Left Termination proof of /tmp/tmp2GsOFm/quicksort.pl

Left Termination of the query pattern qs_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:

↳ Prolog

  ↳ PrologToPiTRSProof

Clauses:

qs(.(X, Xs), Ys) :- ','(part(X, Xs, Littles, Bigs), ','(qs(Littles, Ls), ','(qs(Bigs, Bs), app(Ls, .(X, Bs), Ys)))).
qs([], []).
part(X, .(Y, Xs), .(Y, Ls), Bs) :- ','(gt(X, Y), part(X, Xs, Ls, Bs)).
part(X, .(Y, Xs), Ls, .(Y, Bs)) :- ','(le(X, Y), part(X, Xs, Ls, Bs)).
part(X, [], [], []).
app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).
app([], Ys, Ys).
gt(s(X), s(Y)) :- gt(X, Y).
gt(s(0), 0).
le(s(X), s(Y)) :- le(X, Y).
le(0, s(X)).
le(0, 0).

Queries:

qs(g,a).

We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

qs_in([], []) → qs_out([], [])
qs_in(.(X, Xs), Ys) → U1(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
part_in(X, [], [], []) → part_out(X, [], [], [])
part_in(X, .(Y, Xs), Ls, .(Y, Bs)) → U7(X, Y, Xs, Ls, Bs, le_in(X, Y))
le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U11(X, Y, le_in(X, Y))
U11(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U7(X, Y, Xs, Ls, Bs, le_out(X, Y)) → U8(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
part_in(X, .(Y, Xs), .(Y, Ls), Bs) → U5(X, Y, Xs, Ls, Bs, gt_in(X, Y))
gt_in(s(0), 0) → gt_out(s(0), 0)
gt_in(s(X), s(Y)) → U10(X, Y, gt_in(X, Y))
U10(X, Y, gt_out(X, Y)) → gt_out(s(X), s(Y))
U5(X, Y, Xs, Ls, Bs, gt_out(X, Y)) → U6(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U6(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), .(Y, Ls), Bs)
U8(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), Ls, .(Y, Bs))
U1(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
U2(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U3(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U9(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U9(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Xs, Ys, app_out(Ls, .(X, Bs), Ys)) → qs_out(.(X, Xs), Ys)

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:

qs_in([], []) → qs_out([], [])
qs_in(.(X, Xs), Ys) → U1(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
part_in(X, [], [], []) → part_out(X, [], [], [])
part_in(X, .(Y, Xs), Ls, .(Y, Bs)) → U7(X, Y, Xs, Ls, Bs, le_in(X, Y))
le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U11(X, Y, le_in(X, Y))
U11(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U7(X, Y, Xs, Ls, Bs, le_out(X, Y)) → U8(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
part_in(X, .(Y, Xs), .(Y, Ls), Bs) → U5(X, Y, Xs, Ls, Bs, gt_in(X, Y))
gt_in(s(0), 0) → gt_out(s(0), 0)
gt_in(s(X), s(Y)) → U10(X, Y, gt_in(X, Y))
U10(X, Y, gt_out(X, Y)) → gt_out(s(X), s(Y))
U5(X, Y, Xs, Ls, Bs, gt_out(X, Y)) → U6(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U6(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), .(Y, Ls), Bs)
U8(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), Ls, .(Y, Bs))
U1(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
U2(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U3(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U9(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U9(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Xs, Ys, app_out(Ls, .(X, Bs), Ys)) → qs_out(.(X, Xs), Ys)

QS_IN(.(X, Xs), Ys) → U1¹(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
QS_IN(.(X, Xs), Ys) → PART_IN(X, Xs, Littles, Bigs)
PART_IN(X, .(Y, Xs), Ls, .(Y, Bs)) → U7¹(X, Y, Xs, Ls, Bs, le_in(X, Y))
PART_IN(X, .(Y, Xs), Ls, .(Y, Bs)) → LE_IN(X, Y)
LE_IN(s(X), s(Y)) → U11¹(X, Y, le_in(X, Y))
LE_IN(s(X), s(Y)) → LE_IN(X, Y)
U7¹(X, Y, Xs, Ls, Bs, le_out(X, Y)) → U8¹(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U7¹(X, Y, Xs, Ls, Bs, le_out(X, Y)) → PART_IN(X, Xs, Ls, Bs)
PART_IN(X, .(Y, Xs), .(Y, Ls), Bs) → U5¹(X, Y, Xs, Ls, Bs, gt_in(X, Y))
PART_IN(X, .(Y, Xs), .(Y, Ls), Bs) → GT_IN(X, Y)
GT_IN(s(X), s(Y)) → U10¹(X, Y, gt_in(X, Y))
GT_IN(s(X), s(Y)) → GT_IN(X, Y)
U5¹(X, Y, Xs, Ls, Bs, gt_out(X, Y)) → U6¹(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U5¹(X, Y, Xs, Ls, Bs, gt_out(X, Y)) → PART_IN(X, Xs, Ls, Bs)
U1¹(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2¹(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
U1¹(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → QS_IN(Littles, Ls)
U2¹(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3¹(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U2¹(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → QS_IN(Bigs, Bs)
U3¹(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4¹(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
U3¹(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → APP_IN(Ls, .(X, Bs), Ys)
APP_IN(.(X, Xs), Ys, .(X, Zs)) → U9¹(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

qs_in([], []) → qs_out([], [])
qs_in(.(X, Xs), Ys) → U1(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
part_in(X, [], [], []) → part_out(X, [], [], [])
part_in(X, .(Y, Xs), Ls, .(Y, Bs)) → U7(X, Y, Xs, Ls, Bs, le_in(X, Y))
le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U11(X, Y, le_in(X, Y))
U11(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U7(X, Y, Xs, Ls, Bs, le_out(X, Y)) → U8(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
part_in(X, .(Y, Xs), .(Y, Ls), Bs) → U5(X, Y, Xs, Ls, Bs, gt_in(X, Y))
gt_in(s(0), 0) → gt_out(s(0), 0)
gt_in(s(X), s(Y)) → U10(X, Y, gt_in(X, Y))
U10(X, Y, gt_out(X, Y)) → gt_out(s(X), s(Y))
U5(X, Y, Xs, Ls, Bs, gt_out(X, Y)) → U6(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U6(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), .(Y, Ls), Bs)
U8(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), Ls, .(Y, Bs))
U1(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
U2(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U3(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U9(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U9(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Xs, Ys, app_out(Ls, .(X, Bs), Ys)) → qs_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
qs_in(x1, x2) = qs_in(x1)
[] = []
qs_out(x1, x2) = qs_out(x2)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x1, x4)
part_in(x1, x2, x3, x4) = part_in(x1, x2)
part_out(x1, x2, x3, x4) = part_out(x3, x4)
U7(x1, x2, x3, x4, x5, x6) = U7(x1, x2, x3, x6)
le_in(x1, x2) = le_in(x1, x2)
0 = 0
le_out(x1, x2) = le_out
s(x1) = s(x1)
U11(x1, x2, x3) = U11(x3)
U8(x1, x2, x3, x4, x5, x6) = U8(x2, x6)
U5(x1, x2, x3, x4, x5, x6) = U5(x1, x2, x3, x6)
gt_in(x1, x2) = gt_in(x1, x2)
gt_out(x1, x2) = gt_out
U10(x1, x2, x3) = U10(x3)
U6(x1, x2, x3, x4, x5, x6) = U6(x2, x6)
U2(x1, x2, x3, x4, x5) = U2(x1, x4, x5)
U3(x1, x2, x3, x4, x5) = U3(x1, x4, x5)
U4(x1, x2, x3, x4) = U4(x4)
app_in(x1, x2, x3) = app_in(x1, x2)
app_out(x1, x2, x3) = app_out(x3)
U9(x1, x2, x3, x4, x5) = U9(x1, x5)
U11¹(x1, x2, x3) = U11¹(x3)
U5¹(x1, x2, x3, x4, x5, x6) = U5¹(x1, x2, x3, x6)
U4¹(x1, x2, x3, x4) = U4¹(x4)
U3¹(x1, x2, x3, x4, x5) = U3¹(x1, x4, x5)
U9¹(x1, x2, x3, x4, x5) = U9¹(x1, x5)
PART_IN(x1, x2, x3, x4) = PART_IN(x1, x2)
U8¹(x1, x2, x3, x4, x5, x6) = U8¹(x2, x6)
LE_IN(x1, x2) = LE_IN(x1, x2)
QS_IN(x1, x2) = QS_IN(x1)
APP_IN(x1, x2, x3) = APP_IN(x1, x2)
U7¹(x1, x2, x3, x4, x5, x6) = U7¹(x1, x2, x3, x6)
U10¹(x1, x2, x3) = U10¹(x3)
U2¹(x1, x2, x3, x4, x5) = U2¹(x1, x4, x5)
GT_IN(x1, x2) = GT_IN(x1, x2)
U1¹(x1, x2, x3, x4) = U1¹(x1, x4)
U6¹(x1, x2, x3, x4, x5, x6) = U6¹(x2, 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:

QS_IN(.(X, Xs), Ys) → U1¹(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
QS_IN(.(X, Xs), Ys) → PART_IN(X, Xs, Littles, Bigs)
PART_IN(X, .(Y, Xs), Ls, .(Y, Bs)) → U7¹(X, Y, Xs, Ls, Bs, le_in(X, Y))
PART_IN(X, .(Y, Xs), Ls, .(Y, Bs)) → LE_IN(X, Y)
LE_IN(s(X), s(Y)) → U11¹(X, Y, le_in(X, Y))
LE_IN(s(X), s(Y)) → LE_IN(X, Y)
U7¹(X, Y, Xs, Ls, Bs, le_out(X, Y)) → U8¹(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U7¹(X, Y, Xs, Ls, Bs, le_out(X, Y)) → PART_IN(X, Xs, Ls, Bs)
PART_IN(X, .(Y, Xs), .(Y, Ls), Bs) → U5¹(X, Y, Xs, Ls, Bs, gt_in(X, Y))
PART_IN(X, .(Y, Xs), .(Y, Ls), Bs) → GT_IN(X, Y)
GT_IN(s(X), s(Y)) → U10¹(X, Y, gt_in(X, Y))
GT_IN(s(X), s(Y)) → GT_IN(X, Y)
U5¹(X, Y, Xs, Ls, Bs, gt_out(X, Y)) → U6¹(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U5¹(X, Y, Xs, Ls, Bs, gt_out(X, Y)) → PART_IN(X, Xs, Ls, Bs)
U1¹(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2¹(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
U1¹(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → QS_IN(Littles, Ls)
U2¹(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3¹(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U2¹(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → QS_IN(Bigs, Bs)
U3¹(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4¹(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
U3¹(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → APP_IN(Ls, .(X, Bs), Ys)
APP_IN(.(X, Xs), Ys, .(X, Zs)) → U9¹(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

qs_in([], []) → qs_out([], [])
qs_in(.(X, Xs), Ys) → U1(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
part_in(X, [], [], []) → part_out(X, [], [], [])
part_in(X, .(Y, Xs), Ls, .(Y, Bs)) → U7(X, Y, Xs, Ls, Bs, le_in(X, Y))
le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U11(X, Y, le_in(X, Y))
U11(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U7(X, Y, Xs, Ls, Bs, le_out(X, Y)) → U8(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
part_in(X, .(Y, Xs), .(Y, Ls), Bs) → U5(X, Y, Xs, Ls, Bs, gt_in(X, Y))
gt_in(s(0), 0) → gt_out(s(0), 0)
gt_in(s(X), s(Y)) → U10(X, Y, gt_in(X, Y))
U10(X, Y, gt_out(X, Y)) → gt_out(s(X), s(Y))
U5(X, Y, Xs, Ls, Bs, gt_out(X, Y)) → U6(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U6(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), .(Y, Ls), Bs)
U8(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), Ls, .(Y, Bs))
U1(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
U2(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U3(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U9(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U9(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Xs, Ys, app_out(Ls, .(X, Bs), Ys)) → qs_out(.(X, Xs), Ys)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

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

APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

qs_in([], []) → qs_out([], [])
qs_in(.(X, Xs), Ys) → U1(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
part_in(X, [], [], []) → part_out(X, [], [], [])
part_in(X, .(Y, Xs), Ls, .(Y, Bs)) → U7(X, Y, Xs, Ls, Bs, le_in(X, Y))
le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U11(X, Y, le_in(X, Y))
U11(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U7(X, Y, Xs, Ls, Bs, le_out(X, Y)) → U8(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
part_in(X, .(Y, Xs), .(Y, Ls), Bs) → U5(X, Y, Xs, Ls, Bs, gt_in(X, Y))
gt_in(s(0), 0) → gt_out(s(0), 0)
gt_in(s(X), s(Y)) → U10(X, Y, gt_in(X, Y))
U10(X, Y, gt_out(X, Y)) → gt_out(s(X), s(Y))
U5(X, Y, Xs, Ls, Bs, gt_out(X, Y)) → U6(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U6(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), .(Y, Ls), Bs)
U8(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), Ls, .(Y, Bs))
U1(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
U2(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U3(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U9(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U9(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Xs, Ys, app_out(Ls, .(X, Bs), Ys)) → qs_out(.(X, Xs), Ys)

↳ 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:

APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

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

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] 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:

APP_IN(.(X, Xs), Ys) → APP_IN(Xs, Ys)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:

APP_IN(.(X, Xs), Ys) → APP_IN(Xs, Ys)
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_IN(s(X), s(Y)) → GT_IN(X, Y)

The TRS R consists of the following rules:

qs_in([], []) → qs_out([], [])
qs_in(.(X, Xs), Ys) → U1(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
part_in(X, [], [], []) → part_out(X, [], [], [])
part_in(X, .(Y, Xs), Ls, .(Y, Bs)) → U7(X, Y, Xs, Ls, Bs, le_in(X, Y))
le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U11(X, Y, le_in(X, Y))
U11(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U7(X, Y, Xs, Ls, Bs, le_out(X, Y)) → U8(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
part_in(X, .(Y, Xs), .(Y, Ls), Bs) → U5(X, Y, Xs, Ls, Bs, gt_in(X, Y))
gt_in(s(0), 0) → gt_out(s(0), 0)
gt_in(s(X), s(Y)) → U10(X, Y, gt_in(X, Y))
U10(X, Y, gt_out(X, Y)) → gt_out(s(X), s(Y))
U5(X, Y, Xs, Ls, Bs, gt_out(X, Y)) → U6(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U6(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), .(Y, Ls), Bs)
U8(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), Ls, .(Y, Bs))
U1(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
U2(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U3(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U9(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U9(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Xs, Ys, app_out(Ls, .(X, Bs), Ys)) → qs_out(.(X, Xs), Ys)

↳ 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_IN(s(X), s(Y)) → GT_IN(X, Y)

R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] 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_IN(s(X), s(Y)) → GT_IN(X, Y)

From the DPs we obtained the following set of size-change graphs:

GT_IN(s(X), s(Y)) → GT_IN(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_IN(s(X), s(Y)) → LE_IN(X, Y)

The TRS R consists of the following rules:

qs_in([], []) → qs_out([], [])
qs_in(.(X, Xs), Ys) → U1(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
part_in(X, [], [], []) → part_out(X, [], [], [])
part_in(X, .(Y, Xs), Ls, .(Y, Bs)) → U7(X, Y, Xs, Ls, Bs, le_in(X, Y))
le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U11(X, Y, le_in(X, Y))
U11(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U7(X, Y, Xs, Ls, Bs, le_out(X, Y)) → U8(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
part_in(X, .(Y, Xs), .(Y, Ls), Bs) → U5(X, Y, Xs, Ls, Bs, gt_in(X, Y))
gt_in(s(0), 0) → gt_out(s(0), 0)
gt_in(s(X), s(Y)) → U10(X, Y, gt_in(X, Y))
U10(X, Y, gt_out(X, Y)) → gt_out(s(X), s(Y))
U5(X, Y, Xs, Ls, Bs, gt_out(X, Y)) → U6(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U6(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), .(Y, Ls), Bs)
U8(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), Ls, .(Y, Bs))
U1(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
U2(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U3(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U9(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U9(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Xs, Ys, app_out(Ls, .(X, Bs), Ys)) → qs_out(.(X, Xs), Ys)

↳ 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_IN(s(X), s(Y)) → LE_IN(X, Y)

R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] 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_IN(s(X), s(Y)) → LE_IN(X, Y)

From the DPs we obtained the following set of size-change graphs:

LE_IN(s(X), s(Y)) → LE_IN(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:

PART_IN(X, .(Y, Xs), Ls, .(Y, Bs)) → U7¹(X, Y, Xs, Ls, Bs, le_in(X, Y))
U7¹(X, Y, Xs, Ls, Bs, le_out(X, Y)) → PART_IN(X, Xs, Ls, Bs)
PART_IN(X, .(Y, Xs), .(Y, Ls), Bs) → U5¹(X, Y, Xs, Ls, Bs, gt_in(X, Y))
U5¹(X, Y, Xs, Ls, Bs, gt_out(X, Y)) → PART_IN(X, Xs, Ls, Bs)

The TRS R consists of the following rules:

qs_in([], []) → qs_out([], [])
qs_in(.(X, Xs), Ys) → U1(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
part_in(X, [], [], []) → part_out(X, [], [], [])
part_in(X, .(Y, Xs), Ls, .(Y, Bs)) → U7(X, Y, Xs, Ls, Bs, le_in(X, Y))
le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U11(X, Y, le_in(X, Y))
U11(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U7(X, Y, Xs, Ls, Bs, le_out(X, Y)) → U8(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
part_in(X, .(Y, Xs), .(Y, Ls), Bs) → U5(X, Y, Xs, Ls, Bs, gt_in(X, Y))
gt_in(s(0), 0) → gt_out(s(0), 0)
gt_in(s(X), s(Y)) → U10(X, Y, gt_in(X, Y))
U10(X, Y, gt_out(X, Y)) → gt_out(s(X), s(Y))
U5(X, Y, Xs, Ls, Bs, gt_out(X, Y)) → U6(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U6(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), .(Y, Ls), Bs)
U8(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), Ls, .(Y, Bs))
U1(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
U2(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U3(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U9(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U9(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Xs, Ys, app_out(Ls, .(X, Bs), Ys)) → qs_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
qs_in(x1, x2) = qs_in(x1)
[] = []
qs_out(x1, x2) = qs_out(x2)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x1, x4)
part_in(x1, x2, x3, x4) = part_in(x1, x2)
part_out(x1, x2, x3, x4) = part_out(x3, x4)
U7(x1, x2, x3, x4, x5, x6) = U7(x1, x2, x3, x6)
le_in(x1, x2) = le_in(x1, x2)
0 = 0
le_out(x1, x2) = le_out
s(x1) = s(x1)
U11(x1, x2, x3) = U11(x3)
U8(x1, x2, x3, x4, x5, x6) = U8(x2, x6)
U5(x1, x2, x3, x4, x5, x6) = U5(x1, x2, x3, x6)
gt_in(x1, x2) = gt_in(x1, x2)
gt_out(x1, x2) = gt_out
U10(x1, x2, x3) = U10(x3)
U6(x1, x2, x3, x4, x5, x6) = U6(x2, x6)
U2(x1, x2, x3, x4, x5) = U2(x1, x4, x5)
U3(x1, x2, x3, x4, x5) = U3(x1, x4, x5)
U4(x1, x2, x3, x4) = U4(x4)
app_in(x1, x2, x3) = app_in(x1, x2)
app_out(x1, x2, x3) = app_out(x3)
U9(x1, x2, x3, x4, x5) = U9(x1, x5)
U5¹(x1, x2, x3, x4, x5, x6) = U5¹(x1, x2, x3, x6)
PART_IN(x1, x2, x3, x4) = PART_IN(x1, x2)
U7¹(x1, x2, x3, x4, x5, x6) = U7¹(x1, x2, x3, x6)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ 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:

PART_IN(X, .(Y, Xs), Ls, .(Y, Bs)) → U7¹(X, Y, Xs, Ls, Bs, le_in(X, Y))
U7¹(X, Y, Xs, Ls, Bs, le_out(X, Y)) → PART_IN(X, Xs, Ls, Bs)
PART_IN(X, .(Y, Xs), .(Y, Ls), Bs) → U5¹(X, Y, Xs, Ls, Bs, gt_in(X, Y))
U5¹(X, Y, Xs, Ls, Bs, gt_out(X, Y)) → PART_IN(X, Xs, Ls, Bs)

The TRS R consists of the following rules:

le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U11(X, Y, le_in(X, Y))
gt_in(s(0), 0) → gt_out(s(0), 0)
gt_in(s(X), s(Y)) → U10(X, Y, gt_in(X, Y))
U11(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U10(X, Y, gt_out(X, Y)) → gt_out(s(X), s(Y))

The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
le_in(x1, x2) = le_in(x1, x2)
0 = 0
le_out(x1, x2) = le_out
s(x1) = s(x1)
U11(x1, x2, x3) = U11(x3)
gt_in(x1, x2) = gt_in(x1, x2)
gt_out(x1, x2) = gt_out
U10(x1, x2, x3) = U10(x3)
U5¹(x1, x2, x3, x4, x5, x6) = U5¹(x1, x2, x3, x6)
PART_IN(x1, x2, x3, x4) = PART_IN(x1, x2)
U7¹(x1, x2, x3, x4, x5, x6) = U7¹(x1, x2, x3, x6)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] 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:

PART_IN(X, .(Y, Xs)) → U5¹(X, Y, Xs, gt_in(X, Y))
U7¹(X, Y, Xs, le_out) → PART_IN(X, Xs)
PART_IN(X, .(Y, Xs)) → U7¹(X, Y, Xs, le_in(X, Y))
U5¹(X, Y, Xs, gt_out) → PART_IN(X, Xs)

The TRS R consists of the following rules:

le_in(0, 0) → le_out
le_in(0, s(X)) → le_out
le_in(s(X), s(Y)) → U11(le_in(X, Y))
gt_in(s(0), 0) → gt_out
gt_in(s(X), s(Y)) → U10(gt_in(X, Y))
U11(le_out) → le_out
U10(gt_out) → gt_out

The set Q consists of the following terms:

le_in(x0, x1)
gt_in(x0, x1)
U11(x0)
U10(x0)

We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:

U7¹(X, Y, Xs, le_out) → PART_IN(X, Xs)
The graph contains the following edges 1 >= 1, 3 >= 2
U5¹(X, Y, Xs, gt_out) → PART_IN(X, Xs)
The graph contains the following edges 1 >= 1, 3 >= 2
PART_IN(X, .(Y, Xs)) → U7¹(X, Y, Xs, le_in(X, Y))
The graph contains the following edges 1 >= 1, 2 > 2, 2 > 3
PART_IN(X, .(Y, Xs)) → U5¹(X, Y, Xs, gt_in(X, Y))
The graph contains the following edges 1 >= 1, 2 > 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:

U1¹(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2¹(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
U1¹(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → QS_IN(Littles, Ls)
QS_IN(.(X, Xs), Ys) → U1¹(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
U2¹(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → QS_IN(Bigs, Bs)

The TRS R consists of the following rules:

qs_in([], []) → qs_out([], [])
qs_in(.(X, Xs), Ys) → U1(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
part_in(X, [], [], []) → part_out(X, [], [], [])
part_in(X, .(Y, Xs), Ls, .(Y, Bs)) → U7(X, Y, Xs, Ls, Bs, le_in(X, Y))
le_in(0, 0) → le_out(0, 0)
le_in(0, s(X)) → le_out(0, s(X))
le_in(s(X), s(Y)) → U11(X, Y, le_in(X, Y))
U11(X, Y, le_out(X, Y)) → le_out(s(X), s(Y))
U7(X, Y, Xs, Ls, Bs, le_out(X, Y)) → U8(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
part_in(X, .(Y, Xs), .(Y, Ls), Bs) → U5(X, Y, Xs, Ls, Bs, gt_in(X, Y))
gt_in(s(0), 0) → gt_out(s(0), 0)
gt_in(s(X), s(Y)) → U10(X, Y, gt_in(X, Y))
U10(X, Y, gt_out(X, Y)) → gt_out(s(X), s(Y))
U5(X, Y, Xs, Ls, Bs, gt_out(X, Y)) → U6(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U6(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), .(Y, Ls), Bs)
U8(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), Ls, .(Y, Bs))
U1(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
U2(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U3(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
app_in([], Ys, Ys) → app_out([], Ys, Ys)
app_in(.(X, Xs), Ys, .(X, Zs)) → U9(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
U9(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Xs, Ys, app_out(Ls, .(X, Bs), Ys)) → qs_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
qs_in(x1, x2) = qs_in(x1)
[] = []
qs_out(x1, x2) = qs_out(x2)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4) = U1(x1, x4)
part_in(x1, x2, x3, x4) = part_in(x1, x2)
part_out(x1, x2, x3, x4) = part_out(x3, x4)
U7(x1, x2, x3, x4, x5, x6) = U7(x1, x2, x3, x6)
le_in(x1, x2) = le_in(x1, x2)
0 = 0
le_out(x1, x2) = le_out
s(x1) = s(x1)
U11(x1, x2, x3) = U11(x3)
U8(x1, x2, x3, x4, x5, x6) = U8(x2, x6)
U5(x1, x2, x3, x4, x5, x6) = U5(x1, x2, x3, x6)
gt_in(x1, x2) = gt_in(x1, x2)
gt_out(x1, x2) = gt_out
U10(x1, x2, x3) = U10(x3)
U6(x1, x2, x3, x4, x5, x6) = U6(x2, x6)
U2(x1, x2, x3, x4, x5) = U2(x1, x4, x5)
U3(x1, x2, x3, x4, x5) = U3(x1, x4, x5)
U4(x1, x2, x3, x4) = U4(x4)
app_in(x1, x2, x3) = app_in(x1, x2)
app_out(x1, x2, x3) = app_out(x3)
U9(x1, x2, x3, x4, x5) = U9(x1, x5)
QS_IN(x1, x2) = QS_IN(x1)
U2¹(x1, x2, x3, x4, x5) = U2¹(x1, x4, x5)
U1¹(x1, x2, x3, x4) = U1¹(x1, x4)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPOrderProof

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

U2¹(X, Bigs, qs_out(Ls)) → QS_IN(Bigs)
QS_IN(.(X, Xs)) → U1¹(X, part_in(X, Xs))
U1¹(X, part_out(Littles, Bigs)) → QS_IN(Littles)
U1¹(X, part_out(Littles, Bigs)) → U2¹(X, Bigs, qs_in(Littles))

The TRS R consists of the following rules:

qs_in([]) → qs_out([])
qs_in(.(X, Xs)) → U1(X, part_in(X, Xs))
part_in(X, []) → part_out([], [])
part_in(X, .(Y, Xs)) → U7(X, Y, Xs, le_in(X, Y))
le_in(0, 0) → le_out
le_in(0, s(X)) → le_out
le_in(s(X), s(Y)) → U11(le_in(X, Y))
U11(le_out) → le_out
U7(X, Y, Xs, le_out) → U8(Y, part_in(X, Xs))
part_in(X, .(Y, Xs)) → U5(X, Y, Xs, gt_in(X, Y))
gt_in(s(0), 0) → gt_out
gt_in(s(X), s(Y)) → U10(gt_in(X, Y))
U10(gt_out) → gt_out
U5(X, Y, Xs, gt_out) → U6(Y, part_in(X, Xs))
U6(Y, part_out(Ls, Bs)) → part_out(.(Y, Ls), Bs)
U8(Y, part_out(Ls, Bs)) → part_out(Ls, .(Y, Bs))
U1(X, part_out(Littles, Bigs)) → U2(X, Bigs, qs_in(Littles))
U2(X, Bigs, qs_out(Ls)) → U3(X, Ls, qs_in(Bigs))
U3(X, Ls, qs_out(Bs)) → U4(app_in(Ls, .(X, Bs)))
app_in([], Ys) → app_out(Ys)
app_in(.(X, Xs), Ys) → U9(X, app_in(Xs, Ys))
U9(X, app_out(Zs)) → app_out(.(X, Zs))
U4(app_out(Ys)) → qs_out(Ys)

The set Q consists of the following terms:

qs_in(x0)
part_in(x0, x1)
le_in(x0, x1)
U11(x0)
U7(x0, x1, x2, x3)
gt_in(x0, x1)
U10(x0)
U5(x0, x1, x2, x3)
U6(x0, x1)
U8(x0, x1)
U1(x0, x1)
U2(x0, x1, x2)
U3(x0, x1, x2)
app_in(x0, x1)
U9(x0, x1)
U4(x0)

We have to consider all (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

U1¹(X, part_out(Littles, Bigs)) → QS_IN(Littles)
U1¹(X, part_out(Littles, Bigs)) → U2¹(X, Bigs, qs_in(Littles))

The remaining pairs can at least be oriented weakly.

U2¹(X, Bigs, qs_out(Ls)) → QS_IN(Bigs)
QS_IN(.(X, Xs)) → U1¹(X, part_in(X, Xs))

Used ordering: Polynomial interpretation [25]:

POL(.(x₁, x₂)) = 1 + x₂
POL(0) = 0
POL(QS_IN(x₁)) = x₁
POL(U1(x₁, x₂)) = 0
POL(U10(x₁)) = 0
POL(U11(x₁)) = 0
POL(U1¹(x₁, x₂)) = 1 + x₂
POL(U2(x₁, x₂, x₃)) = 0
POL(U2¹(x₁, x₂, x₃)) = x₂
POL(U3(x₁, x₂, x₃)) = 0
POL(U4(x₁)) = 0
POL(U5(x₁, x₂, x₃, x₄)) = 1 + x₃
POL(U6(x₁, x₂)) = 1 + x₂
POL(U7(x₁, x₂, x₃, x₄)) = 1 + x₃
POL(U8(x₁, x₂)) = 1 + x₂
POL(U9(x₁, x₂)) = x₂
POL([]) = 0
POL(app_in(x₁, x₂)) = 1 + x₁
POL(app_out(x₁)) = 0
POL(gt_in(x₁, x₂)) = 0
POL(gt_out) = 0
POL(le_in(x₁, x₂)) = 0
POL(le_out) = 0
POL(part_in(x₁, x₂)) = x₂
POL(part_out(x₁, x₂)) = x₁ + x₂
POL(qs_in(x₁)) = 0
POL(qs_out(x₁)) = 0
POL(s(x₁)) = 0

The following usable rules [17] were oriented:

U7(X, Y, Xs, le_out) → U8(Y, part_in(X, Xs))
U6(Y, part_out(Ls, Bs)) → part_out(.(Y, Ls), Bs)
part_in(X, .(Y, Xs)) → U7(X, Y, Xs, le_in(X, Y))
U5(X, Y, Xs, gt_out) → U6(Y, part_in(X, Xs))
part_in(X, .(Y, Xs)) → U5(X, Y, Xs, gt_in(X, Y))
part_in(X, []) → part_out([], [])
U8(Y, part_out(Ls, Bs)) → part_out(Ls, .(Y, Bs))

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

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

U2¹(X, Bigs, qs_out(Ls)) → QS_IN(Bigs)
QS_IN(.(X, Xs)) → U1¹(X, part_in(X, Xs))

The TRS R consists of the following rules:

qs_in([]) → qs_out([])
qs_in(.(X, Xs)) → U1(X, part_in(X, Xs))
part_in(X, []) → part_out([], [])
part_in(X, .(Y, Xs)) → U7(X, Y, Xs, le_in(X, Y))
le_in(0, 0) → le_out
le_in(0, s(X)) → le_out
le_in(s(X), s(Y)) → U11(le_in(X, Y))
U11(le_out) → le_out
U7(X, Y, Xs, le_out) → U8(Y, part_in(X, Xs))
part_in(X, .(Y, Xs)) → U5(X, Y, Xs, gt_in(X, Y))
gt_in(s(0), 0) → gt_out
gt_in(s(X), s(Y)) → U10(gt_in(X, Y))
U10(gt_out) → gt_out
U5(X, Y, Xs, gt_out) → U6(Y, part_in(X, Xs))
U6(Y, part_out(Ls, Bs)) → part_out(.(Y, Ls), Bs)
U8(Y, part_out(Ls, Bs)) → part_out(Ls, .(Y, Bs))
U1(X, part_out(Littles, Bigs)) → U2(X, Bigs, qs_in(Littles))
U2(X, Bigs, qs_out(Ls)) → U3(X, Ls, qs_in(Bigs))
U3(X, Ls, qs_out(Bs)) → U4(app_in(Ls, .(X, Bs)))
app_in([], Ys) → app_out(Ys)
app_in(.(X, Xs), Ys) → U9(X, app_in(Xs, Ys))
U9(X, app_out(Zs)) → app_out(.(X, Zs))
U4(app_out(Ys)) → qs_out(Ys)

The set Q consists of the following terms:

qs_in(x0)
part_in(x0, x1)
le_in(x0, x1)
U11(x0)
U7(x0, x1, x2, x3)
gt_in(x0, x1)
U10(x0)
U5(x0, x1, x2, x3)
U6(x0, x1)
U8(x0, x1)
U1(x0, x1)
U2(x0, x1, x2)
U3(x0, x1, x2)
app_in(x0, x1)
U9(x0, x1)
U4(x0)

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.