(0) Obligation:

Clauses:

palindrome(Xs) :- reverse(Xs, Xs).
reverse(X1s, X2s) :- reverse(X1s, [], X2s).
reverse([], Xs, Xs).
reverse(.(X, X1s), X2s, Ys) :- reverse(X1s, .(X, X2s), Ys).

Query: palindrome(g)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

(2) Obligation:

Triples:

reverseA(.(X1, X2), X3, X4, X5, X6) :- reverseA(X2, X1, .(X3, X4), X5, X6).
palindromeB(.(X1, .(X2, .(X3, .(X4, .(X5, .(X6, .(X7, .(X8, X9))))))))) :- reverseA(X9, X8, .(X7, .(X6, .(X5, .(X4, .(X3, .(X2, .(X1, []))))))), X1, .(X2, .(X3, .(X4, .(X5, .(X6, .(X7, .(X8, X9)))))))).

Clauses:

reversecA([], X1, X2, X1, X2).
reversecA(.(X1, X2), X3, X4, X5, X6) :- reversecA(X2, X1, .(X3, X4), X5, X6).

Afs:

palindromeB(x1)  =  palindromeB(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:
palindromeB_in: (b)
reverseA_in: (b,b,b,b,b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

PALINDROMEB_IN_G(.(X1, .(X2, .(X3, .(X4, .(X5, .(X6, .(X7, .(X8, X9))))))))) → U2_G(X1, X2, X3, X4, X5, X6, X7, X8, X9, reverseA_in_ggggg(X9, X8, .(X7, .(X6, .(X5, .(X4, .(X3, .(X2, .(X1, []))))))), X1, .(X2, .(X3, .(X4, .(X5, .(X6, .(X7, .(X8, X9)))))))))
PALINDROMEB_IN_G(.(X1, .(X2, .(X3, .(X4, .(X5, .(X6, .(X7, .(X8, X9))))))))) → REVERSEA_IN_GGGGG(X9, X8, .(X7, .(X6, .(X5, .(X4, .(X3, .(X2, .(X1, []))))))), X1, .(X2, .(X3, .(X4, .(X5, .(X6, .(X7, .(X8, X9))))))))
REVERSEA_IN_GGGGG(.(X1, X2), X3, X4, X5, X6) → U1_GGGGG(X1, X2, X3, X4, X5, X6, reverseA_in_ggggg(X2, X1, .(X3, X4), X5, X6))
REVERSEA_IN_GGGGG(.(X1, X2), X3, X4, X5, X6) → REVERSEA_IN_GGGGG(X2, X1, .(X3, X4), X5, X6)

R is empty.
Pi is empty.
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:

PALINDROMEB_IN_G(.(X1, .(X2, .(X3, .(X4, .(X5, .(X6, .(X7, .(X8, X9))))))))) → U2_G(X1, X2, X3, X4, X5, X6, X7, X8, X9, reverseA_in_ggggg(X9, X8, .(X7, .(X6, .(X5, .(X4, .(X3, .(X2, .(X1, []))))))), X1, .(X2, .(X3, .(X4, .(X5, .(X6, .(X7, .(X8, X9)))))))))
PALINDROMEB_IN_G(.(X1, .(X2, .(X3, .(X4, .(X5, .(X6, .(X7, .(X8, X9))))))))) → REVERSEA_IN_GGGGG(X9, X8, .(X7, .(X6, .(X5, .(X4, .(X3, .(X2, .(X1, []))))))), X1, .(X2, .(X3, .(X4, .(X5, .(X6, .(X7, .(X8, X9))))))))
REVERSEA_IN_GGGGG(.(X1, X2), X3, X4, X5, X6) → U1_GGGGG(X1, X2, X3, X4, X5, X6, reverseA_in_ggggg(X2, X1, .(X3, X4), X5, X6))
REVERSEA_IN_GGGGG(.(X1, X2), X3, X4, X5, X6) → REVERSEA_IN_GGGGG(X2, X1, .(X3, X4), X5, X6)

R is empty.
Pi is empty.
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_GGGGG(.(X1, X2), X3, X4, X5, X6) → REVERSEA_IN_GGGGG(X2, X1, .(X3, X4), X5, X6)

R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains

(7) PiDPToQDPProof (EQUIVALENT 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_GGGGG(.(X1, X2), X3, X4, X5, X6) → REVERSEA_IN_GGGGG(X2, X1, .(X3, X4), X5, X6)

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_GGGGG(.(X1, X2), X3, X4, X5, X6) → REVERSEA_IN_GGGGG(X2, X1, .(X3, X4), X5, X6)
    The graph contains the following edges 1 > 1, 1 > 2, 4 >= 4, 5 >= 5

(10) YES