(0) Obligation:
Clauses:
reverse([], X, X).
reverse(.(X, Y), Z, U) :- reverse(Y, Z, .(X, U)).
Query: reverse(g,a,g)
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph DT10.
(2) Obligation:
Triples:
reverseA(.(X1, .(X2, X3)), X4, X5) :- reverseA(X3, X4, .(X2, .(X1, X5))).
Clauses:
reversecA([], X1, X1).
reversecA(.(X1, []), .(X1, X2), X2).
reversecA(.(X1, .(X2, X3)), X4, X5) :- reversecA(X3, X4, .(X2, .(X1, X5))).
Afs:
reverseA(x1, x2, x3) = reverseA(x1, x3)
(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:
reverseA_in: (b,f,b)
Transforming
TRIPLES into the following
Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
REVERSEA_IN_GAG(.(X1, .(X2, X3)), X4, X5) → U1_GAG(X1, X2, X3, X4, X5, reverseA_in_gag(X3, X4, .(X2, .(X1, X5))))
REVERSEA_IN_GAG(.(X1, .(X2, X3)), X4, X5) → REVERSEA_IN_GAG(X3, X4, .(X2, .(X1, X5)))
R is empty.
The argument filtering Pi contains the following mapping:
reverseA_in_gag(
x1,
x2,
x3) =
reverseA_in_gag(
x1,
x3)
.(
x1,
x2) =
.(
x1,
x2)
REVERSEA_IN_GAG(
x1,
x2,
x3) =
REVERSEA_IN_GAG(
x1,
x3)
U1_GAG(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_GAG(
x1,
x2,
x3,
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:
REVERSEA_IN_GAG(.(X1, .(X2, X3)), X4, X5) → U1_GAG(X1, X2, X3, X4, X5, reverseA_in_gag(X3, X4, .(X2, .(X1, X5))))
REVERSEA_IN_GAG(.(X1, .(X2, X3)), X4, X5) → REVERSEA_IN_GAG(X3, X4, .(X2, .(X1, X5)))
R is empty.
The argument filtering Pi contains the following mapping:
reverseA_in_gag(
x1,
x2,
x3) =
reverseA_in_gag(
x1,
x3)
.(
x1,
x2) =
.(
x1,
x2)
REVERSEA_IN_GAG(
x1,
x2,
x3) =
REVERSEA_IN_GAG(
x1,
x3)
U1_GAG(
x1,
x2,
x3,
x4,
x5,
x6) =
U1_GAG(
x1,
x2,
x3,
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 1 less node.
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
REVERSEA_IN_GAG(.(X1, .(X2, X3)), X4, X5) → REVERSEA_IN_GAG(X3, X4, .(X2, .(X1, X5)))
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
REVERSEA_IN_GAG(
x1,
x2,
x3) =
REVERSEA_IN_GAG(
x1,
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_GAG(.(X1, .(X2, X3)), X5) → REVERSEA_IN_GAG(X3, .(X2, .(X1, X5)))
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_GAG(.(X1, .(X2, X3)), X5) → REVERSEA_IN_GAG(X3, .(X2, .(X1, X5)))
The graph contains the following edges 1 > 1
(10) YES