(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).

Queries:

palindrome(g).

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph.

(2) Obligation:

Triples:

reverse70(.(T379, T380), T381, T382, T383, T384) :- reverse70(T380, T379, .(T381, T382), T383, T384).
palindrome1(.(T334, .(T333, .(T332, .(T331, .(T330, .(T329, .(T328, .(T326, T327))))))))) :- reverse70(T327, T326, .(T328, .(T329, .(T330, .(T331, .(T332, .(T333, .(T334, []))))))), T334, .(T333, .(T332, .(T331, .(T330, .(T329, .(T328, .(T326, T327)))))))).

Clauses:

reversec70([], T365, T366, T365, T366).
reversec70(.(T379, T380), T381, T382, T383, T384) :- reversec70(T380, T379, .(T381, T382), T383, T384).

Afs:

palindrome1(x1)  =  palindrome1(x1)

(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:
palindrome1_in: (b)
reverse70_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:

PALINDROME1_IN_G(.(T334, .(T333, .(T332, .(T331, .(T330, .(T329, .(T328, .(T326, T327))))))))) → U2_G(T334, T333, T332, T331, T330, T329, T328, T326, T327, reverse70_in_ggggg(T327, T326, .(T328, .(T329, .(T330, .(T331, .(T332, .(T333, .(T334, []))))))), T334, .(T333, .(T332, .(T331, .(T330, .(T329, .(T328, .(T326, T327)))))))))
PALINDROME1_IN_G(.(T334, .(T333, .(T332, .(T331, .(T330, .(T329, .(T328, .(T326, T327))))))))) → REVERSE70_IN_GGGGG(T327, T326, .(T328, .(T329, .(T330, .(T331, .(T332, .(T333, .(T334, []))))))), T334, .(T333, .(T332, .(T331, .(T330, .(T329, .(T328, .(T326, T327))))))))
REVERSE70_IN_GGGGG(.(T379, T380), T381, T382, T383, T384) → U1_GGGGG(T379, T380, T381, T382, T383, T384, reverse70_in_ggggg(T380, T379, .(T381, T382), T383, T384))
REVERSE70_IN_GGGGG(.(T379, T380), T381, T382, T383, T384) → REVERSE70_IN_GGGGG(T380, T379, .(T381, T382), T383, T384)

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:

PALINDROME1_IN_G(.(T334, .(T333, .(T332, .(T331, .(T330, .(T329, .(T328, .(T326, T327))))))))) → U2_G(T334, T333, T332, T331, T330, T329, T328, T326, T327, reverse70_in_ggggg(T327, T326, .(T328, .(T329, .(T330, .(T331, .(T332, .(T333, .(T334, []))))))), T334, .(T333, .(T332, .(T331, .(T330, .(T329, .(T328, .(T326, T327)))))))))
PALINDROME1_IN_G(.(T334, .(T333, .(T332, .(T331, .(T330, .(T329, .(T328, .(T326, T327))))))))) → REVERSE70_IN_GGGGG(T327, T326, .(T328, .(T329, .(T330, .(T331, .(T332, .(T333, .(T334, []))))))), T334, .(T333, .(T332, .(T331, .(T330, .(T329, .(T328, .(T326, T327))))))))
REVERSE70_IN_GGGGG(.(T379, T380), T381, T382, T383, T384) → U1_GGGGG(T379, T380, T381, T382, T383, T384, reverse70_in_ggggg(T380, T379, .(T381, T382), T383, T384))
REVERSE70_IN_GGGGG(.(T379, T380), T381, T382, T383, T384) → REVERSE70_IN_GGGGG(T380, T379, .(T381, T382), T383, T384)

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:

REVERSE70_IN_GGGGG(.(T379, T380), T381, T382, T383, T384) → REVERSE70_IN_GGGGG(T380, T379, .(T381, T382), T383, T384)

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:

REVERSE70_IN_GGGGG(.(T379, T380), T381, T382, T383, T384) → REVERSE70_IN_GGGGG(T380, T379, .(T381, T382), T383, T384)

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:

  • REVERSE70_IN_GGGGG(.(T379, T380), T381, T382, T383, T384) → REVERSE70_IN_GGGGG(T380, T379, .(T381, T382), T383, T384)
    The graph contains the following edges 1 > 1, 1 > 2, 4 >= 4, 5 >= 5

(10) YES