(0) Obligation:
Clauses:
reverse(X1s, X2s) :- reverse(X1s, [], X2s).
reverse([], Xs, Xs).
reverse(.(X, X1s), X2s, Ys) :- reverse(X1s, .(X, X2s), Ys).
Queries:
reverse(a,g).
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph.
(2) Obligation:
Triples:
reverse68(.(T451, T450), T452, T453, T449) :- reverse68(T450, T451, .(T452, T453), T449).
reverse1(.(T403, .(T402, .(T401, .(T400, .(T399, .(T398, .(T397, .(T396, T395)))))))), T394) :- reverse68(T395, T396, .(T397, .(T398, .(T399, .(T400, .(T401, .(T402, .(T403, []))))))), T394).
Clauses:
reversec68([], T433, T434, .(T433, T434)).
reversec68(.(T451, T450), T452, T453, T449) :- reversec68(T450, T451, .(T452, T453), T449).
Afs:
reverse1(x1, x2) = reverse1(x2)
(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:
reverse1_in: (f,b)
reverse68_in: (f,f,b,b)
Transforming
TRIPLES into the following
Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
REVERSE1_IN_AG(.(T403, .(T402, .(T401, .(T400, .(T399, .(T398, .(T397, .(T396, T395)))))))), T394) → U2_AG(T403, T402, T401, T400, T399, T398, T397, T396, T395, T394, reverse68_in_aagg(T395, T396, .(T397, .(T398, .(T399, .(T400, .(T401, .(T402, .(T403, []))))))), T394))
REVERSE1_IN_AG(.(T403, .(T402, .(T401, .(T400, .(T399, .(T398, .(T397, .(T396, T395)))))))), T394) → REVERSE68_IN_AAGG(T395, T396, .(T397, .(T398, .(T399, .(T400, .(T401, .(T402, .(T403, []))))))), T394)
REVERSE68_IN_AAGG(.(T451, T450), T452, T453, T449) → U1_AAGG(T451, T450, T452, T453, T449, reverse68_in_aagg(T450, T451, .(T452, T453), T449))
REVERSE68_IN_AAGG(.(T451, T450), T452, T453, T449) → REVERSE68_IN_AAGG(T450, T451, .(T452, T453), T449)
R is empty.
The argument filtering Pi contains the following mapping:
reverse68_in_aagg(
x1,
x2,
x3,
x4) =
reverse68_in_aagg(
x3,
x4)
.(
x1,
x2) =
.(
x2)
[] =
[]
REVERSE1_IN_AG(
x1,
x2) =
REVERSE1_IN_AG(
x2)
U2_AG(
x1,
x2,
x3,
x4,
x5,
x6,
x7,
x8,
x9,
x10,
x11) =
U2_AG(
x10,
x11)
REVERSE68_IN_AAGG(
x1,
x2,
x3,
x4) =
REVERSE68_IN_AAGG(
x3,
x4)
U1_AAGG(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_AAGG(
x4,
x5,
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:
REVERSE1_IN_AG(.(T403, .(T402, .(T401, .(T400, .(T399, .(T398, .(T397, .(T396, T395)))))))), T394) → U2_AG(T403, T402, T401, T400, T399, T398, T397, T396, T395, T394, reverse68_in_aagg(T395, T396, .(T397, .(T398, .(T399, .(T400, .(T401, .(T402, .(T403, []))))))), T394))
REVERSE1_IN_AG(.(T403, .(T402, .(T401, .(T400, .(T399, .(T398, .(T397, .(T396, T395)))))))), T394) → REVERSE68_IN_AAGG(T395, T396, .(T397, .(T398, .(T399, .(T400, .(T401, .(T402, .(T403, []))))))), T394)
REVERSE68_IN_AAGG(.(T451, T450), T452, T453, T449) → U1_AAGG(T451, T450, T452, T453, T449, reverse68_in_aagg(T450, T451, .(T452, T453), T449))
REVERSE68_IN_AAGG(.(T451, T450), T452, T453, T449) → REVERSE68_IN_AAGG(T450, T451, .(T452, T453), T449)
R is empty.
The argument filtering Pi contains the following mapping:
reverse68_in_aagg(
x1,
x2,
x3,
x4) =
reverse68_in_aagg(
x3,
x4)
.(
x1,
x2) =
.(
x2)
[] =
[]
REVERSE1_IN_AG(
x1,
x2) =
REVERSE1_IN_AG(
x2)
U2_AG(
x1,
x2,
x3,
x4,
x5,
x6,
x7,
x8,
x9,
x10,
x11) =
U2_AG(
x10,
x11)
REVERSE68_IN_AAGG(
x1,
x2,
x3,
x4) =
REVERSE68_IN_AAGG(
x3,
x4)
U1_AAGG(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_AAGG(
x4,
x5,
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:
REVERSE68_IN_AAGG(.(T451, T450), T452, T453, T449) → REVERSE68_IN_AAGG(T450, T451, .(T452, T453), T449)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x2)
REVERSE68_IN_AAGG(
x1,
x2,
x3,
x4) =
REVERSE68_IN_AAGG(
x3,
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:
REVERSE68_IN_AAGG(T453, T449) → REVERSE68_IN_AAGG(.(T453), T449)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(9) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
REVERSE68_IN_AAGG(
T453,
T449) →
REVERSE68_IN_AAGG(
.(
T453),
T449) we obtained the following new rules [LPAR04]:
REVERSE68_IN_AAGG(.(z0), z1) → REVERSE68_IN_AAGG(.(.(z0)), z1)
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
REVERSE68_IN_AAGG(.(z0), z1) → REVERSE68_IN_AAGG(.(.(z0)), z1)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(11) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
REVERSE68_IN_AAGG(
.(
z0),
z1) →
REVERSE68_IN_AAGG(
.(
.(
z0)),
z1) we obtained the following new rules [LPAR04]:
REVERSE68_IN_AAGG(.(.(z0)), z1) → REVERSE68_IN_AAGG(.(.(.(z0))), z1)
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
REVERSE68_IN_AAGG(.(.(z0)), z1) → REVERSE68_IN_AAGG(.(.(.(z0))), z1)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(13) 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 =
REVERSE68_IN_AAGG(
.(
.(
z0)),
z1) evaluates to t =
REVERSE68_IN_AAGG(
.(
.(
.(
z0))),
z1)
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Matcher: [z0 / .(z0)]
- Semiunifier: [ ]
Rewriting sequenceThe DP semiunifies directly so there is only one rewrite step from REVERSE68_IN_AAGG(.(.(z0)), z1) to REVERSE68_IN_AAGG(.(.(.(z0))), z1).
(14) NO