_GAAGA(x1, x2, x5, x7)
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:
FL1_IN_GAA(.(T8, T9), T12, s(T13)) → U3_GAA(T8, T9, T12, T13, p7_in_gaaga(T8, X13, T12, T9, T13))
FL1_IN_GAA(.(T8, T9), T12, s(T13)) → P7_IN_GAAGA(T8, X13, T12, T9, T13)
P7_IN_GAAGA([], T19, T19, T9, T20) → U1_GAAGA(T19, T9, T20, fl1_in_gaa(T9, T19, T20))
P7_IN_GAAGA([], T19, T19, T9, T20) → FL1_IN_GAA(T9, T19, T20)
P7_IN_GAAGA(.(T27, T28), X43, .(T27, T30), T9, T31) → U2_GAAGA(T27, T28, X43, T30, T9, T31, p7_in_gaaga(T28, X43, T30, T9, T31))
P7_IN_GAAGA(.(T27, T28), X43, .(T27, T30), T9, T31) → P7_IN_GAAGA(T28, X43, T30, T9, T31)
R is empty.
The argument filtering Pi contains the following mapping:
fl1_in_gaa(x1, x2, x3) = fl1_in_gaa(x1)
.(x1, x2) = .(x1, x2)
p7_in_gaaga(x1, x2, x3, x4, x5) = p7_in_gaaga(x1, x4)
[] = []
s(x1) = s(x1)
FL1_IN_GAA(x1, x2, x3) = FL1_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4, x5) = U3_GAA(x1, x2, x5)
P7_IN_GAAGA(x1, x2, x3, x4, x5) = P7_IN_GAAGA(x1, x4)
U1_GAAGA(x1, x2, x3, x4) = U1_GAAGA(x2, x4)
U2_GAAGA(x1, x2, x3, x4, x5, x6, x7) = U2_GAAGA(x1, x2, x5, 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:
FL1_IN_GAA(.(T8, T9), T12, s(T13)) → P7_IN_GAAGA(T8, X13, T12, T9, T13)
P7_IN_GAAGA([], T19, T19, T9, T20) → FL1_IN_GAA(T9, T19, T20)
P7_IN_GAAGA(.(T27, T28), X43, .(T27, T30), T9, T31) → P7_IN_GAAGA(T28, X43, T30, T9, T31)
R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
[] = []
s(x1) = s(x1)
FL1_IN_GAA(x1, x2, x3) = FL1_IN_GAA(x1)
P7_IN_GAAGA(x1, x2, x3, x4, x5) = P7_IN_GAAGA(x1, x4)
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:
FL1_IN_GAA(.(T8, T9)) → P7_IN_GAAGA(T8, T9)
P7_IN_GAAGA([], T9) → FL1_IN_GAA(T9)
P7_IN_GAAGA(.(T27, T28), T9) → P7_IN_GAAGA(T28, T9)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(9) 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:
- P7_IN_GAAGA([], T9) → FL1_IN_GAA(T9)
The graph contains the following edges 2 >= 1 - P7_IN_GAAGA(.(T27, T28), T9) → P7_IN_GAAGA(T28, T9)
The graph contains the following edges 1 > 1, 2 >= 2 - FL1_IN_GAA(.(T8, T9)) → P7_IN_GAAGA(T8, T9)
The graph contains the following edges 1 > 1, 1 > 2