Left Termination proof of /tmp/tmpeFtXo6/pl6.1.1.pl

Left Termination of the query pattern qsort(b,f) w.r.t. the given Prolog program could successfully be proven:

↳ PROLOG

  ↳ PrologToPiTRSProof

qsort₂({}₀, {}₀).
qsort₂(.₂(H, L), S) :- split₄(L, H, A, B), qsort₂(A, A1), qsort₂(B, B1), append₃(A1, .₂(H, B1), S).
split₄({}₀, Y, {}₀, {}₀).
split₄(.₂(X, Xs), Y, .₂(X, Ls), Bs) :- le₂(X, Y), split₄(Xs, Y, Ls, Bs).
split₄(.₂(X, Xs), Y, Ls, .₂(X, Bs)) :- gt₂(X, Y), split₄(Xs, Y, Ls, Bs).
append₃({}₀, L, L).
append₃(.₂(H, L1), L2, .₂(H, L3)) :- append₃(L1, L2, L3).
gt₂(s₁(X), s₁(Y)) :- gt₂(X, Y).
gt₂(s₁(X), 0₀).
le₂(s₁(X), s₁(Y)) :- le₂(X, Y).
le₂(0₀, s₁(Y)).
le₂(0₀, 0₀).

With regard to the inferred argument filtering the predicates were used in the following modes:
qsort₂: (b,f)
split₄: (b,b,f,f)
le₂: (b,b)
gt₂: (b,b)
append₃: (b,b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

qsort_2_in_ga₂([]_0, []_0) -> qsort_2_out_ga₂([]_0, []_0)
qsort_2_in_ga₂(._2₂(H, L), S) -> if_qsort_2_in_1_ga₄(H, L, S, split_4_in_ggaa₄(L, H, A, B))
split_4_in_ggaa₄([]_0, Y, []_0, []_0) -> split_4_out_ggaa₄([]_0, Y, []_0, []_0)
split_4_in_ggaa₄(._2₂(X, Xs), Y, ._2₂(X, Ls), Bs) -> if_split_4_in_1_ggaa₆(X, Xs, Y, Ls, Bs, le_2_in_gg₂(X, Y))
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(Y)) -> le_2_out_gg₂(0_0, s_1₁(Y))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_1_ggaa₆(X, Xs, Y, Ls, Bs, le_2_out_gg₂(X, Y)) -> if_split_4_in_2_ggaa₆(X, Xs, Y, Ls, Bs, split_4_in_ggaa₄(Xs, Y, Ls, Bs))
split_4_in_ggaa₄(._2₂(X, Xs), Y, Ls, ._2₂(X, Bs)) -> if_split_4_in_3_ggaa₆(X, Xs, Y, Ls, Bs, gt_2_in_gg₂(X, Y))
gt_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_gt_2_in_1_gg₃(X, Y, gt_2_in_gg₂(X, Y))
gt_2_in_gg₂(s_1₁(X), 0_0) -> gt_2_out_gg₂(s_1₁(X), 0_0)
if_gt_2_in_1_gg₃(X, Y, gt_2_out_gg₂(X, Y)) -> gt_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_3_ggaa₆(X, Xs, Y, Ls, Bs, gt_2_out_gg₂(X, Y)) -> if_split_4_in_4_ggaa₆(X, Xs, Y, Ls, Bs, split_4_in_ggaa₄(Xs, Y, Ls, Bs))
if_split_4_in_4_ggaa₆(X, Xs, Y, Ls, Bs, split_4_out_ggaa₄(Xs, Y, Ls, Bs)) -> split_4_out_ggaa₄(._2₂(X, Xs), Y, Ls, ._2₂(X, Bs))
if_split_4_in_2_ggaa₆(X, Xs, Y, Ls, Bs, split_4_out_ggaa₄(Xs, Y, Ls, Bs)) -> split_4_out_ggaa₄(._2₂(X, Xs), Y, ._2₂(X, Ls), Bs)
if_qsort_2_in_1_ga₄(H, L, S, split_4_out_ggaa₄(L, H, A, B)) -> if_qsort_2_in_2_ga₆(H, L, S, A, B, qsort_2_in_ga₂(A, A1))
if_qsort_2_in_2_ga₆(H, L, S, A, B, qsort_2_out_ga₂(A, A1)) -> if_qsort_2_in_3_ga₆(H, L, S, B, A1, qsort_2_in_ga₂(B, B1))
if_qsort_2_in_3_ga₆(H, L, S, B, A1, qsort_2_out_ga₂(B, B1)) -> if_qsort_2_in_4_ga₆(H, L, S, A1, B1, append_3_in_gga₃(A1, ._2₂(H, B1), S))
append_3_in_gga₃([]_0, L, L) -> append_3_out_gga₃([]_0, L, L)
append_3_in_gga₃(._2₂(H, L1), L2, ._2₂(H, L3)) -> if_append_3_in_1_gga₅(H, L1, L2, L3, append_3_in_gga₃(L1, L2, L3))
if_append_3_in_1_gga₅(H, L1, L2, L3, append_3_out_gga₃(L1, L2, L3)) -> append_3_out_gga₃(._2₂(H, L1), L2, ._2₂(H, L3))
if_qsort_2_in_4_ga₆(H, L, S, A1, B1, append_3_out_gga₃(A1, ._2₂(H, B1), S)) -> qsort_2_out_ga₂(._2₂(H, L), S)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

qsort_2_in_ga₂([]_0, []_0) -> qsort_2_out_ga₂([]_0, []_0)
qsort_2_in_ga₂(._2₂(H, L), S) -> if_qsort_2_in_1_ga₄(H, L, S, split_4_in_ggaa₄(L, H, A, B))
split_4_in_ggaa₄([]_0, Y, []_0, []_0) -> split_4_out_ggaa₄([]_0, Y, []_0, []_0)
split_4_in_ggaa₄(._2₂(X, Xs), Y, ._2₂(X, Ls), Bs) -> if_split_4_in_1_ggaa₆(X, Xs, Y, Ls, Bs, le_2_in_gg₂(X, Y))
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(Y)) -> le_2_out_gg₂(0_0, s_1₁(Y))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_1_ggaa₆(X, Xs, Y, Ls, Bs, le_2_out_gg₂(X, Y)) -> if_split_4_in_2_ggaa₆(X, Xs, Y, Ls, Bs, split_4_in_ggaa₄(Xs, Y, Ls, Bs))
split_4_in_ggaa₄(._2₂(X, Xs), Y, Ls, ._2₂(X, Bs)) -> if_split_4_in_3_ggaa₆(X, Xs, Y, Ls, Bs, gt_2_in_gg₂(X, Y))
gt_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_gt_2_in_1_gg₃(X, Y, gt_2_in_gg₂(X, Y))
gt_2_in_gg₂(s_1₁(X), 0_0) -> gt_2_out_gg₂(s_1₁(X), 0_0)
if_gt_2_in_1_gg₃(X, Y, gt_2_out_gg₂(X, Y)) -> gt_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_3_ggaa₆(X, Xs, Y, Ls, Bs, gt_2_out_gg₂(X, Y)) -> if_split_4_in_4_ggaa₆(X, Xs, Y, Ls, Bs, split_4_in_ggaa₄(Xs, Y, Ls, Bs))
if_split_4_in_4_ggaa₆(X, Xs, Y, Ls, Bs, split_4_out_ggaa₄(Xs, Y, Ls, Bs)) -> split_4_out_ggaa₄(._2₂(X, Xs), Y, Ls, ._2₂(X, Bs))
if_split_4_in_2_ggaa₆(X, Xs, Y, Ls, Bs, split_4_out_ggaa₄(Xs, Y, Ls, Bs)) -> split_4_out_ggaa₄(._2₂(X, Xs), Y, ._2₂(X, Ls), Bs)
if_qsort_2_in_1_ga₄(H, L, S, split_4_out_ggaa₄(L, H, A, B)) -> if_qsort_2_in_2_ga₆(H, L, S, A, B, qsort_2_in_ga₂(A, A1))
if_qsort_2_in_2_ga₆(H, L, S, A, B, qsort_2_out_ga₂(A, A1)) -> if_qsort_2_in_3_ga₆(H, L, S, B, A1, qsort_2_in_ga₂(B, B1))
if_qsort_2_in_3_ga₆(H, L, S, B, A1, qsort_2_out_ga₂(B, B1)) -> if_qsort_2_in_4_ga₆(H, L, S, A1, B1, append_3_in_gga₃(A1, ._2₂(H, B1), S))
append_3_in_gga₃([]_0, L, L) -> append_3_out_gga₃([]_0, L, L)
append_3_in_gga₃(._2₂(H, L1), L2, ._2₂(H, L3)) -> if_append_3_in_1_gga₅(H, L1, L2, L3, append_3_in_gga₃(L1, L2, L3))
if_append_3_in_1_gga₅(H, L1, L2, L3, append_3_out_gga₃(L1, L2, L3)) -> append_3_out_gga₃(._2₂(H, L1), L2, ._2₂(H, L3))
if_qsort_2_in_4_ga₆(H, L, S, A1, B1, append_3_out_gga₃(A1, ._2₂(H, B1), S)) -> qsort_2_out_ga₂(._2₂(H, L), S)

QSORT_2_IN_GA₂(._2₂(H, L), S) -> IF_QSORT_2_IN_1_GA₄(H, L, S, split_4_in_ggaa₄(L, H, A, B))
QSORT_2_IN_GA₂(._2₂(H, L), S) -> SPLIT_4_IN_GGAA₄(L, H, A, B)
SPLIT_4_IN_GGAA₄(._2₂(X, Xs), Y, ._2₂(X, Ls), Bs) -> IF_SPLIT_4_IN_1_GGAA₆(X, Xs, Y, Ls, Bs, le_2_in_gg₂(X, Y))
SPLIT_4_IN_GGAA₄(._2₂(X, Xs), Y, ._2₂(X, Ls), Bs) -> LE_2_IN_GG₂(X, Y)
LE_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> IF_LE_2_IN_1_GG₃(X, Y, le_2_in_gg₂(X, Y))
LE_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> LE_2_IN_GG₂(X, Y)
IF_SPLIT_4_IN_1_GGAA₆(X, Xs, Y, Ls, Bs, le_2_out_gg₂(X, Y)) -> IF_SPLIT_4_IN_2_GGAA₆(X, Xs, Y, Ls, Bs, split_4_in_ggaa₄(Xs, Y, Ls, Bs))
IF_SPLIT_4_IN_1_GGAA₆(X, Xs, Y, Ls, Bs, le_2_out_gg₂(X, Y)) -> SPLIT_4_IN_GGAA₄(Xs, Y, Ls, Bs)
SPLIT_4_IN_GGAA₄(._2₂(X, Xs), Y, Ls, ._2₂(X, Bs)) -> IF_SPLIT_4_IN_3_GGAA₆(X, Xs, Y, Ls, Bs, gt_2_in_gg₂(X, Y))
SPLIT_4_IN_GGAA₄(._2₂(X, Xs), Y, Ls, ._2₂(X, Bs)) -> GT_2_IN_GG₂(X, Y)
GT_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> IF_GT_2_IN_1_GG₃(X, Y, gt_2_in_gg₂(X, Y))
GT_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> GT_2_IN_GG₂(X, Y)
IF_SPLIT_4_IN_3_GGAA₆(X, Xs, Y, Ls, Bs, gt_2_out_gg₂(X, Y)) -> IF_SPLIT_4_IN_4_GGAA₆(X, Xs, Y, Ls, Bs, split_4_in_ggaa₄(Xs, Y, Ls, Bs))
IF_SPLIT_4_IN_3_GGAA₆(X, Xs, Y, Ls, Bs, gt_2_out_gg₂(X, Y)) -> SPLIT_4_IN_GGAA₄(Xs, Y, Ls, Bs)
IF_QSORT_2_IN_1_GA₄(H, L, S, split_4_out_ggaa₄(L, H, A, B)) -> IF_QSORT_2_IN_2_GA₆(H, L, S, A, B, qsort_2_in_ga₂(A, A1))
IF_QSORT_2_IN_1_GA₄(H, L, S, split_4_out_ggaa₄(L, H, A, B)) -> QSORT_2_IN_GA₂(A, A1)
IF_QSORT_2_IN_2_GA₆(H, L, S, A, B, qsort_2_out_ga₂(A, A1)) -> IF_QSORT_2_IN_3_GA₆(H, L, S, B, A1, qsort_2_in_ga₂(B, B1))
IF_QSORT_2_IN_2_GA₆(H, L, S, A, B, qsort_2_out_ga₂(A, A1)) -> QSORT_2_IN_GA₂(B, B1)
IF_QSORT_2_IN_3_GA₆(H, L, S, B, A1, qsort_2_out_ga₂(B, B1)) -> IF_QSORT_2_IN_4_GA₆(H, L, S, A1, B1, append_3_in_gga₃(A1, ._2₂(H, B1), S))
IF_QSORT_2_IN_3_GA₆(H, L, S, B, A1, qsort_2_out_ga₂(B, B1)) -> APPEND_3_IN_GGA₃(A1, ._2₂(H, B1), S)
APPEND_3_IN_GGA₃(._2₂(H, L1), L2, ._2₂(H, L3)) -> IF_APPEND_3_IN_1_GGA₅(H, L1, L2, L3, append_3_in_gga₃(L1, L2, L3))
APPEND_3_IN_GGA₃(._2₂(H, L1), L2, ._2₂(H, L3)) -> APPEND_3_IN_GGA₃(L1, L2, L3)

The TRS R consists of the following rules:

qsort_2_in_ga₂([]_0, []_0) -> qsort_2_out_ga₂([]_0, []_0)
qsort_2_in_ga₂(._2₂(H, L), S) -> if_qsort_2_in_1_ga₄(H, L, S, split_4_in_ggaa₄(L, H, A, B))
split_4_in_ggaa₄([]_0, Y, []_0, []_0) -> split_4_out_ggaa₄([]_0, Y, []_0, []_0)
split_4_in_ggaa₄(._2₂(X, Xs), Y, ._2₂(X, Ls), Bs) -> if_split_4_in_1_ggaa₆(X, Xs, Y, Ls, Bs, le_2_in_gg₂(X, Y))
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(Y)) -> le_2_out_gg₂(0_0, s_1₁(Y))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_1_ggaa₆(X, Xs, Y, Ls, Bs, le_2_out_gg₂(X, Y)) -> if_split_4_in_2_ggaa₆(X, Xs, Y, Ls, Bs, split_4_in_ggaa₄(Xs, Y, Ls, Bs))
split_4_in_ggaa₄(._2₂(X, Xs), Y, Ls, ._2₂(X, Bs)) -> if_split_4_in_3_ggaa₆(X, Xs, Y, Ls, Bs, gt_2_in_gg₂(X, Y))
gt_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_gt_2_in_1_gg₃(X, Y, gt_2_in_gg₂(X, Y))
gt_2_in_gg₂(s_1₁(X), 0_0) -> gt_2_out_gg₂(s_1₁(X), 0_0)
if_gt_2_in_1_gg₃(X, Y, gt_2_out_gg₂(X, Y)) -> gt_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_3_ggaa₆(X, Xs, Y, Ls, Bs, gt_2_out_gg₂(X, Y)) -> if_split_4_in_4_ggaa₆(X, Xs, Y, Ls, Bs, split_4_in_ggaa₄(Xs, Y, Ls, Bs))
if_split_4_in_4_ggaa₆(X, Xs, Y, Ls, Bs, split_4_out_ggaa₄(Xs, Y, Ls, Bs)) -> split_4_out_ggaa₄(._2₂(X, Xs), Y, Ls, ._2₂(X, Bs))
if_split_4_in_2_ggaa₆(X, Xs, Y, Ls, Bs, split_4_out_ggaa₄(Xs, Y, Ls, Bs)) -> split_4_out_ggaa₄(._2₂(X, Xs), Y, ._2₂(X, Ls), Bs)
if_qsort_2_in_1_ga₄(H, L, S, split_4_out_ggaa₄(L, H, A, B)) -> if_qsort_2_in_2_ga₆(H, L, S, A, B, qsort_2_in_ga₂(A, A1))
if_qsort_2_in_2_ga₆(H, L, S, A, B, qsort_2_out_ga₂(A, A1)) -> if_qsort_2_in_3_ga₆(H, L, S, B, A1, qsort_2_in_ga₂(B, B1))
if_qsort_2_in_3_ga₆(H, L, S, B, A1, qsort_2_out_ga₂(B, B1)) -> if_qsort_2_in_4_ga₆(H, L, S, A1, B1, append_3_in_gga₃(A1, ._2₂(H, B1), S))
append_3_in_gga₃([]_0, L, L) -> append_3_out_gga₃([]_0, L, L)
append_3_in_gga₃(._2₂(H, L1), L2, ._2₂(H, L3)) -> if_append_3_in_1_gga₅(H, L1, L2, L3, append_3_in_gga₃(L1, L2, L3))
if_append_3_in_1_gga₅(H, L1, L2, L3, append_3_out_gga₃(L1, L2, L3)) -> append_3_out_gga₃(._2₂(H, L1), L2, ._2₂(H, L3))
if_qsort_2_in_4_ga₆(H, L, S, A1, B1, append_3_out_gga₃(A1, ._2₂(H, B1), S)) -> qsort_2_out_ga₂(._2₂(H, L), S)

The argument filtering Pi contains the following mapping:
qsort_2_in_ga₂(x1, x2) = qsort_2_in_ga₁(x1)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
qsort_2_out_ga₂(x1, x2) = qsort_2_out_ga₁(x2)
if_qsort_2_in_1_ga₄(x1, x2, x3, x4) = if_qsort_2_in_1_ga₂(x1, x4)
split_4_in_ggaa₄(x1, x2, x3, x4) = split_4_in_ggaa₂(x1, x2)
split_4_out_ggaa₄(x1, x2, x3, x4) = split_4_out_ggaa₂(x3, x4)
if_split_4_in_1_ggaa₆(x1, x2, x3, x4, x5, x6) = if_split_4_in_1_ggaa₄(x1, x2, x3, x6)
le_2_in_gg₂(x1, x2) = le_2_in_gg₂(x1, x2)
if_le_2_in_1_gg₃(x1, x2, x3) = if_le_2_in_1_gg₁(x3)
le_2_out_gg₂(x1, x2) = le_2_out_gg
if_split_4_in_2_ggaa₆(x1, x2, x3, x4, x5, x6) = if_split_4_in_2_ggaa₂(x1, x6)
if_split_4_in_3_ggaa₆(x1, x2, x3, x4, x5, x6) = if_split_4_in_3_ggaa₄(x1, x2, x3, x6)
gt_2_in_gg₂(x1, x2) = gt_2_in_gg₂(x1, x2)
if_gt_2_in_1_gg₃(x1, x2, x3) = if_gt_2_in_1_gg₁(x3)
gt_2_out_gg₂(x1, x2) = gt_2_out_gg
if_split_4_in_4_ggaa₆(x1, x2, x3, x4, x5, x6) = if_split_4_in_4_ggaa₂(x1, x6)
if_qsort_2_in_2_ga₆(x1, x2, x3, x4, x5, x6) = if_qsort_2_in_2_ga₃(x1, x5, x6)
if_qsort_2_in_3_ga₆(x1, x2, x3, x4, x5, x6) = if_qsort_2_in_3_ga₃(x1, x5, x6)
if_qsort_2_in_4_ga₆(x1, x2, x3, x4, x5, x6) = if_qsort_2_in_4_ga₁(x6)
append_3_in_gga₃(x1, x2, x3) = append_3_in_gga₂(x1, x2)
append_3_out_gga₃(x1, x2, x3) = append_3_out_gga₁(x3)
if_append_3_in_1_gga₅(x1, x2, x3, x4, x5) = if_append_3_in_1_gga₂(x1, x5)
IF_QSORT_2_IN_1_GA₄(x1, x2, x3, x4) = IF_QSORT_2_IN_1_GA₂(x1, x4)
GT_2_IN_GG₂(x1, x2) = GT_2_IN_GG₂(x1, x2)
IF_SPLIT_4_IN_1_GGAA₆(x1, x2, x3, x4, x5, x6) = IF_SPLIT_4_IN_1_GGAA₄(x1, x2, x3, x6)
IF_QSORT_2_IN_4_GA₆(x1, x2, x3, x4, x5, x6) = IF_QSORT_2_IN_4_GA₁(x6)
IF_GT_2_IN_1_GG₃(x1, x2, x3) = IF_GT_2_IN_1_GG₁(x3)
SPLIT_4_IN_GGAA₄(x1, x2, x3, x4) = SPLIT_4_IN_GGAA₂(x1, x2)
IF_QSORT_2_IN_3_GA₆(x1, x2, x3, x4, x5, x6) = IF_QSORT_2_IN_3_GA₃(x1, x5, x6)
APPEND_3_IN_GGA₃(x1, x2, x3) = APPEND_3_IN_GGA₂(x1, x2)
IF_LE_2_IN_1_GG₃(x1, x2, x3) = IF_LE_2_IN_1_GG₁(x3)
QSORT_2_IN_GA₂(x1, x2) = QSORT_2_IN_GA₁(x1)
IF_QSORT_2_IN_2_GA₆(x1, x2, x3, x4, x5, x6) = IF_QSORT_2_IN_2_GA₃(x1, x5, x6)
IF_SPLIT_4_IN_4_GGAA₆(x1, x2, x3, x4, x5, x6) = IF_SPLIT_4_IN_4_GGAA₂(x1, x6)
IF_APPEND_3_IN_1_GGA₅(x1, x2, x3, x4, x5) = IF_APPEND_3_IN_1_GGA₂(x1, x5)
LE_2_IN_GG₂(x1, x2) = LE_2_IN_GG₂(x1, x2)
IF_SPLIT_4_IN_3_GGAA₆(x1, x2, x3, x4, x5, x6) = IF_SPLIT_4_IN_3_GGAA₄(x1, x2, x3, x6)
IF_SPLIT_4_IN_2_GGAA₆(x1, x2, x3, x4, x5, x6) = IF_SPLIT_4_IN_2_GGAA₂(x1, x6)

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

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

Pi DP problem:
The TRS P consists of the following rules:

QSORT_2_IN_GA₂(._2₂(H, L), S) -> IF_QSORT_2_IN_1_GA₄(H, L, S, split_4_in_ggaa₄(L, H, A, B))
QSORT_2_IN_GA₂(._2₂(H, L), S) -> SPLIT_4_IN_GGAA₄(L, H, A, B)
SPLIT_4_IN_GGAA₄(._2₂(X, Xs), Y, ._2₂(X, Ls), Bs) -> IF_SPLIT_4_IN_1_GGAA₆(X, Xs, Y, Ls, Bs, le_2_in_gg₂(X, Y))
SPLIT_4_IN_GGAA₄(._2₂(X, Xs), Y, ._2₂(X, Ls), Bs) -> LE_2_IN_GG₂(X, Y)
LE_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> IF_LE_2_IN_1_GG₃(X, Y, le_2_in_gg₂(X, Y))
LE_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> LE_2_IN_GG₂(X, Y)
IF_SPLIT_4_IN_1_GGAA₆(X, Xs, Y, Ls, Bs, le_2_out_gg₂(X, Y)) -> IF_SPLIT_4_IN_2_GGAA₆(X, Xs, Y, Ls, Bs, split_4_in_ggaa₄(Xs, Y, Ls, Bs))
IF_SPLIT_4_IN_1_GGAA₆(X, Xs, Y, Ls, Bs, le_2_out_gg₂(X, Y)) -> SPLIT_4_IN_GGAA₄(Xs, Y, Ls, Bs)
SPLIT_4_IN_GGAA₄(._2₂(X, Xs), Y, Ls, ._2₂(X, Bs)) -> IF_SPLIT_4_IN_3_GGAA₆(X, Xs, Y, Ls, Bs, gt_2_in_gg₂(X, Y))
SPLIT_4_IN_GGAA₄(._2₂(X, Xs), Y, Ls, ._2₂(X, Bs)) -> GT_2_IN_GG₂(X, Y)
GT_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> IF_GT_2_IN_1_GG₃(X, Y, gt_2_in_gg₂(X, Y))
GT_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> GT_2_IN_GG₂(X, Y)
IF_SPLIT_4_IN_3_GGAA₆(X, Xs, Y, Ls, Bs, gt_2_out_gg₂(X, Y)) -> IF_SPLIT_4_IN_4_GGAA₆(X, Xs, Y, Ls, Bs, split_4_in_ggaa₄(Xs, Y, Ls, Bs))
IF_SPLIT_4_IN_3_GGAA₆(X, Xs, Y, Ls, Bs, gt_2_out_gg₂(X, Y)) -> SPLIT_4_IN_GGAA₄(Xs, Y, Ls, Bs)
IF_QSORT_2_IN_1_GA₄(H, L, S, split_4_out_ggaa₄(L, H, A, B)) -> IF_QSORT_2_IN_2_GA₆(H, L, S, A, B, qsort_2_in_ga₂(A, A1))
IF_QSORT_2_IN_1_GA₄(H, L, S, split_4_out_ggaa₄(L, H, A, B)) -> QSORT_2_IN_GA₂(A, A1)
IF_QSORT_2_IN_2_GA₆(H, L, S, A, B, qsort_2_out_ga₂(A, A1)) -> IF_QSORT_2_IN_3_GA₆(H, L, S, B, A1, qsort_2_in_ga₂(B, B1))
IF_QSORT_2_IN_2_GA₆(H, L, S, A, B, qsort_2_out_ga₂(A, A1)) -> QSORT_2_IN_GA₂(B, B1)
IF_QSORT_2_IN_3_GA₆(H, L, S, B, A1, qsort_2_out_ga₂(B, B1)) -> IF_QSORT_2_IN_4_GA₆(H, L, S, A1, B1, append_3_in_gga₃(A1, ._2₂(H, B1), S))
IF_QSORT_2_IN_3_GA₆(H, L, S, B, A1, qsort_2_out_ga₂(B, B1)) -> APPEND_3_IN_GGA₃(A1, ._2₂(H, B1), S)
APPEND_3_IN_GGA₃(._2₂(H, L1), L2, ._2₂(H, L3)) -> IF_APPEND_3_IN_1_GGA₅(H, L1, L2, L3, append_3_in_gga₃(L1, L2, L3))
APPEND_3_IN_GGA₃(._2₂(H, L1), L2, ._2₂(H, L3)) -> APPEND_3_IN_GGA₃(L1, L2, L3)

The TRS R consists of the following rules:

qsort_2_in_ga₂([]_0, []_0) -> qsort_2_out_ga₂([]_0, []_0)
qsort_2_in_ga₂(._2₂(H, L), S) -> if_qsort_2_in_1_ga₄(H, L, S, split_4_in_ggaa₄(L, H, A, B))
split_4_in_ggaa₄([]_0, Y, []_0, []_0) -> split_4_out_ggaa₄([]_0, Y, []_0, []_0)
split_4_in_ggaa₄(._2₂(X, Xs), Y, ._2₂(X, Ls), Bs) -> if_split_4_in_1_ggaa₆(X, Xs, Y, Ls, Bs, le_2_in_gg₂(X, Y))
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(Y)) -> le_2_out_gg₂(0_0, s_1₁(Y))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_1_ggaa₆(X, Xs, Y, Ls, Bs, le_2_out_gg₂(X, Y)) -> if_split_4_in_2_ggaa₆(X, Xs, Y, Ls, Bs, split_4_in_ggaa₄(Xs, Y, Ls, Bs))
split_4_in_ggaa₄(._2₂(X, Xs), Y, Ls, ._2₂(X, Bs)) -> if_split_4_in_3_ggaa₆(X, Xs, Y, Ls, Bs, gt_2_in_gg₂(X, Y))
gt_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_gt_2_in_1_gg₃(X, Y, gt_2_in_gg₂(X, Y))
gt_2_in_gg₂(s_1₁(X), 0_0) -> gt_2_out_gg₂(s_1₁(X), 0_0)
if_gt_2_in_1_gg₃(X, Y, gt_2_out_gg₂(X, Y)) -> gt_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_3_ggaa₆(X, Xs, Y, Ls, Bs, gt_2_out_gg₂(X, Y)) -> if_split_4_in_4_ggaa₆(X, Xs, Y, Ls, Bs, split_4_in_ggaa₄(Xs, Y, Ls, Bs))
if_split_4_in_4_ggaa₆(X, Xs, Y, Ls, Bs, split_4_out_ggaa₄(Xs, Y, Ls, Bs)) -> split_4_out_ggaa₄(._2₂(X, Xs), Y, Ls, ._2₂(X, Bs))
if_split_4_in_2_ggaa₆(X, Xs, Y, Ls, Bs, split_4_out_ggaa₄(Xs, Y, Ls, Bs)) -> split_4_out_ggaa₄(._2₂(X, Xs), Y, ._2₂(X, Ls), Bs)
if_qsort_2_in_1_ga₄(H, L, S, split_4_out_ggaa₄(L, H, A, B)) -> if_qsort_2_in_2_ga₆(H, L, S, A, B, qsort_2_in_ga₂(A, A1))
if_qsort_2_in_2_ga₆(H, L, S, A, B, qsort_2_out_ga₂(A, A1)) -> if_qsort_2_in_3_ga₆(H, L, S, B, A1, qsort_2_in_ga₂(B, B1))
if_qsort_2_in_3_ga₆(H, L, S, B, A1, qsort_2_out_ga₂(B, B1)) -> if_qsort_2_in_4_ga₆(H, L, S, A1, B1, append_3_in_gga₃(A1, ._2₂(H, B1), S))
append_3_in_gga₃([]_0, L, L) -> append_3_out_gga₃([]_0, L, L)
append_3_in_gga₃(._2₂(H, L1), L2, ._2₂(H, L3)) -> if_append_3_in_1_gga₅(H, L1, L2, L3, append_3_in_gga₃(L1, L2, L3))
if_append_3_in_1_gga₅(H, L1, L2, L3, append_3_out_gga₃(L1, L2, L3)) -> append_3_out_gga₃(._2₂(H, L1), L2, ._2₂(H, L3))
if_qsort_2_in_4_ga₆(H, L, S, A1, B1, append_3_out_gga₃(A1, ._2₂(H, B1), S)) -> qsort_2_out_ga₂(._2₂(H, L), S)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

APPEND_3_IN_GGA₃(._2₂(H, L1), L2, ._2₂(H, L3)) -> APPEND_3_IN_GGA₃(L1, L2, L3)

The TRS R consists of the following rules:

qsort_2_in_ga₂([]_0, []_0) -> qsort_2_out_ga₂([]_0, []_0)
qsort_2_in_ga₂(._2₂(H, L), S) -> if_qsort_2_in_1_ga₄(H, L, S, split_4_in_ggaa₄(L, H, A, B))
split_4_in_ggaa₄([]_0, Y, []_0, []_0) -> split_4_out_ggaa₄([]_0, Y, []_0, []_0)
split_4_in_ggaa₄(._2₂(X, Xs), Y, ._2₂(X, Ls), Bs) -> if_split_4_in_1_ggaa₆(X, Xs, Y, Ls, Bs, le_2_in_gg₂(X, Y))
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(Y)) -> le_2_out_gg₂(0_0, s_1₁(Y))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_1_ggaa₆(X, Xs, Y, Ls, Bs, le_2_out_gg₂(X, Y)) -> if_split_4_in_2_ggaa₆(X, Xs, Y, Ls, Bs, split_4_in_ggaa₄(Xs, Y, Ls, Bs))
split_4_in_ggaa₄(._2₂(X, Xs), Y, Ls, ._2₂(X, Bs)) -> if_split_4_in_3_ggaa₆(X, Xs, Y, Ls, Bs, gt_2_in_gg₂(X, Y))
gt_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_gt_2_in_1_gg₃(X, Y, gt_2_in_gg₂(X, Y))
gt_2_in_gg₂(s_1₁(X), 0_0) -> gt_2_out_gg₂(s_1₁(X), 0_0)
if_gt_2_in_1_gg₃(X, Y, gt_2_out_gg₂(X, Y)) -> gt_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_3_ggaa₆(X, Xs, Y, Ls, Bs, gt_2_out_gg₂(X, Y)) -> if_split_4_in_4_ggaa₆(X, Xs, Y, Ls, Bs, split_4_in_ggaa₄(Xs, Y, Ls, Bs))
if_split_4_in_4_ggaa₆(X, Xs, Y, Ls, Bs, split_4_out_ggaa₄(Xs, Y, Ls, Bs)) -> split_4_out_ggaa₄(._2₂(X, Xs), Y, Ls, ._2₂(X, Bs))
if_split_4_in_2_ggaa₆(X, Xs, Y, Ls, Bs, split_4_out_ggaa₄(Xs, Y, Ls, Bs)) -> split_4_out_ggaa₄(._2₂(X, Xs), Y, ._2₂(X, Ls), Bs)
if_qsort_2_in_1_ga₄(H, L, S, split_4_out_ggaa₄(L, H, A, B)) -> if_qsort_2_in_2_ga₆(H, L, S, A, B, qsort_2_in_ga₂(A, A1))
if_qsort_2_in_2_ga₆(H, L, S, A, B, qsort_2_out_ga₂(A, A1)) -> if_qsort_2_in_3_ga₆(H, L, S, B, A1, qsort_2_in_ga₂(B, B1))
if_qsort_2_in_3_ga₆(H, L, S, B, A1, qsort_2_out_ga₂(B, B1)) -> if_qsort_2_in_4_ga₆(H, L, S, A1, B1, append_3_in_gga₃(A1, ._2₂(H, B1), S))
append_3_in_gga₃([]_0, L, L) -> append_3_out_gga₃([]_0, L, L)
append_3_in_gga₃(._2₂(H, L1), L2, ._2₂(H, L3)) -> if_append_3_in_1_gga₅(H, L1, L2, L3, append_3_in_gga₃(L1, L2, L3))
if_append_3_in_1_gga₅(H, L1, L2, L3, append_3_out_gga₃(L1, L2, L3)) -> append_3_out_gga₃(._2₂(H, L1), L2, ._2₂(H, L3))
if_qsort_2_in_4_ga₆(H, L, S, A1, B1, append_3_out_gga₃(A1, ._2₂(H, B1), S)) -> qsort_2_out_ga₂(._2₂(H, L), S)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

APPEND_3_IN_GGA₃(._2₂(H, L1), L2, ._2₂(H, L3)) -> APPEND_3_IN_GGA₃(L1, L2, L3)

R is empty.
The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
APPEND_3_IN_GGA₃(x1, x2, x3) = APPEND_3_IN_GGA₂(x1, x2)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

Q DP problem:
The TRS P consists of the following rules:

APPEND_3_IN_GGA₂(._2₂(H, L1), L2) -> APPEND_3_IN_GGA₂(L1, L2)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {APPEND_3_IN_GGA₂}.
By using the subterm criterion together with the size-change analysis 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_3_IN_GGA₂(._2₂(H, L1), L2) -> APPEND_3_IN_GGA₂(L1, L2)
The graph contains the following edges 1 > 1, 2 >= 2

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

GT_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> GT_2_IN_GG₂(X, Y)

The TRS R consists of the following rules:

qsort_2_in_ga₂([]_0, []_0) -> qsort_2_out_ga₂([]_0, []_0)
qsort_2_in_ga₂(._2₂(H, L), S) -> if_qsort_2_in_1_ga₄(H, L, S, split_4_in_ggaa₄(L, H, A, B))
split_4_in_ggaa₄([]_0, Y, []_0, []_0) -> split_4_out_ggaa₄([]_0, Y, []_0, []_0)
split_4_in_ggaa₄(._2₂(X, Xs), Y, ._2₂(X, Ls), Bs) -> if_split_4_in_1_ggaa₆(X, Xs, Y, Ls, Bs, le_2_in_gg₂(X, Y))
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(Y)) -> le_2_out_gg₂(0_0, s_1₁(Y))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_1_ggaa₆(X, Xs, Y, Ls, Bs, le_2_out_gg₂(X, Y)) -> if_split_4_in_2_ggaa₆(X, Xs, Y, Ls, Bs, split_4_in_ggaa₄(Xs, Y, Ls, Bs))
split_4_in_ggaa₄(._2₂(X, Xs), Y, Ls, ._2₂(X, Bs)) -> if_split_4_in_3_ggaa₆(X, Xs, Y, Ls, Bs, gt_2_in_gg₂(X, Y))
gt_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_gt_2_in_1_gg₃(X, Y, gt_2_in_gg₂(X, Y))
gt_2_in_gg₂(s_1₁(X), 0_0) -> gt_2_out_gg₂(s_1₁(X), 0_0)
if_gt_2_in_1_gg₃(X, Y, gt_2_out_gg₂(X, Y)) -> gt_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_3_ggaa₆(X, Xs, Y, Ls, Bs, gt_2_out_gg₂(X, Y)) -> if_split_4_in_4_ggaa₆(X, Xs, Y, Ls, Bs, split_4_in_ggaa₄(Xs, Y, Ls, Bs))
if_split_4_in_4_ggaa₆(X, Xs, Y, Ls, Bs, split_4_out_ggaa₄(Xs, Y, Ls, Bs)) -> split_4_out_ggaa₄(._2₂(X, Xs), Y, Ls, ._2₂(X, Bs))
if_split_4_in_2_ggaa₆(X, Xs, Y, Ls, Bs, split_4_out_ggaa₄(Xs, Y, Ls, Bs)) -> split_4_out_ggaa₄(._2₂(X, Xs), Y, ._2₂(X, Ls), Bs)
if_qsort_2_in_1_ga₄(H, L, S, split_4_out_ggaa₄(L, H, A, B)) -> if_qsort_2_in_2_ga₆(H, L, S, A, B, qsort_2_in_ga₂(A, A1))
if_qsort_2_in_2_ga₆(H, L, S, A, B, qsort_2_out_ga₂(A, A1)) -> if_qsort_2_in_3_ga₆(H, L, S, B, A1, qsort_2_in_ga₂(B, B1))
if_qsort_2_in_3_ga₆(H, L, S, B, A1, qsort_2_out_ga₂(B, B1)) -> if_qsort_2_in_4_ga₆(H, L, S, A1, B1, append_3_in_gga₃(A1, ._2₂(H, B1), S))
append_3_in_gga₃([]_0, L, L) -> append_3_out_gga₃([]_0, L, L)
append_3_in_gga₃(._2₂(H, L1), L2, ._2₂(H, L3)) -> if_append_3_in_1_gga₅(H, L1, L2, L3, append_3_in_gga₃(L1, L2, L3))
if_append_3_in_1_gga₅(H, L1, L2, L3, append_3_out_gga₃(L1, L2, L3)) -> append_3_out_gga₃(._2₂(H, L1), L2, ._2₂(H, L3))
if_qsort_2_in_4_ga₆(H, L, S, A1, B1, append_3_out_gga₃(A1, ._2₂(H, B1), S)) -> qsort_2_out_ga₂(._2₂(H, L), S)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

GT_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> GT_2_IN_GG₂(X, Y)

R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

Q DP problem:
The TRS P consists of the following rules:

GT_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> GT_2_IN_GG₂(X, Y)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {GT_2_IN_GG₂}.
By using the subterm criterion together with the size-change analysis 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:

GT_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> GT_2_IN_GG₂(X, Y)
The graph contains the following edges 1 > 1, 2 > 2

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

LE_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> LE_2_IN_GG₂(X, Y)

The TRS R consists of the following rules:

qsort_2_in_ga₂([]_0, []_0) -> qsort_2_out_ga₂([]_0, []_0)
qsort_2_in_ga₂(._2₂(H, L), S) -> if_qsort_2_in_1_ga₄(H, L, S, split_4_in_ggaa₄(L, H, A, B))
split_4_in_ggaa₄([]_0, Y, []_0, []_0) -> split_4_out_ggaa₄([]_0, Y, []_0, []_0)
split_4_in_ggaa₄(._2₂(X, Xs), Y, ._2₂(X, Ls), Bs) -> if_split_4_in_1_ggaa₆(X, Xs, Y, Ls, Bs, le_2_in_gg₂(X, Y))
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(Y)) -> le_2_out_gg₂(0_0, s_1₁(Y))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_1_ggaa₆(X, Xs, Y, Ls, Bs, le_2_out_gg₂(X, Y)) -> if_split_4_in_2_ggaa₆(X, Xs, Y, Ls, Bs, split_4_in_ggaa₄(Xs, Y, Ls, Bs))
split_4_in_ggaa₄(._2₂(X, Xs), Y, Ls, ._2₂(X, Bs)) -> if_split_4_in_3_ggaa₆(X, Xs, Y, Ls, Bs, gt_2_in_gg₂(X, Y))
gt_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_gt_2_in_1_gg₃(X, Y, gt_2_in_gg₂(X, Y))
gt_2_in_gg₂(s_1₁(X), 0_0) -> gt_2_out_gg₂(s_1₁(X), 0_0)
if_gt_2_in_1_gg₃(X, Y, gt_2_out_gg₂(X, Y)) -> gt_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_3_ggaa₆(X, Xs, Y, Ls, Bs, gt_2_out_gg₂(X, Y)) -> if_split_4_in_4_ggaa₆(X, Xs, Y, Ls, Bs, split_4_in_ggaa₄(Xs, Y, Ls, Bs))
if_split_4_in_4_ggaa₆(X, Xs, Y, Ls, Bs, split_4_out_ggaa₄(Xs, Y, Ls, Bs)) -> split_4_out_ggaa₄(._2₂(X, Xs), Y, Ls, ._2₂(X, Bs))
if_split_4_in_2_ggaa₆(X, Xs, Y, Ls, Bs, split_4_out_ggaa₄(Xs, Y, Ls, Bs)) -> split_4_out_ggaa₄(._2₂(X, Xs), Y, ._2₂(X, Ls), Bs)
if_qsort_2_in_1_ga₄(H, L, S, split_4_out_ggaa₄(L, H, A, B)) -> if_qsort_2_in_2_ga₆(H, L, S, A, B, qsort_2_in_ga₂(A, A1))
if_qsort_2_in_2_ga₆(H, L, S, A, B, qsort_2_out_ga₂(A, A1)) -> if_qsort_2_in_3_ga₆(H, L, S, B, A1, qsort_2_in_ga₂(B, B1))
if_qsort_2_in_3_ga₆(H, L, S, B, A1, qsort_2_out_ga₂(B, B1)) -> if_qsort_2_in_4_ga₆(H, L, S, A1, B1, append_3_in_gga₃(A1, ._2₂(H, B1), S))
append_3_in_gga₃([]_0, L, L) -> append_3_out_gga₃([]_0, L, L)
append_3_in_gga₃(._2₂(H, L1), L2, ._2₂(H, L3)) -> if_append_3_in_1_gga₅(H, L1, L2, L3, append_3_in_gga₃(L1, L2, L3))
if_append_3_in_1_gga₅(H, L1, L2, L3, append_3_out_gga₃(L1, L2, L3)) -> append_3_out_gga₃(._2₂(H, L1), L2, ._2₂(H, L3))
if_qsort_2_in_4_ga₆(H, L, S, A1, B1, append_3_out_gga₃(A1, ._2₂(H, B1), S)) -> qsort_2_out_ga₂(._2₂(H, L), S)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

LE_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> LE_2_IN_GG₂(X, Y)

R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

              ↳ PiDP

Q DP problem:
The TRS P consists of the following rules:

LE_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> LE_2_IN_GG₂(X, Y)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {LE_2_IN_GG₂}.
By using the subterm criterion together with the size-change analysis 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:

LE_2_IN_GG₂(s_1₁(X), s_1₁(Y)) -> LE_2_IN_GG₂(X, Y)
The graph contains the following edges 1 > 1, 2 > 2

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

SPLIT_4_IN_GGAA₄(._2₂(X, Xs), Y, Ls, ._2₂(X, Bs)) -> IF_SPLIT_4_IN_3_GGAA₆(X, Xs, Y, Ls, Bs, gt_2_in_gg₂(X, Y))
IF_SPLIT_4_IN_1_GGAA₆(X, Xs, Y, Ls, Bs, le_2_out_gg₂(X, Y)) -> SPLIT_4_IN_GGAA₄(Xs, Y, Ls, Bs)
IF_SPLIT_4_IN_3_GGAA₆(X, Xs, Y, Ls, Bs, gt_2_out_gg₂(X, Y)) -> SPLIT_4_IN_GGAA₄(Xs, Y, Ls, Bs)
SPLIT_4_IN_GGAA₄(._2₂(X, Xs), Y, ._2₂(X, Ls), Bs) -> IF_SPLIT_4_IN_1_GGAA₆(X, Xs, Y, Ls, Bs, le_2_in_gg₂(X, Y))

The TRS R consists of the following rules:

qsort_2_in_ga₂([]_0, []_0) -> qsort_2_out_ga₂([]_0, []_0)
qsort_2_in_ga₂(._2₂(H, L), S) -> if_qsort_2_in_1_ga₄(H, L, S, split_4_in_ggaa₄(L, H, A, B))
split_4_in_ggaa₄([]_0, Y, []_0, []_0) -> split_4_out_ggaa₄([]_0, Y, []_0, []_0)
split_4_in_ggaa₄(._2₂(X, Xs), Y, ._2₂(X, Ls), Bs) -> if_split_4_in_1_ggaa₆(X, Xs, Y, Ls, Bs, le_2_in_gg₂(X, Y))
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(Y)) -> le_2_out_gg₂(0_0, s_1₁(Y))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_1_ggaa₆(X, Xs, Y, Ls, Bs, le_2_out_gg₂(X, Y)) -> if_split_4_in_2_ggaa₆(X, Xs, Y, Ls, Bs, split_4_in_ggaa₄(Xs, Y, Ls, Bs))
split_4_in_ggaa₄(._2₂(X, Xs), Y, Ls, ._2₂(X, Bs)) -> if_split_4_in_3_ggaa₆(X, Xs, Y, Ls, Bs, gt_2_in_gg₂(X, Y))
gt_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_gt_2_in_1_gg₃(X, Y, gt_2_in_gg₂(X, Y))
gt_2_in_gg₂(s_1₁(X), 0_0) -> gt_2_out_gg₂(s_1₁(X), 0_0)
if_gt_2_in_1_gg₃(X, Y, gt_2_out_gg₂(X, Y)) -> gt_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_3_ggaa₆(X, Xs, Y, Ls, Bs, gt_2_out_gg₂(X, Y)) -> if_split_4_in_4_ggaa₆(X, Xs, Y, Ls, Bs, split_4_in_ggaa₄(Xs, Y, Ls, Bs))
if_split_4_in_4_ggaa₆(X, Xs, Y, Ls, Bs, split_4_out_ggaa₄(Xs, Y, Ls, Bs)) -> split_4_out_ggaa₄(._2₂(X, Xs), Y, Ls, ._2₂(X, Bs))
if_split_4_in_2_ggaa₆(X, Xs, Y, Ls, Bs, split_4_out_ggaa₄(Xs, Y, Ls, Bs)) -> split_4_out_ggaa₄(._2₂(X, Xs), Y, ._2₂(X, Ls), Bs)
if_qsort_2_in_1_ga₄(H, L, S, split_4_out_ggaa₄(L, H, A, B)) -> if_qsort_2_in_2_ga₆(H, L, S, A, B, qsort_2_in_ga₂(A, A1))
if_qsort_2_in_2_ga₆(H, L, S, A, B, qsort_2_out_ga₂(A, A1)) -> if_qsort_2_in_3_ga₆(H, L, S, B, A1, qsort_2_in_ga₂(B, B1))
if_qsort_2_in_3_ga₆(H, L, S, B, A1, qsort_2_out_ga₂(B, B1)) -> if_qsort_2_in_4_ga₆(H, L, S, A1, B1, append_3_in_gga₃(A1, ._2₂(H, B1), S))
append_3_in_gga₃([]_0, L, L) -> append_3_out_gga₃([]_0, L, L)
append_3_in_gga₃(._2₂(H, L1), L2, ._2₂(H, L3)) -> if_append_3_in_1_gga₅(H, L1, L2, L3, append_3_in_gga₃(L1, L2, L3))
if_append_3_in_1_gga₅(H, L1, L2, L3, append_3_out_gga₃(L1, L2, L3)) -> append_3_out_gga₃(._2₂(H, L1), L2, ._2₂(H, L3))
if_qsort_2_in_4_ga₆(H, L, S, A1, B1, append_3_out_gga₃(A1, ._2₂(H, B1), S)) -> qsort_2_out_ga₂(._2₂(H, L), S)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

SPLIT_4_IN_GGAA₄(._2₂(X, Xs), Y, Ls, ._2₂(X, Bs)) -> IF_SPLIT_4_IN_3_GGAA₆(X, Xs, Y, Ls, Bs, gt_2_in_gg₂(X, Y))
IF_SPLIT_4_IN_1_GGAA₆(X, Xs, Y, Ls, Bs, le_2_out_gg₂(X, Y)) -> SPLIT_4_IN_GGAA₄(Xs, Y, Ls, Bs)
IF_SPLIT_4_IN_3_GGAA₆(X, Xs, Y, Ls, Bs, gt_2_out_gg₂(X, Y)) -> SPLIT_4_IN_GGAA₄(Xs, Y, Ls, Bs)
SPLIT_4_IN_GGAA₄(._2₂(X, Xs), Y, ._2₂(X, Ls), Bs) -> IF_SPLIT_4_IN_1_GGAA₆(X, Xs, Y, Ls, Bs, le_2_in_gg₂(X, Y))

The TRS R consists of the following rules:

gt_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_gt_2_in_1_gg₃(X, Y, gt_2_in_gg₂(X, Y))
gt_2_in_gg₂(s_1₁(X), 0_0) -> gt_2_out_gg₂(s_1₁(X), 0_0)
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(Y)) -> le_2_out_gg₂(0_0, s_1₁(Y))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_gt_2_in_1_gg₃(X, Y, gt_2_out_gg₂(X, Y)) -> gt_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))

The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
le_2_in_gg₂(x1, x2) = le_2_in_gg₂(x1, x2)
if_le_2_in_1_gg₃(x1, x2, x3) = if_le_2_in_1_gg₁(x3)
le_2_out_gg₂(x1, x2) = le_2_out_gg
gt_2_in_gg₂(x1, x2) = gt_2_in_gg₂(x1, x2)
if_gt_2_in_1_gg₃(x1, x2, x3) = if_gt_2_in_1_gg₁(x3)
gt_2_out_gg₂(x1, x2) = gt_2_out_gg
IF_SPLIT_4_IN_1_GGAA₆(x1, x2, x3, x4, x5, x6) = IF_SPLIT_4_IN_1_GGAA₄(x1, x2, x3, x6)
SPLIT_4_IN_GGAA₄(x1, x2, x3, x4) = SPLIT_4_IN_GGAA₂(x1, x2)
IF_SPLIT_4_IN_3_GGAA₆(x1, x2, x3, x4, x5, x6) = IF_SPLIT_4_IN_3_GGAA₄(x1, x2, x3, x6)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

Q DP problem:
The TRS P consists of the following rules:

SPLIT_4_IN_GGAA₂(._2₂(X, Xs), Y) -> IF_SPLIT_4_IN_3_GGAA₄(X, Xs, Y, gt_2_in_gg₂(X, Y))
IF_SPLIT_4_IN_1_GGAA₄(X, Xs, Y, le_2_out_gg) -> SPLIT_4_IN_GGAA₂(Xs, Y)
IF_SPLIT_4_IN_3_GGAA₄(X, Xs, Y, gt_2_out_gg) -> SPLIT_4_IN_GGAA₂(Xs, Y)
SPLIT_4_IN_GGAA₂(._2₂(X, Xs), Y) -> IF_SPLIT_4_IN_1_GGAA₄(X, Xs, Y, le_2_in_gg₂(X, Y))

The TRS R consists of the following rules:

gt_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_gt_2_in_1_gg₁(gt_2_in_gg₂(X, Y))
gt_2_in_gg₂(s_1₁(X), 0_0) -> gt_2_out_gg
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₁(le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(Y)) -> le_2_out_gg
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg
if_gt_2_in_1_gg₁(gt_2_out_gg) -> gt_2_out_gg
if_le_2_in_1_gg₁(le_2_out_gg) -> le_2_out_gg

The set Q consists of the following terms:

gt_2_in_gg₂(x0, x1)
le_2_in_gg₂(x0, x1)
if_gt_2_in_1_gg₁(x0)
if_le_2_in_1_gg₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_SPLIT_4_IN_3_GGAA₄, SPLIT_4_IN_GGAA₂, IF_SPLIT_4_IN_1_GGAA₄}.
By using the subterm criterion together with the size-change analysis 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:

IF_SPLIT_4_IN_3_GGAA₄(X, Xs, Y, gt_2_out_gg) -> SPLIT_4_IN_GGAA₂(Xs, Y)
The graph contains the following edges 2 >= 1, 3 >= 2
IF_SPLIT_4_IN_1_GGAA₄(X, Xs, Y, le_2_out_gg) -> SPLIT_4_IN_GGAA₂(Xs, Y)
The graph contains the following edges 2 >= 1, 3 >= 2
SPLIT_4_IN_GGAA₂(._2₂(X, Xs), Y) -> IF_SPLIT_4_IN_1_GGAA₄(X, Xs, Y, le_2_in_gg₂(X, Y))
The graph contains the following edges 1 > 1, 1 > 2, 2 >= 3
SPLIT_4_IN_GGAA₂(._2₂(X, Xs), Y) -> IF_SPLIT_4_IN_3_GGAA₄(X, Xs, Y, gt_2_in_gg₂(X, Y))
The graph contains the following edges 1 > 1, 1 > 2, 2 >= 3

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

Pi DP problem:
The TRS P consists of the following rules:

QSORT_2_IN_GA₂(._2₂(H, L), S) -> IF_QSORT_2_IN_1_GA₄(H, L, S, split_4_in_ggaa₄(L, H, A, B))
IF_QSORT_2_IN_2_GA₆(H, L, S, A, B, qsort_2_out_ga₂(A, A1)) -> QSORT_2_IN_GA₂(B, B1)
IF_QSORT_2_IN_1_GA₄(H, L, S, split_4_out_ggaa₄(L, H, A, B)) -> QSORT_2_IN_GA₂(A, A1)
IF_QSORT_2_IN_1_GA₄(H, L, S, split_4_out_ggaa₄(L, H, A, B)) -> IF_QSORT_2_IN_2_GA₆(H, L, S, A, B, qsort_2_in_ga₂(A, A1))

The TRS R consists of the following rules:

qsort_2_in_ga₂([]_0, []_0) -> qsort_2_out_ga₂([]_0, []_0)
qsort_2_in_ga₂(._2₂(H, L), S) -> if_qsort_2_in_1_ga₄(H, L, S, split_4_in_ggaa₄(L, H, A, B))
split_4_in_ggaa₄([]_0, Y, []_0, []_0) -> split_4_out_ggaa₄([]_0, Y, []_0, []_0)
split_4_in_ggaa₄(._2₂(X, Xs), Y, ._2₂(X, Ls), Bs) -> if_split_4_in_1_ggaa₆(X, Xs, Y, Ls, Bs, le_2_in_gg₂(X, Y))
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₃(X, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(Y)) -> le_2_out_gg₂(0_0, s_1₁(Y))
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg₂(0_0, 0_0)
if_le_2_in_1_gg₃(X, Y, le_2_out_gg₂(X, Y)) -> le_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_1_ggaa₆(X, Xs, Y, Ls, Bs, le_2_out_gg₂(X, Y)) -> if_split_4_in_2_ggaa₆(X, Xs, Y, Ls, Bs, split_4_in_ggaa₄(Xs, Y, Ls, Bs))
split_4_in_ggaa₄(._2₂(X, Xs), Y, Ls, ._2₂(X, Bs)) -> if_split_4_in_3_ggaa₆(X, Xs, Y, Ls, Bs, gt_2_in_gg₂(X, Y))
gt_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_gt_2_in_1_gg₃(X, Y, gt_2_in_gg₂(X, Y))
gt_2_in_gg₂(s_1₁(X), 0_0) -> gt_2_out_gg₂(s_1₁(X), 0_0)
if_gt_2_in_1_gg₃(X, Y, gt_2_out_gg₂(X, Y)) -> gt_2_out_gg₂(s_1₁(X), s_1₁(Y))
if_split_4_in_3_ggaa₆(X, Xs, Y, Ls, Bs, gt_2_out_gg₂(X, Y)) -> if_split_4_in_4_ggaa₆(X, Xs, Y, Ls, Bs, split_4_in_ggaa₄(Xs, Y, Ls, Bs))
if_split_4_in_4_ggaa₆(X, Xs, Y, Ls, Bs, split_4_out_ggaa₄(Xs, Y, Ls, Bs)) -> split_4_out_ggaa₄(._2₂(X, Xs), Y, Ls, ._2₂(X, Bs))
if_split_4_in_2_ggaa₆(X, Xs, Y, Ls, Bs, split_4_out_ggaa₄(Xs, Y, Ls, Bs)) -> split_4_out_ggaa₄(._2₂(X, Xs), Y, ._2₂(X, Ls), Bs)
if_qsort_2_in_1_ga₄(H, L, S, split_4_out_ggaa₄(L, H, A, B)) -> if_qsort_2_in_2_ga₆(H, L, S, A, B, qsort_2_in_ga₂(A, A1))
if_qsort_2_in_2_ga₆(H, L, S, A, B, qsort_2_out_ga₂(A, A1)) -> if_qsort_2_in_3_ga₆(H, L, S, B, A1, qsort_2_in_ga₂(B, B1))
if_qsort_2_in_3_ga₆(H, L, S, B, A1, qsort_2_out_ga₂(B, B1)) -> if_qsort_2_in_4_ga₆(H, L, S, A1, B1, append_3_in_gga₃(A1, ._2₂(H, B1), S))
append_3_in_gga₃([]_0, L, L) -> append_3_out_gga₃([]_0, L, L)
append_3_in_gga₃(._2₂(H, L1), L2, ._2₂(H, L3)) -> if_append_3_in_1_gga₅(H, L1, L2, L3, append_3_in_gga₃(L1, L2, L3))
if_append_3_in_1_gga₅(H, L1, L2, L3, append_3_out_gga₃(L1, L2, L3)) -> append_3_out_gga₃(._2₂(H, L1), L2, ._2₂(H, L3))
if_qsort_2_in_4_ga₆(H, L, S, A1, B1, append_3_out_gga₃(A1, ._2₂(H, B1), S)) -> qsort_2_out_ga₂(._2₂(H, L), S)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPPoloProof

Q DP problem:
The TRS P consists of the following rules:

QSORT_2_IN_GA₁(._2₂(H, L)) -> IF_QSORT_2_IN_1_GA₂(H, split_4_in_ggaa₂(L, H))
IF_QSORT_2_IN_2_GA₃(H, B, qsort_2_out_ga₁(A1)) -> QSORT_2_IN_GA₁(B)
IF_QSORT_2_IN_1_GA₂(H, split_4_out_ggaa₂(A, B)) -> QSORT_2_IN_GA₁(A)
IF_QSORT_2_IN_1_GA₂(H, split_4_out_ggaa₂(A, B)) -> IF_QSORT_2_IN_2_GA₃(H, B, qsort_2_in_ga₁(A))

The TRS R consists of the following rules:

qsort_2_in_ga₁([]_0) -> qsort_2_out_ga₁([]_0)
qsort_2_in_ga₁(._2₂(H, L)) -> if_qsort_2_in_1_ga₂(H, split_4_in_ggaa₂(L, H))
split_4_in_ggaa₂([]_0, Y) -> split_4_out_ggaa₂([]_0, []_0)
split_4_in_ggaa₂(._2₂(X, Xs), Y) -> if_split_4_in_1_ggaa₄(X, Xs, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₁(le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(Y)) -> le_2_out_gg
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg
if_le_2_in_1_gg₁(le_2_out_gg) -> le_2_out_gg
if_split_4_in_1_ggaa₄(X, Xs, Y, le_2_out_gg) -> if_split_4_in_2_ggaa₂(X, split_4_in_ggaa₂(Xs, Y))
split_4_in_ggaa₂(._2₂(X, Xs), Y) -> if_split_4_in_3_ggaa₄(X, Xs, Y, gt_2_in_gg₂(X, Y))
gt_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_gt_2_in_1_gg₁(gt_2_in_gg₂(X, Y))
gt_2_in_gg₂(s_1₁(X), 0_0) -> gt_2_out_gg
if_gt_2_in_1_gg₁(gt_2_out_gg) -> gt_2_out_gg
if_split_4_in_3_ggaa₄(X, Xs, Y, gt_2_out_gg) -> if_split_4_in_4_ggaa₂(X, split_4_in_ggaa₂(Xs, Y))
if_split_4_in_4_ggaa₂(X, split_4_out_ggaa₂(Ls, Bs)) -> split_4_out_ggaa₂(Ls, ._2₂(X, Bs))
if_split_4_in_2_ggaa₂(X, split_4_out_ggaa₂(Ls, Bs)) -> split_4_out_ggaa₂(._2₂(X, Ls), Bs)
if_qsort_2_in_1_ga₂(H, split_4_out_ggaa₂(A, B)) -> if_qsort_2_in_2_ga₃(H, B, qsort_2_in_ga₁(A))
if_qsort_2_in_2_ga₃(H, B, qsort_2_out_ga₁(A1)) -> if_qsort_2_in_3_ga₃(H, A1, qsort_2_in_ga₁(B))
if_qsort_2_in_3_ga₃(H, A1, qsort_2_out_ga₁(B1)) -> if_qsort_2_in_4_ga₁(append_3_in_gga₂(A1, ._2₂(H, B1)))
append_3_in_gga₂([]_0, L) -> append_3_out_gga₁(L)
append_3_in_gga₂(._2₂(H, L1), L2) -> if_append_3_in_1_gga₂(H, append_3_in_gga₂(L1, L2))
if_append_3_in_1_gga₂(H, append_3_out_gga₁(L3)) -> append_3_out_gga₁(._2₂(H, L3))
if_qsort_2_in_4_ga₁(append_3_out_gga₁(S)) -> qsort_2_out_ga₁(S)

The set Q consists of the following terms:

qsort_2_in_ga₁(x0)
split_4_in_ggaa₂(x0, x1)
le_2_in_gg₂(x0, x1)
if_le_2_in_1_gg₁(x0)
if_split_4_in_1_ggaa₄(x0, x1, x2, x3)
gt_2_in_gg₂(x0, x1)
if_gt_2_in_1_gg₁(x0)
if_split_4_in_3_ggaa₄(x0, x1, x2, x3)
if_split_4_in_4_ggaa₂(x0, x1)
if_split_4_in_2_ggaa₂(x0, x1)
if_qsort_2_in_1_ga₂(x0, x1)
if_qsort_2_in_2_ga₃(x0, x1, x2)
if_qsort_2_in_3_ga₃(x0, x1, x2)
append_3_in_gga₂(x0, x1)
if_append_3_in_1_gga₂(x0, x1)
if_qsort_2_in_4_ga₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_QSORT_2_IN_1_GA₂, QSORT_2_IN_GA₁, IF_QSORT_2_IN_2_GA₃}.
By using a polynomial ordering, the following set of Dependency Pairs of this DP problem can be strictly oriented.

QSORT_2_IN_GA₁(._2₂(H, L)) -> IF_QSORT_2_IN_1_GA₂(H, split_4_in_ggaa₂(L, H))

The remaining Dependency Pairs were at least non-strictly be oriented.

IF_QSORT_2_IN_2_GA₃(H, B, qsort_2_out_ga₁(A1)) -> QSORT_2_IN_GA₁(B)
IF_QSORT_2_IN_1_GA₂(H, split_4_out_ggaa₂(A, B)) -> QSORT_2_IN_GA₁(A)
IF_QSORT_2_IN_1_GA₂(H, split_4_out_ggaa₂(A, B)) -> IF_QSORT_2_IN_2_GA₃(H, B, qsort_2_in_ga₁(A))

With the implicit AFS we had to orient the following set of usable rules non-strictly.

split_4_in_ggaa₂(._2₂(X, Xs), Y) -> if_split_4_in_1_ggaa₄(X, Xs, Y, le_2_in_gg₂(X, Y))
if_split_4_in_4_ggaa₂(X, split_4_out_ggaa₂(Ls, Bs)) -> split_4_out_ggaa₂(Ls, ._2₂(X, Bs))
split_4_in_ggaa₂([]_0, Y) -> split_4_out_ggaa₂([]_0, []_0)
if_split_4_in_2_ggaa₂(X, split_4_out_ggaa₂(Ls, Bs)) -> split_4_out_ggaa₂(._2₂(X, Ls), Bs)
split_4_in_ggaa₂(._2₂(X, Xs), Y) -> if_split_4_in_3_ggaa₄(X, Xs, Y, gt_2_in_gg₂(X, Y))
if_split_4_in_3_ggaa₄(X, Xs, Y, gt_2_out_gg) -> if_split_4_in_4_ggaa₂(X, split_4_in_ggaa₂(Xs, Y))
if_split_4_in_1_ggaa₄(X, Xs, Y, le_2_out_gg) -> if_split_4_in_2_ggaa₂(X, split_4_in_ggaa₂(Xs, Y))

Used ordering: POLO with Polynomial interpretation:

POL(0_0) = 0
POL(gt_2_in_gg₂(x₁, x₂)) = 0
POL(split_4_in_ggaa₂(x₁, x₂)) = x₁
POL(if_append_3_in_1_gga₂(x₁, x₂)) = 0
POL(qsort_2_out_ga₁(x₁)) = 0
POL(if_qsort_2_in_2_ga₃(x₁, x₂, x₃)) = 0
POL(append_3_in_gga₂(x₁, x₂)) = 0
POL(if_split_4_in_1_ggaa₄(x₁, x₂, x₃, x₄)) = 1 + x₂
POL(append_3_out_gga₁(x₁)) = 0
POL(if_gt_2_in_1_gg₁(x₁)) = 0
POL(if_qsort_2_in_1_ga₂(x₁, x₂)) = 0
POL(IF_QSORT_2_IN_1_GA₂(x₁, x₂)) = x₂
POL(le_2_out_gg) = 0
POL(qsort_2_in_ga₁(x₁)) = 0
POL([]_0) = 0
POL(if_le_2_in_1_gg₁(x₁)) = 0
POL(split_4_out_ggaa₂(x₁, x₂)) = x₁ + x₂
POL(._2₂(x₁, x₂)) = 1 + x₂
POL(if_split_4_in_2_ggaa₂(x₁, x₂)) = 1 + x₂
POL(QSORT_2_IN_GA₁(x₁)) = x₁
POL(if_qsort_2_in_4_ga₁(x₁)) = 0
POL(IF_QSORT_2_IN_2_GA₃(x₁, x₂, x₃)) = x₂
POL(le_2_in_gg₂(x₁, x₂)) = 0
POL(if_qsort_2_in_3_ga₃(x₁, x₂, x₃)) = 0
POL(gt_2_out_gg) = 0
POL(s_1₁(x₁)) = 0
POL(if_split_4_in_4_ggaa₂(x₁, x₂)) = 1 + x₂
POL(if_split_4_in_3_ggaa₄(x₁, x₂, x₃, x₄)) = 1 + x₂

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPPoloProof

                      ↳ QDP

                        ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

IF_QSORT_2_IN_2_GA₃(H, B, qsort_2_out_ga₁(A1)) -> QSORT_2_IN_GA₁(B)
IF_QSORT_2_IN_1_GA₂(H, split_4_out_ggaa₂(A, B)) -> QSORT_2_IN_GA₁(A)
IF_QSORT_2_IN_1_GA₂(H, split_4_out_ggaa₂(A, B)) -> IF_QSORT_2_IN_2_GA₃(H, B, qsort_2_in_ga₁(A))

The TRS R consists of the following rules:

qsort_2_in_ga₁([]_0) -> qsort_2_out_ga₁([]_0)
qsort_2_in_ga₁(._2₂(H, L)) -> if_qsort_2_in_1_ga₂(H, split_4_in_ggaa₂(L, H))
split_4_in_ggaa₂([]_0, Y) -> split_4_out_ggaa₂([]_0, []_0)
split_4_in_ggaa₂(._2₂(X, Xs), Y) -> if_split_4_in_1_ggaa₄(X, Xs, Y, le_2_in_gg₂(X, Y))
le_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_le_2_in_1_gg₁(le_2_in_gg₂(X, Y))
le_2_in_gg₂(0_0, s_1₁(Y)) -> le_2_out_gg
le_2_in_gg₂(0_0, 0_0) -> le_2_out_gg
if_le_2_in_1_gg₁(le_2_out_gg) -> le_2_out_gg
if_split_4_in_1_ggaa₄(X, Xs, Y, le_2_out_gg) -> if_split_4_in_2_ggaa₂(X, split_4_in_ggaa₂(Xs, Y))
split_4_in_ggaa₂(._2₂(X, Xs), Y) -> if_split_4_in_3_ggaa₄(X, Xs, Y, gt_2_in_gg₂(X, Y))
gt_2_in_gg₂(s_1₁(X), s_1₁(Y)) -> if_gt_2_in_1_gg₁(gt_2_in_gg₂(X, Y))
gt_2_in_gg₂(s_1₁(X), 0_0) -> gt_2_out_gg
if_gt_2_in_1_gg₁(gt_2_out_gg) -> gt_2_out_gg
if_split_4_in_3_ggaa₄(X, Xs, Y, gt_2_out_gg) -> if_split_4_in_4_ggaa₂(X, split_4_in_ggaa₂(Xs, Y))
if_split_4_in_4_ggaa₂(X, split_4_out_ggaa₂(Ls, Bs)) -> split_4_out_ggaa₂(Ls, ._2₂(X, Bs))
if_split_4_in_2_ggaa₂(X, split_4_out_ggaa₂(Ls, Bs)) -> split_4_out_ggaa₂(._2₂(X, Ls), Bs)
if_qsort_2_in_1_ga₂(H, split_4_out_ggaa₂(A, B)) -> if_qsort_2_in_2_ga₃(H, B, qsort_2_in_ga₁(A))
if_qsort_2_in_2_ga₃(H, B, qsort_2_out_ga₁(A1)) -> if_qsort_2_in_3_ga₃(H, A1, qsort_2_in_ga₁(B))
if_qsort_2_in_3_ga₃(H, A1, qsort_2_out_ga₁(B1)) -> if_qsort_2_in_4_ga₁(append_3_in_gga₂(A1, ._2₂(H, B1)))
append_3_in_gga₂([]_0, L) -> append_3_out_gga₁(L)
append_3_in_gga₂(._2₂(H, L1), L2) -> if_append_3_in_1_gga₂(H, append_3_in_gga₂(L1, L2))
if_append_3_in_1_gga₂(H, append_3_out_gga₁(L3)) -> append_3_out_gga₁(._2₂(H, L3))
if_qsort_2_in_4_ga₁(append_3_out_gga₁(S)) -> qsort_2_out_ga₁(S)

The set Q consists of the following terms:

qsort_2_in_ga₁(x0)
split_4_in_ggaa₂(x0, x1)
le_2_in_gg₂(x0, x1)
if_le_2_in_1_gg₁(x0)
if_split_4_in_1_ggaa₄(x0, x1, x2, x3)
gt_2_in_gg₂(x0, x1)
if_gt_2_in_1_gg₁(x0)
if_split_4_in_3_ggaa₄(x0, x1, x2, x3)
if_split_4_in_4_ggaa₂(x0, x1)
if_split_4_in_2_ggaa₂(x0, x1)
if_qsort_2_in_1_ga₂(x0, x1)
if_qsort_2_in_2_ga₃(x0, x1, x2)
if_qsort_2_in_3_ga₃(x0, x1, x2)
append_3_in_gga₂(x0, x1)
if_append_3_in_1_gga₂(x0, x1)
if_qsort_2_in_4_ga₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {QSORT_2_IN_GA₁, IF_QSORT_2_IN_2_GA₃, IF_QSORT_2_IN_1_GA₂}.
The approximation of the Dependency Graph contains 0 SCCs with 3 less nodes.