(0) Obligation:
Clauses:
prefix(Xs, Ys) :- app(Xs, X1, Ys).
app([], X, X).
app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).
Query: prefix(g,a)
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph DT10.
(2) Obligation:
Triples:
appA(.(X1, X2), X3, .(X1, X4)) :- appA(X2, X3, X4).
prefixB(X1, X2) :- appA(X1, X3, X2).
Clauses:
appcA([], X1, X1).
appcA(.(X1, X2), X3, .(X1, X4)) :- appcA(X2, X3, X4).
Afs:
prefixB(x1, x2) = prefixB(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:
prefixB_in: (b,f)
appA_in: (b,f,f)
Transforming
TRIPLES into the following
Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
PREFIXB_IN_GA(X1, X2) → U2_GA(X1, X2, appA_in_gaa(X1, X3, X2))
PREFIXB_IN_GA(X1, X2) → APPA_IN_GAA(X1, X3, X2)
APPA_IN_GAA(.(X1, X2), X3, .(X1, X4)) → U1_GAA(X1, X2, X3, X4, appA_in_gaa(X2, X3, X4))
APPA_IN_GAA(.(X1, X2), X3, .(X1, X4)) → APPA_IN_GAA(X2, X3, X4)
R is empty.
The argument filtering Pi contains the following mapping:
appA_in_gaa(
x1,
x2,
x3) =
appA_in_gaa(
x1)
.(
x1,
x2) =
.(
x1,
x2)
PREFIXB_IN_GA(
x1,
x2) =
PREFIXB_IN_GA(
x1)
U2_GA(
x1,
x2,
x3) =
U2_GA(
x1,
x3)
APPA_IN_GAA(
x1,
x2,
x3) =
APPA_IN_GAA(
x1)
U1_GAA(
x1,
x2,
x3,
x4,
x5) =
U1_GAA(
x1,
x2,
x5)
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:
PREFIXB_IN_GA(X1, X2) → U2_GA(X1, X2, appA_in_gaa(X1, X3, X2))
PREFIXB_IN_GA(X1, X2) → APPA_IN_GAA(X1, X3, X2)
APPA_IN_GAA(.(X1, X2), X3, .(X1, X4)) → U1_GAA(X1, X2, X3, X4, appA_in_gaa(X2, X3, X4))
APPA_IN_GAA(.(X1, X2), X3, .(X1, X4)) → APPA_IN_GAA(X2, X3, X4)
R is empty.
The argument filtering Pi contains the following mapping:
appA_in_gaa(
x1,
x2,
x3) =
appA_in_gaa(
x1)
.(
x1,
x2) =
.(
x1,
x2)
PREFIXB_IN_GA(
x1,
x2) =
PREFIXB_IN_GA(
x1)
U2_GA(
x1,
x2,
x3) =
U2_GA(
x1,
x3)
APPA_IN_GAA(
x1,
x2,
x3) =
APPA_IN_GAA(
x1)
U1_GAA(
x1,
x2,
x3,
x4,
x5) =
U1_GAA(
x1,
x2,
x5)
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:
APPA_IN_GAA(.(X1, X2), X3, .(X1, X4)) → APPA_IN_GAA(X2, X3, X4)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
APPA_IN_GAA(
x1,
x2,
x3) =
APPA_IN_GAA(
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:
APPA_IN_GAA(.(X1, X2)) → APPA_IN_GAA(X2)
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:
- APPA_IN_GAA(.(X1, X2)) → APPA_IN_GAA(X2)
The graph contains the following edges 1 > 1
(10) YES