(0) Obligation:
Clauses:
fl([], [], 0).
fl(.(E, X), R, s(Z)) :- ','(append(E, Y, R), fl(X, Y, Z)).
append([], X, X).
append(.(X, Xs), Ys, .(X, Zs)) :- append(Xs, Ys, Zs).
Queries:
fl(g,a,a).
(1) PrologToPiTRSProof (SOUND transformation)
We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
fl_in: (b,f,f)
append_in: (b,f,f)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))
The argument filtering Pi contains the following mapping:
fl_in_gaa(
x1,
x2,
x3) =
fl_in_gaa(
x1)
[] =
[]
fl_out_gaa(
x1,
x2,
x3) =
fl_out_gaa(
x1,
x3)
.(
x1,
x2) =
.(
x1,
x2)
U1_gaa(
x1,
x2,
x3,
x4,
x5) =
U1_gaa(
x1,
x2,
x5)
append_in_gaa(
x1,
x2,
x3) =
append_in_gaa(
x1)
append_out_gaa(
x1,
x2,
x3) =
append_out_gaa(
x1)
U3_gaa(
x1,
x2,
x3,
x4,
x5) =
U3_gaa(
x1,
x2,
x5)
U2_gaa(
x1,
x2,
x3,
x4,
x5) =
U2_gaa(
x1,
x2,
x5)
s(
x1) =
s(
x1)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(2) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))
The argument filtering Pi contains the following mapping:
fl_in_gaa(
x1,
x2,
x3) =
fl_in_gaa(
x1)
[] =
[]
fl_out_gaa(
x1,
x2,
x3) =
fl_out_gaa(
x1,
x3)
.(
x1,
x2) =
.(
x1,
x2)
U1_gaa(
x1,
x2,
x3,
x4,
x5) =
U1_gaa(
x1,
x2,
x5)
append_in_gaa(
x1,
x2,
x3) =
append_in_gaa(
x1)
append_out_gaa(
x1,
x2,
x3) =
append_out_gaa(
x1)
U3_gaa(
x1,
x2,
x3,
x4,
x5) =
U3_gaa(
x1,
x2,
x5)
U2_gaa(
x1,
x2,
x3,
x4,
x5) =
U2_gaa(
x1,
x2,
x5)
s(
x1) =
s(
x1)
(3) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:
FL_IN_GAA(.(E, X), R, s(Z)) → U1_GAA(E, X, R, Z, append_in_gaa(E, Y, R))
FL_IN_GAA(.(E, X), R, s(Z)) → APPEND_IN_GAA(E, Y, R)
APPEND_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → U3_GAA(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
APPEND_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_GAA(Xs, Ys, Zs)
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_GAA(E, X, R, Z, fl_in_gaa(X, Y, Z))
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → FL_IN_GAA(X, Y, Z)
The TRS R consists of the following rules:
fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))
The argument filtering Pi contains the following mapping:
fl_in_gaa(
x1,
x2,
x3) =
fl_in_gaa(
x1)
[] =
[]
fl_out_gaa(
x1,
x2,
x3) =
fl_out_gaa(
x1,
x3)
.(
x1,
x2) =
.(
x1,
x2)
U1_gaa(
x1,
x2,
x3,
x4,
x5) =
U1_gaa(
x1,
x2,
x5)
append_in_gaa(
x1,
x2,
x3) =
append_in_gaa(
x1)
append_out_gaa(
x1,
x2,
x3) =
append_out_gaa(
x1)
U3_gaa(
x1,
x2,
x3,
x4,
x5) =
U3_gaa(
x1,
x2,
x5)
U2_gaa(
x1,
x2,
x3,
x4,
x5) =
U2_gaa(
x1,
x2,
x5)
s(
x1) =
s(
x1)
FL_IN_GAA(
x1,
x2,
x3) =
FL_IN_GAA(
x1)
U1_GAA(
x1,
x2,
x3,
x4,
x5) =
U1_GAA(
x1,
x2,
x5)
APPEND_IN_GAA(
x1,
x2,
x3) =
APPEND_IN_GAA(
x1)
U3_GAA(
x1,
x2,
x3,
x4,
x5) =
U3_GAA(
x1,
x2,
x5)
U2_GAA(
x1,
x2,
x3,
x4,
x5) =
U2_GAA(
x1,
x2,
x5)
We have to consider all (P,R,Pi)-chains
(4) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
FL_IN_GAA(.(E, X), R, s(Z)) → U1_GAA(E, X, R, Z, append_in_gaa(E, Y, R))
FL_IN_GAA(.(E, X), R, s(Z)) → APPEND_IN_GAA(E, Y, R)
APPEND_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → U3_GAA(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
APPEND_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_GAA(Xs, Ys, Zs)
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_GAA(E, X, R, Z, fl_in_gaa(X, Y, Z))
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → FL_IN_GAA(X, Y, Z)
The TRS R consists of the following rules:
fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))
The argument filtering Pi contains the following mapping:
fl_in_gaa(
x1,
x2,
x3) =
fl_in_gaa(
x1)
[] =
[]
fl_out_gaa(
x1,
x2,
x3) =
fl_out_gaa(
x1,
x3)
.(
x1,
x2) =
.(
x1,
x2)
U1_gaa(
x1,
x2,
x3,
x4,
x5) =
U1_gaa(
x1,
x2,
x5)
append_in_gaa(
x1,
x2,
x3) =
append_in_gaa(
x1)
append_out_gaa(
x1,
x2,
x3) =
append_out_gaa(
x1)
U3_gaa(
x1,
x2,
x3,
x4,
x5) =
U3_gaa(
x1,
x2,
x5)
U2_gaa(
x1,
x2,
x3,
x4,
x5) =
U2_gaa(
x1,
x2,
x5)
s(
x1) =
s(
x1)
FL_IN_GAA(
x1,
x2,
x3) =
FL_IN_GAA(
x1)
U1_GAA(
x1,
x2,
x3,
x4,
x5) =
U1_GAA(
x1,
x2,
x5)
APPEND_IN_GAA(
x1,
x2,
x3) =
APPEND_IN_GAA(
x1)
U3_GAA(
x1,
x2,
x3,
x4,
x5) =
U3_GAA(
x1,
x2,
x5)
U2_GAA(
x1,
x2,
x3,
x4,
x5) =
U2_GAA(
x1,
x2,
x5)
We have to consider all (P,R,Pi)-chains
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 3 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
APPEND_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_GAA(Xs, Ys, Zs)
The TRS R consists of the following rules:
fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))
The argument filtering Pi contains the following mapping:
fl_in_gaa(
x1,
x2,
x3) =
fl_in_gaa(
x1)
[] =
[]
fl_out_gaa(
x1,
x2,
x3) =
fl_out_gaa(
x1,
x3)
.(
x1,
x2) =
.(
x1,
x2)
U1_gaa(
x1,
x2,
x3,
x4,
x5) =
U1_gaa(
x1,
x2,
x5)
append_in_gaa(
x1,
x2,
x3) =
append_in_gaa(
x1)
append_out_gaa(
x1,
x2,
x3) =
append_out_gaa(
x1)
U3_gaa(
x1,
x2,
x3,
x4,
x5) =
U3_gaa(
x1,
x2,
x5)
U2_gaa(
x1,
x2,
x3,
x4,
x5) =
U2_gaa(
x1,
x2,
x5)
s(
x1) =
s(
x1)
APPEND_IN_GAA(
x1,
x2,
x3) =
APPEND_IN_GAA(
x1)
We have to consider all (P,R,Pi)-chains
(8) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(9) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
APPEND_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_GAA(Xs, Ys, Zs)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
APPEND_IN_GAA(
x1,
x2,
x3) =
APPEND_IN_GAA(
x1)
We have to consider all (P,R,Pi)-chains
(10) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
APPEND_IN_GAA(.(X, Xs)) → APPEND_IN_GAA(Xs)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(12) 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:
- APPEND_IN_GAA(.(X, Xs)) → APPEND_IN_GAA(Xs)
The graph contains the following edges 1 > 1
(13) TRUE
(14) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → FL_IN_GAA(X, Y, Z)
FL_IN_GAA(.(E, X), R, s(Z)) → U1_GAA(E, X, R, Z, append_in_gaa(E, Y, R))
The TRS R consists of the following rules:
fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))
The argument filtering Pi contains the following mapping:
fl_in_gaa(
x1,
x2,
x3) =
fl_in_gaa(
x1)
[] =
[]
fl_out_gaa(
x1,
x2,
x3) =
fl_out_gaa(
x1,
x3)
.(
x1,
x2) =
.(
x1,
x2)
U1_gaa(
x1,
x2,
x3,
x4,
x5) =
U1_gaa(
x1,
x2,
x5)
append_in_gaa(
x1,
x2,
x3) =
append_in_gaa(
x1)
append_out_gaa(
x1,
x2,
x3) =
append_out_gaa(
x1)
U3_gaa(
x1,
x2,
x3,
x4,
x5) =
U3_gaa(
x1,
x2,
x5)
U2_gaa(
x1,
x2,
x3,
x4,
x5) =
U2_gaa(
x1,
x2,
x5)
s(
x1) =
s(
x1)
FL_IN_GAA(
x1,
x2,
x3) =
FL_IN_GAA(
x1)
U1_GAA(
x1,
x2,
x3,
x4,
x5) =
U1_GAA(
x1,
x2,
x5)
We have to consider all (P,R,Pi)-chains
(15) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(16) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → FL_IN_GAA(X, Y, Z)
FL_IN_GAA(.(E, X), R, s(Z)) → U1_GAA(E, X, R, Z, append_in_gaa(E, Y, R))
The TRS R consists of the following rules:
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
The argument filtering Pi contains the following mapping:
[] =
[]
.(
x1,
x2) =
.(
x1,
x2)
append_in_gaa(
x1,
x2,
x3) =
append_in_gaa(
x1)
append_out_gaa(
x1,
x2,
x3) =
append_out_gaa(
x1)
U3_gaa(
x1,
x2,
x3,
x4,
x5) =
U3_gaa(
x1,
x2,
x5)
s(
x1) =
s(
x1)
FL_IN_GAA(
x1,
x2,
x3) =
FL_IN_GAA(
x1)
U1_GAA(
x1,
x2,
x3,
x4,
x5) =
U1_GAA(
x1,
x2,
x5)
We have to consider all (P,R,Pi)-chains
(17) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
U1_GAA(E, X, append_out_gaa(E)) → FL_IN_GAA(X)
FL_IN_GAA(.(E, X)) → U1_GAA(E, X, append_in_gaa(E))
The TRS R consists of the following rules:
append_in_gaa([]) → append_out_gaa([])
append_in_gaa(.(X, Xs)) → U3_gaa(X, Xs, append_in_gaa(Xs))
U3_gaa(X, Xs, append_out_gaa(Xs)) → append_out_gaa(.(X, Xs))
The set Q consists of the following terms:
append_in_gaa(x0)
U3_gaa(x0, x1, x2)
We have to consider all (P,Q,R)-chains.
(19) PrologToPiTRSProof (SOUND transformation)
We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
fl_in: (b,f,f)
append_in: (b,f,f)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))
The argument filtering Pi contains the following mapping:
fl_in_gaa(
x1,
x2,
x3) =
fl_in_gaa(
x1)
[] =
[]
fl_out_gaa(
x1,
x2,
x3) =
fl_out_gaa(
x3)
.(
x1,
x2) =
.(
x1,
x2)
U1_gaa(
x1,
x2,
x3,
x4,
x5) =
U1_gaa(
x2,
x5)
append_in_gaa(
x1,
x2,
x3) =
append_in_gaa(
x1)
append_out_gaa(
x1,
x2,
x3) =
append_out_gaa
U3_gaa(
x1,
x2,
x3,
x4,
x5) =
U3_gaa(
x5)
U2_gaa(
x1,
x2,
x3,
x4,
x5) =
U2_gaa(
x5)
s(
x1) =
s(
x1)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(20) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))
The argument filtering Pi contains the following mapping:
fl_in_gaa(
x1,
x2,
x3) =
fl_in_gaa(
x1)
[] =
[]
fl_out_gaa(
x1,
x2,
x3) =
fl_out_gaa(
x3)
.(
x1,
x2) =
.(
x1,
x2)
U1_gaa(
x1,
x2,
x3,
x4,
x5) =
U1_gaa(
x2,
x5)
append_in_gaa(
x1,
x2,
x3) =
append_in_gaa(
x1)
append_out_gaa(
x1,
x2,
x3) =
append_out_gaa
U3_gaa(
x1,
x2,
x3,
x4,
x5) =
U3_gaa(
x5)
U2_gaa(
x1,
x2,
x3,
x4,
x5) =
U2_gaa(
x5)
s(
x1) =
s(
x1)
(21) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:
FL_IN_GAA(.(E, X), R, s(Z)) → U1_GAA(E, X, R, Z, append_in_gaa(E, Y, R))
FL_IN_GAA(.(E, X), R, s(Z)) → APPEND_IN_GAA(E, Y, R)
APPEND_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → U3_GAA(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
APPEND_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_GAA(Xs, Ys, Zs)
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_GAA(E, X, R, Z, fl_in_gaa(X, Y, Z))
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → FL_IN_GAA(X, Y, Z)
The TRS R consists of the following rules:
fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))
The argument filtering Pi contains the following mapping:
fl_in_gaa(
x1,
x2,
x3) =
fl_in_gaa(
x1)
[] =
[]
fl_out_gaa(
x1,
x2,
x3) =
fl_out_gaa(
x3)
.(
x1,
x2) =
.(
x1,
x2)
U1_gaa(
x1,
x2,
x3,
x4,
x5) =
U1_gaa(
x2,
x5)
append_in_gaa(
x1,
x2,
x3) =
append_in_gaa(
x1)
append_out_gaa(
x1,
x2,
x3) =
append_out_gaa
U3_gaa(
x1,
x2,
x3,
x4,
x5) =
U3_gaa(
x5)
U2_gaa(
x1,
x2,
x3,
x4,
x5) =
U2_gaa(
x5)
s(
x1) =
s(
x1)
FL_IN_GAA(
x1,
x2,
x3) =
FL_IN_GAA(
x1)
U1_GAA(
x1,
x2,
x3,
x4,
x5) =
U1_GAA(
x2,
x5)
APPEND_IN_GAA(
x1,
x2,
x3) =
APPEND_IN_GAA(
x1)
U3_GAA(
x1,
x2,
x3,
x4,
x5) =
U3_GAA(
x5)
U2_GAA(
x1,
x2,
x3,
x4,
x5) =
U2_GAA(
x5)
We have to consider all (P,R,Pi)-chains
(22) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
FL_IN_GAA(.(E, X), R, s(Z)) → U1_GAA(E, X, R, Z, append_in_gaa(E, Y, R))
FL_IN_GAA(.(E, X), R, s(Z)) → APPEND_IN_GAA(E, Y, R)
APPEND_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → U3_GAA(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
APPEND_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_GAA(Xs, Ys, Zs)
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_GAA(E, X, R, Z, fl_in_gaa(X, Y, Z))
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → FL_IN_GAA(X, Y, Z)
The TRS R consists of the following rules:
fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))
The argument filtering Pi contains the following mapping:
fl_in_gaa(
x1,
x2,
x3) =
fl_in_gaa(
x1)
[] =
[]
fl_out_gaa(
x1,
x2,
x3) =
fl_out_gaa(
x3)
.(
x1,
x2) =
.(
x1,
x2)
U1_gaa(
x1,
x2,
x3,
x4,
x5) =
U1_gaa(
x2,
x5)
append_in_gaa(
x1,
x2,
x3) =
append_in_gaa(
x1)
append_out_gaa(
x1,
x2,
x3) =
append_out_gaa
U3_gaa(
x1,
x2,
x3,
x4,
x5) =
U3_gaa(
x5)
U2_gaa(
x1,
x2,
x3,
x4,
x5) =
U2_gaa(
x5)
s(
x1) =
s(
x1)
FL_IN_GAA(
x1,
x2,
x3) =
FL_IN_GAA(
x1)
U1_GAA(
x1,
x2,
x3,
x4,
x5) =
U1_GAA(
x2,
x5)
APPEND_IN_GAA(
x1,
x2,
x3) =
APPEND_IN_GAA(
x1)
U3_GAA(
x1,
x2,
x3,
x4,
x5) =
U3_GAA(
x5)
U2_GAA(
x1,
x2,
x3,
x4,
x5) =
U2_GAA(
x5)
We have to consider all (P,R,Pi)-chains
(23) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 3 less nodes.
(24) Complex Obligation (AND)
(25) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
APPEND_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_GAA(Xs, Ys, Zs)
The TRS R consists of the following rules:
fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))
The argument filtering Pi contains the following mapping:
fl_in_gaa(
x1,
x2,
x3) =
fl_in_gaa(
x1)
[] =
[]
fl_out_gaa(
x1,
x2,
x3) =
fl_out_gaa(
x3)
.(
x1,
x2) =
.(
x1,
x2)
U1_gaa(
x1,
x2,
x3,
x4,
x5) =
U1_gaa(
x2,
x5)
append_in_gaa(
x1,
x2,
x3) =
append_in_gaa(
x1)
append_out_gaa(
x1,
x2,
x3) =
append_out_gaa
U3_gaa(
x1,
x2,
x3,
x4,
x5) =
U3_gaa(
x5)
U2_gaa(
x1,
x2,
x3,
x4,
x5) =
U2_gaa(
x5)
s(
x1) =
s(
x1)
APPEND_IN_GAA(
x1,
x2,
x3) =
APPEND_IN_GAA(
x1)
We have to consider all (P,R,Pi)-chains
(26) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(27) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
APPEND_IN_GAA(.(X, Xs), Ys, .(X, Zs)) → APPEND_IN_GAA(Xs, Ys, Zs)
R is empty.
The argument filtering Pi contains the following mapping:
.(
x1,
x2) =
.(
x1,
x2)
APPEND_IN_GAA(
x1,
x2,
x3) =
APPEND_IN_GAA(
x1)
We have to consider all (P,R,Pi)-chains
(28) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(29) Obligation:
Q DP problem:
The TRS P consists of the following rules:
APPEND_IN_GAA(.(X, Xs)) → APPEND_IN_GAA(Xs)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(30) 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:
- APPEND_IN_GAA(.(X, Xs)) → APPEND_IN_GAA(Xs)
The graph contains the following edges 1 > 1
(31) TRUE
(32) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → FL_IN_GAA(X, Y, Z)
FL_IN_GAA(.(E, X), R, s(Z)) → U1_GAA(E, X, R, Z, append_in_gaa(E, Y, R))
The TRS R consists of the following rules:
fl_in_gaa([], [], 0) → fl_out_gaa([], [], 0)
fl_in_gaa(.(E, X), R, s(Z)) → U1_gaa(E, X, R, Z, append_in_gaa(E, Y, R))
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
U1_gaa(E, X, R, Z, append_out_gaa(E, Y, R)) → U2_gaa(E, X, R, Z, fl_in_gaa(X, Y, Z))
U2_gaa(E, X, R, Z, fl_out_gaa(X, Y, Z)) → fl_out_gaa(.(E, X), R, s(Z))
The argument filtering Pi contains the following mapping:
fl_in_gaa(
x1,
x2,
x3) =
fl_in_gaa(
x1)
[] =
[]
fl_out_gaa(
x1,
x2,
x3) =
fl_out_gaa(
x3)
.(
x1,
x2) =
.(
x1,
x2)
U1_gaa(
x1,
x2,
x3,
x4,
x5) =
U1_gaa(
x2,
x5)
append_in_gaa(
x1,
x2,
x3) =
append_in_gaa(
x1)
append_out_gaa(
x1,
x2,
x3) =
append_out_gaa
U3_gaa(
x1,
x2,
x3,
x4,
x5) =
U3_gaa(
x5)
U2_gaa(
x1,
x2,
x3,
x4,
x5) =
U2_gaa(
x5)
s(
x1) =
s(
x1)
FL_IN_GAA(
x1,
x2,
x3) =
FL_IN_GAA(
x1)
U1_GAA(
x1,
x2,
x3,
x4,
x5) =
U1_GAA(
x2,
x5)
We have to consider all (P,R,Pi)-chains
(33) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(34) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
U1_GAA(E, X, R, Z, append_out_gaa(E, Y, R)) → FL_IN_GAA(X, Y, Z)
FL_IN_GAA(.(E, X), R, s(Z)) → U1_GAA(E, X, R, Z, append_in_gaa(E, Y, R))
The TRS R consists of the following rules:
append_in_gaa([], X, X) → append_out_gaa([], X, X)
append_in_gaa(.(X, Xs), Ys, .(X, Zs)) → U3_gaa(X, Xs, Ys, Zs, append_in_gaa(Xs, Ys, Zs))
U3_gaa(X, Xs, Ys, Zs, append_out_gaa(Xs, Ys, Zs)) → append_out_gaa(.(X, Xs), Ys, .(X, Zs))
The argument filtering Pi contains the following mapping:
[] =
[]
.(
x1,
x2) =
.(
x1,
x2)
append_in_gaa(
x1,
x2,
x3) =
append_in_gaa(
x1)
append_out_gaa(
x1,
x2,
x3) =
append_out_gaa
U3_gaa(
x1,
x2,
x3,
x4,
x5) =
U3_gaa(
x5)
s(
x1) =
s(
x1)
FL_IN_GAA(
x1,
x2,
x3) =
FL_IN_GAA(
x1)
U1_GAA(
x1,
x2,
x3,
x4,
x5) =
U1_GAA(
x2,
x5)
We have to consider all (P,R,Pi)-chains
(35) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(36) Obligation:
Q DP problem:
The TRS P consists of the following rules:
U1_GAA(X, append_out_gaa) → FL_IN_GAA(X)
FL_IN_GAA(.(E, X)) → U1_GAA(X, append_in_gaa(E))
The TRS R consists of the following rules:
append_in_gaa([]) → append_out_gaa
append_in_gaa(.(X, Xs)) → U3_gaa(append_in_gaa(Xs))
U3_gaa(append_out_gaa) → append_out_gaa
The set Q consists of the following terms:
append_in_gaa(x0)
U3_gaa(x0)
We have to consider all (P,Q,R)-chains.
(37) 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:
- FL_IN_GAA(.(E, X)) → U1_GAA(X, append_in_gaa(E))
The graph contains the following edges 1 > 1
- U1_GAA(X, append_out_gaa) → FL_IN_GAA(X)
The graph contains the following edges 1 >= 1
(38) TRUE