(0) Obligation:

Clauses:

reverse(L, LR) :- revacc(L, LR, []).
revacc([], L, L).
revacc(.(EL, T), R, A) :- revacc(T, R, .(EL, A)).

Query: reverse(g,a)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

(2) Obligation:

Triples:

revaccA(.(X1, X2), X3, X4, X5) :- revaccA(X2, X3, X1, .(X4, X5)).
reverseB(.(X1, .(X2, .(X3, .(X4, .(X5, .(X6, .(X7, .(X8, X9)))))))), X10) :- revaccA(X9, X10, X8, .(X7, .(X6, .(X5, .(X4, .(X3, .(X2, .(X1, [])))))))).

Clauses:

revacccA([], .(X1, X2), X1, X2).
revacccA(.(X1, X2), X3, X4, X5) :- revacccA(X2, X3, X1, .(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)
revaccA_in: (b,f,b,b)
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, revaccA_in_gagg(X9, X10, X8, .(X7, .(X6, .(X5, .(X4, .(X3, .(X2, .(X1, [])))))))))
REVERSEB_IN_GA(.(X1, .(X2, .(X3, .(X4, .(X5, .(X6, .(X7, .(X8, X9)))))))), X10) → REVACCA_IN_GAGG(X9, X10, X8, .(X7, .(X6, .(X5, .(X4, .(X3, .(X2, .(X1, []))))))))
REVACCA_IN_GAGG(.(X1, X2), X3, X4, X5) → U1_GAGG(X1, X2, X3, X4, X5, revaccA_in_gagg(X2, X3, X1, .(X4, X5)))
REVACCA_IN_GAGG(.(X1, X2), X3, X4, X5) → REVACCA_IN_GAGG(X2, X3, X1, .(X4, X5))

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
revaccA_in_gagg(x1, x2, x3, x4)  =  revaccA_in_gagg(x1, x3, x4)
[]  =  []
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)
REVACCA_IN_GAGG(x1, x2, x3, x4)  =  REVACCA_IN_GAGG(x1, x3, x4)
U1_GAGG(x1, x2, x3, x4, x5, x6)  =  U1_GAGG(x1, x2, 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:

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, revaccA_in_gagg(X9, X10, X8, .(X7, .(X6, .(X5, .(X4, .(X3, .(X2, .(X1, [])))))))))
REVERSEB_IN_GA(.(X1, .(X2, .(X3, .(X4, .(X5, .(X6, .(X7, .(X8, X9)))))))), X10) → REVACCA_IN_GAGG(X9, X10, X8, .(X7, .(X6, .(X5, .(X4, .(X3, .(X2, .(X1, []))))))))
REVACCA_IN_GAGG(.(X1, X2), X3, X4, X5) → U1_GAGG(X1, X2, X3, X4, X5, revaccA_in_gagg(X2, X3, X1, .(X4, X5)))
REVACCA_IN_GAGG(.(X1, X2), X3, X4, X5) → REVACCA_IN_GAGG(X2, X3, X1, .(X4, X5))

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
revaccA_in_gagg(x1, x2, x3, x4)  =  revaccA_in_gagg(x1, x3, x4)
[]  =  []
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)
REVACCA_IN_GAGG(x1, x2, x3, x4)  =  REVACCA_IN_GAGG(x1, x3, x4)
U1_GAGG(x1, x2, x3, x4, x5, x6)  =  U1_GAGG(x1, x2, 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:

REVACCA_IN_GAGG(.(X1, X2), X3, X4, X5) → REVACCA_IN_GAGG(X2, X3, X1, .(X4, X5))

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
REVACCA_IN_GAGG(x1, x2, x3, x4)  =  REVACCA_IN_GAGG(x1, 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:

REVACCA_IN_GAGG(.(X1, X2), X4, X5) → REVACCA_IN_GAGG(X2, X1, .(X4, 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:

  • REVACCA_IN_GAGG(.(X1, X2), X4, X5) → REVACCA_IN_GAGG(X2, X1, .(X4, X5))
    The graph contains the following edges 1 > 1, 1 > 2

(10) YES