(0) Obligation:
Clauses:sublist(Xs, Ys) :- ','(app(X1, Zs, Ys), app(Xs, X2, Zs)).
app([], X, X).
app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).
Queries:
sublist(g,a).
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph.
(2) Obligation:
Triples:app7(.(T33, T34), X57, .(T33, T36)) :- app7(T34, X57, T36).
p3([], T19, T19, T5, X9) :- app7(T5, X9, T19).
p3(.(X84, X85), X86, .(X84, T42), T5, X9) :- p3(X85, X86, T42, T5, X9).
sublist1(T5, T7) :- p3(X7, X8, T7, T5, X9).
Clauses:
appc7([], T26, T26).
appc7(.(T33, T34), X57, .(T33, T36)) :- appc7(T34, X57, T36).
qc3([], T19, T19, T5, X9) :- appc7(T5, X9, T19).
qc3(.(X84, X85), X86, .(X84, T42), T5, X9) :- qc3(X85, X86, T42, T5, X9).
Afs:
sublist1(x1, x2) = sublist1(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:sublist1_in: (b,f)
p3_in: (f,f,f,b,f)
app7_in: (b,f,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
SUBLIST1_IN_GA(T5, T7) → U4_GA(T5, T7, p3_in_aaaga(X7, X8, T7, T5, X9))
SUBLIST1_IN_GA(T5, T7) → P3_IN_AAAGA(X7, X8, T7, T5, X9)
P3_IN_AAAGA([], T19, T19, T5, X9) → U2_AAAGA(T19, T5, X9, app7_in_gaa(T5, X9, T19))
P3_IN_AAAGA([], T19, T19, T5, X9) → APP7_IN_GAA(T5, X9, T19)
APP7_IN_GAA(.(T33, T34), X57, .(T33, T36)) → U1_GAA(T33, T34, X57, T36, app7_in_gaa(T34, X57, T36))
APP7_IN_GAA(.(T33, T34), X57, .(T33, T36)) → APP7_IN_GAA(T34, X57, T36)
P3_IN_AAAGA(.(X84, X85), X86, .(X84, T42), T5, X9) → U3_AAAGA(X84, X85, X86, T42, T5, X9, p3_in_aaaga(X85, X86, T42, T5, X9))
P3_IN_AAAGA(.(X84, X85), X86, .(X84, T42), T5, X9) → P3_IN_AAAGA(X85, X86, T42, T5, X9)
R is empty.
The argument filtering Pi contains the following mapping:
p3_in_aaaga(x1, x2, x3, x4, x5) = p3_in_aaaga(x4)
app7_in_gaa(x1, x2, x3) = app7_in_gaa(x1)
.(x1, x2) = .(x2)
[] = []
SUBLIST1_IN_GA(x1, x2) = SUBLIST1_IN_GA(x1)
U4_GA(x1, x2, x3) = U4_GA(x1, x3)
P3_IN_AAAGA(x1, x2, x3, x4, x5) = P3_IN_AAAGA(x4)
U2_AAAGA(x1, x2, x3, x4) = U2_AAAGA(x2, x4)
APP7_IN_GAA(x1, x2, x3) = APP7_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5) = U1_GAA(x2, x5)
U3_AAAGA(x1, x2, x3, x4, x5, x6, x7) = U3_AAAGA(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:
SUBLIST1_IN_GA(T5, T7) → U4_GA(T5, T7, p3_in_aaaga(X7, X8, T7, T5, X9))
SUBLIST1_IN_GA(T5, T7) → P3_IN_AAAGA(X7, X8, T7, T5, X9)
P3_IN_AAAGA([], T19, T19, T5, X9) → U2_AAAGA(T19, T5, X9, app7_in_gaa(T5, X9, T19))
P3_IN_AAAGA([], T19, T19, T5, X9) → APP7_IN_GAA(T5, X9, T19)
APP7_IN_GAA(.(T33, T34), X57, .(T33, T36)) → U1_GAA(T33, T34, X57, T36, app7_in_gaa(T34, X57, T36))
APP7_IN_GAA(.(T33, T34), X57, .(T33, T36)) → APP7_IN_GAA(T34, X57, T36)
P3_IN_AAAGA(.(X84, X85), X86, .(X84, T42), T5, X9) → U3_AAAGA(X84, X85, X86, T42, T5, X9, p3_in_aaaga(X85, X86, T42, T5, X9))
P3_IN_AAAGA(.(X84, X85), X86, .(X84, T42), T5, X9) → P3_IN_AAAGA(X85, X86, T42, T5, X9)
R is empty.
The argument filtering Pi contains the following mapping:
p3_in_aaaga(x1, x2, x3, x4, x5) = p3_in_aaaga(x4)
app7_in_gaa(x1, x2, x3) = app7_in_gaa(x1)
.(x1, x2) = .(x2)
[] = []
SUBLIST1_IN_GA(x1, x2) = SUBLIST1_IN_GA(x1)
U4_GA(x1, x2, x3) = U4_GA(x1, x3)
P3_IN_AAAGA(x1, x2, x3, x4, x5) = P3_IN_AAAGA(x4)
U2_AAAGA(x1, x2, x3, x4) = U2_AAAGA(x2, x4)
APP7_IN_GAA(x1, x2, x3) = APP7_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5) = U1_GAA(x2, x5)
U3_AAAGA(x1, x2, x3, x4, x5, x6, x7) = U3_AAAGA(x5, x7)
We have to consider all (P,R,Pi)-chains
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 6 less nodes.(6) Complex Obligation (AND)
(7) Obligation:
Pi DP problem:The TRS P consists of the following rules:
APP7_IN_GAA(.(T33, T34), X57, .(T33, T36)) → APP7_IN_GAA(T34, X57, T36)
R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x2)
APP7_IN_GAA(x1, x2, x3) = APP7_IN_GAA(x1)
We have to consider all (P,R,Pi)-chains
(8) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.(9) Obligation:
Q DP problem:The TRS P consists of the following rules:
APP7_IN_GAA(.(T34)) → APP7_IN_GAA(T34)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(10) 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:
- APP7_IN_GAA(.(T34)) → APP7_IN_GAA(T34)
The graph contains the following edges 1 > 1
(11) YES
(12) Obligation:
Pi DP problem:The TRS P consists of the following rules:
P3_IN_AAAGA(.(X84, X85), X86, .(X84, T42), T5, X9) → P3_IN_AAAGA(X85, X86, T42, T5, X9)
R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x2)
P3_IN_AAAGA(x1, x2, x3, x4, x5) = P3_IN_AAAGA(x4)
We have to consider all (P,R,Pi)-chains
(13) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.(14) Obligation:
Q DP problem:The TRS P consists of the following rules:
P3_IN_AAAGA(T5) → P3_IN_AAAGA(T5)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(15) 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 = P3_IN_AAAGA(T5) evaluates to t =P3_IN_AAAGA(T5)
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 P3_IN_AAAGA(T5) to P3_IN_AAAGA(T5).