(0) Obligation:
Clauses:
reverse(X1s, X2s) :- reverse(X1s, [], X2s).
reverse([], Xs, Xs).
reverse(.(X, X1s), X2s, Ys) :- reverse(X1s, .(X, X2s), Ys).
Query: reverse(g,a)
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph DT10.
(2) Obligation:
Triples:
reverseA(.(X1, X2), X3, X4, X5) :- reverseA(X2, X1, .(X3, X4), X5).
reverseB(.(X1, .(X2, .(X3, .(X4, .(X5, .(X6, .(X7, .(X8, X9)))))))), X10) :- reverseA(X9, X8, .(X7, .(X6, .(X5, .(X4, .(X3, .(X2, .(X1, []))))))), X10).
Clauses:
reversecA([], X1, X2, .(X1, X2)).
reversecA(.(X1, X2), X3, X4, X5) :- reversecA(X2, X1, .(X3, X4), X5).
Afs:
reverseB(x1, x2) = reverseB(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:
reverseB_in: (b,f)
reverseA_in: (b,b,b,f)
Transforming
TRIPLES into the following
Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
REVERSEB_IN_GA(.(X1, .(X2, .(X3, .(X4, .(X5, .(X6, .(X7, .(X8, X9)))))))), X10) → U2_GA(X1, X2, X3, X4, X5, X6, X7, X8, X9, X10, reverseA_in_ggga(X9, X8, .(X7, .(X6, .(X5, .(X4, .(X3, .(X2, .(X1, []))))))), X10))
REVERSEB_IN_GA(.(X1, .(X2, .(X3, .(X4, .(X5, .(X6, .(X7, .(X8, X9)))))))), X10) → REVERSEA_IN_GGGA(X9, X8, .(X7, .(X6, .(X5, .(X4, .(X3, .(X2, .(X1, []))))))), X10)
REVERSEA_IN_GGGA(.(X1, X2), X3, X4, X5) → U1_GGGA(X1, X2, X3, X4, X5, reverseA_in_ggga(X2, X1, .(X3, X4), X5))
REVERSEA_IN_GGGA(.(X1, X2), X3, X4, X5) → REVERSEA_IN_GGGA(X2, X1, .(X3, X4), X5)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
reverseA_in_ggga(
x1,
x2,
x3,
x4) =
reverseA_in_ggga(
x1,
x2,
x3)
[] =
[]
REVERSEB_IN_GA(
x1,
x2) =
REVERSEB_IN_GA(
x1)
U2_GA(
x1,
x2,
x3,
x4,
x5,
x6,
x7,
x8,
x9,
x10,
x11) =
U2_GA(
x1,
x2,
x3,
x4,
x5,
x6,
x7,
x8,
x9,
x11)
REVERSEA_IN_GGGA(
x1,
x2,
x3,
x4) =
REVERSEA_IN_GGGA(
x1,
x2,
x3)
U1_GGGA(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_GGGA(
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:
REVERSEB_IN_GA(.(X1, .(X2, .(X3, .(X4, .(X5, .(X6, .(X7, .(X8, X9)))))))), X10) → U2_GA(X1, X2, X3, X4, X5, X6, X7, X8, X9, X10, reverseA_in_ggga(X9, X8, .(X7, .(X6, .(X5, .(X4, .(X3, .(X2, .(X1, []))))))), X10))
REVERSEB_IN_GA(.(X1, .(X2, .(X3, .(X4, .(X5, .(X6, .(X7, .(X8, X9)))))))), X10) → REVERSEA_IN_GGGA(X9, X8, .(X7, .(X6, .(X5, .(X4, .(X3, .(X2, .(X1, []))))))), X10)
REVERSEA_IN_GGGA(.(X1, X2), X3, X4, X5) → U1_GGGA(X1, X2, X3, X4, X5, reverseA_in_ggga(X2, X1, .(X3, X4), X5))
REVERSEA_IN_GGGA(.(X1, X2), X3, X4, X5) → REVERSEA_IN_GGGA(X2, X1, .(X3, X4), X5)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
reverseA_in_ggga(
x1,
x2,
x3,
x4) =
reverseA_in_ggga(
x1,
x2,
x3)
[] =
[]
REVERSEB_IN_GA(
x1,
x2) =
REVERSEB_IN_GA(
x1)
U2_GA(
x1,
x2,
x3,
x4,
x5,
x6,
x7,
x8,
x9,
x10,
x11) =
U2_GA(
x1,
x2,
x3,
x4,
x5,
x6,
x7,
x8,
x9,
x11)
REVERSEA_IN_GGGA(
x1,
x2,
x3,
x4) =
REVERSEA_IN_GGGA(
x1,
x2,
x3)
U1_GGGA(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_GGGA(
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 3 less nodes.
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
REVERSEA_IN_GGGA(.(X1, X2), X3, X4, X5) → REVERSEA_IN_GGGA(X2, X1, .(X3, X4), X5)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
REVERSEA_IN_GGGA(
x1,
x2,
x3,
x4) =
REVERSEA_IN_GGGA(
x1,
x2,
x3)
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:
REVERSEA_IN_GGGA(.(X1, X2), X3, X4) → REVERSEA_IN_GGGA(X2, X1, .(X3, X4))
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:
- REVERSEA_IN_GGGA(.(X1, X2), X3, X4) → REVERSEA_IN_GGGA(X2, X1, .(X3, X4))
The graph contains the following edges 1 > 1, 1 > 2
(10) YES