(0) Obligation:
Clauses:
f(RES, [], RES).
f([], .(Head, Tail), RES) :- f(.(Head, Tail), Tail, RES).
f(.(Head, Tail), Y, RES) :- f(Y, Tail, RES).
Query: f(g,g,a)
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph DT10.
(2) Obligation:
Triples:
fA(.(X1, .(X2, X3)), [], X4) :- fA(.(X2, X3), X3, X4).
fA([], .(X1, .(X2, X3)), X4) :- fA(.(X2, X3), X3, X4).
fA(.(X1, .(X2, X3)), [], X4) :- fA(.(X2, X3), X3, X4).
fA(.(X1, X2), .(X3, X4), X5) :- fA(X2, X4, X5).
Clauses:
fcA(X1, [], X1).
fcA(.(X1, []), [], []).
fcA(.(X1, .(X2, X3)), [], X4) :- fcA(.(X2, X3), X3, X4).
fcA([], .(X1, []), .(X1, [])).
fcA([], .(X1, []), []).
fcA([], .(X1, .(X2, X3)), X4) :- fcA(.(X2, X3), X3, X4).
fcA(.(X1, []), X2, X2).
fcA(.(X1, .(X2, X3)), [], X4) :- fcA(.(X2, X3), X3, X4).
fcA(.(X1, X2), .(X3, X4), X5) :- fcA(X2, X4, X5).
Afs:
fA(x1, x2, x3) = fA(x1, x2)
(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:
fA_in: (b,b,f)
Transforming
TRIPLES into the following
Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
FA_IN_GGA(.(X1, .(X2, X3)), [], X4) → U1_GGA(X1, X2, X3, X4, fA_in_gga(.(X2, X3), X3, X4))
FA_IN_GGA(.(X1, .(X2, X3)), [], X4) → FA_IN_GGA(.(X2, X3), X3, X4)
FA_IN_GGA([], .(X1, .(X2, X3)), X4) → U2_GGA(X1, X2, X3, X4, fA_in_gga(.(X2, X3), X3, X4))
FA_IN_GGA([], .(X1, .(X2, X3)), X4) → FA_IN_GGA(.(X2, X3), X3, X4)
FA_IN_GGA(.(X1, X2), .(X3, X4), X5) → U3_GGA(X1, X2, X3, X4, X5, fA_in_gga(X2, X4, X5))
FA_IN_GGA(.(X1, X2), .(X3, X4), X5) → FA_IN_GGA(X2, X4, X5)
R is empty.
The argument filtering Pi contains the following mapping:
fA_in_gga(
x1,
x2,
x3) =
fA_in_gga(
x1,
x2)
.(
x1,
x2) =
.(
x1,
x2)
[] =
[]
FA_IN_GGA(
x1,
x2,
x3) =
FA_IN_GGA(
x1,
x2)
U1_GGA(
x1,
x2,
x3,
x4,
x5) =
U1_GGA(
x1,
x2,
x3,
x5)
U2_GGA(
x1,
x2,
x3,
x4,
x5) =
U2_GGA(
x1,
x2,
x3,
x5)
U3_GGA(
x1,
x2,
x3,
x4,
x5,
x6) =
U3_GGA(
x1,
x2,
x3,
x4,
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:
FA_IN_GGA(.(X1, .(X2, X3)), [], X4) → U1_GGA(X1, X2, X3, X4, fA_in_gga(.(X2, X3), X3, X4))
FA_IN_GGA(.(X1, .(X2, X3)), [], X4) → FA_IN_GGA(.(X2, X3), X3, X4)
FA_IN_GGA([], .(X1, .(X2, X3)), X4) → U2_GGA(X1, X2, X3, X4, fA_in_gga(.(X2, X3), X3, X4))
FA_IN_GGA([], .(X1, .(X2, X3)), X4) → FA_IN_GGA(.(X2, X3), X3, X4)
FA_IN_GGA(.(X1, X2), .(X3, X4), X5) → U3_GGA(X1, X2, X3, X4, X5, fA_in_gga(X2, X4, X5))
FA_IN_GGA(.(X1, X2), .(X3, X4), X5) → FA_IN_GGA(X2, X4, X5)
R is empty.
The argument filtering Pi contains the following mapping:
fA_in_gga(
x1,
x2,
x3) =
fA_in_gga(
x1,
x2)
.(
x1,
x2) =
.(
x1,
x2)
[] =
[]
FA_IN_GGA(
x1,
x2,
x3) =
FA_IN_GGA(
x1,
x2)
U1_GGA(
x1,
x2,
x3,
x4,
x5) =
U1_GGA(
x1,
x2,
x3,
x5)
U2_GGA(
x1,
x2,
x3,
x4,
x5) =
U2_GGA(
x1,
x2,
x3,
x5)
U3_GGA(
x1,
x2,
x3,
x4,
x5,
x6) =
U3_GGA(
x1,
x2,
x3,
x4,
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:
FA_IN_GGA(.(X1, X2), .(X3, X4), X5) → FA_IN_GGA(X2, X4, X5)
FA_IN_GGA(.(X1, .(X2, X3)), [], X4) → FA_IN_GGA(.(X2, X3), X3, X4)
FA_IN_GGA([], .(X1, .(X2, X3)), X4) → FA_IN_GGA(.(X2, X3), X3, X4)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
[] =
[]
FA_IN_GGA(
x1,
x2,
x3) =
FA_IN_GGA(
x1,
x2)
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:
FA_IN_GGA(.(X1, X2), .(X3, X4)) → FA_IN_GGA(X2, X4)
FA_IN_GGA(.(X1, .(X2, X3)), []) → FA_IN_GGA(.(X2, X3), X3)
FA_IN_GGA([], .(X1, .(X2, X3))) → FA_IN_GGA(.(X2, X3), X3)
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:
- FA_IN_GGA(.(X1, X2), .(X3, X4)) → FA_IN_GGA(X2, X4)
The graph contains the following edges 1 > 1, 2 > 2
- FA_IN_GGA(.(X1, .(X2, X3)), []) → FA_IN_GGA(.(X2, X3), X3)
The graph contains the following edges 1 > 1, 1 > 2
- FA_IN_GGA([], .(X1, .(X2, X3))) → FA_IN_GGA(.(X2, X3), X3)
The graph contains the following edges 2 > 1, 2 > 2
(10) YES