Left Termination proof of /tmp/tmpz8KEDM/pl4.4.3.pl

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

↳ PROLOG

  ↳ PrologToPiTRSProof

merge₃(X, {}₀, X).
merge₃({}₀, X, X).
merge₃(.₂(A, X), .₂(B, Y), .₂(A, Z)) :- le₂(A, B), merge₃(X, .₂(B, Y), Z).
merge₃(.₂(A, X), .₂(B, Y), .₂(B, Z)) :- gt₂(A, B), merge₃(.₂(A, X), Y, Z).
gt₂(s₁(X), s₁(Y)) :- gt₂(X, Y).
gt₂(s₁(X), zero₀).
le₂(s₁(X), s₁(Y)) :- le₂(X, Y).
le₂(zero₀, s₁(Y)).
le₂(zero₀, zero₀).

With regard to the inferred argument filtering the predicates were used in the following modes:
merge₃: (b,b,f)
le₂: (b,b)
gt₂: (b,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

merge_3_in_gga₃(X, []_0, X) -> merge_3_out_gga₃(X, []_0, X)
merge_3_in_gga₃([]_0, X, X) -> merge_3_out_gga₃([]_0, X, X)
merge_3_in_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(A, Z)) -> if_merge_3_in_1_gga₆(A, X, B, Y, Z, le_2_in_gg₂(A, B))
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(zero_0, s_1₁(Y)) -> le_2_out_gg₂(zero_0, s_1₁(Y))
le_2_in_gg₂(zero_0, zero_0) -> le_2_out_gg₂(zero_0, zero_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_merge_3_in_1_gga₆(A, X, B, Y, Z, le_2_out_gg₂(A, B)) -> if_merge_3_in_2_gga₆(A, X, B, Y, Z, merge_3_in_gga₃(X, ._2₂(B, Y), Z))
merge_3_in_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(B, Z)) -> if_merge_3_in_3_gga₆(A, X, B, Y, Z, gt_2_in_gg₂(A, B))
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), zero_0) -> gt_2_out_gg₂(s_1₁(X), zero_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_merge_3_in_3_gga₆(A, X, B, Y, Z, gt_2_out_gg₂(A, B)) -> if_merge_3_in_4_gga₆(A, X, B, Y, Z, merge_3_in_gga₃(._2₂(A, X), Y, Z))
if_merge_3_in_4_gga₆(A, X, B, Y, Z, merge_3_out_gga₃(._2₂(A, X), Y, Z)) -> merge_3_out_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(B, Z))
if_merge_3_in_2_gga₆(A, X, B, Y, Z, merge_3_out_gga₃(X, ._2₂(B, Y), Z)) -> merge_3_out_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(A, Z))

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:

merge_3_in_gga₃(X, []_0, X) -> merge_3_out_gga₃(X, []_0, X)
merge_3_in_gga₃([]_0, X, X) -> merge_3_out_gga₃([]_0, X, X)
merge_3_in_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(A, Z)) -> if_merge_3_in_1_gga₆(A, X, B, Y, Z, le_2_in_gg₂(A, B))
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(zero_0, s_1₁(Y)) -> le_2_out_gg₂(zero_0, s_1₁(Y))
le_2_in_gg₂(zero_0, zero_0) -> le_2_out_gg₂(zero_0, zero_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_merge_3_in_1_gga₆(A, X, B, Y, Z, le_2_out_gg₂(A, B)) -> if_merge_3_in_2_gga₆(A, X, B, Y, Z, merge_3_in_gga₃(X, ._2₂(B, Y), Z))
merge_3_in_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(B, Z)) -> if_merge_3_in_3_gga₆(A, X, B, Y, Z, gt_2_in_gg₂(A, B))
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), zero_0) -> gt_2_out_gg₂(s_1₁(X), zero_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_merge_3_in_3_gga₆(A, X, B, Y, Z, gt_2_out_gg₂(A, B)) -> if_merge_3_in_4_gga₆(A, X, B, Y, Z, merge_3_in_gga₃(._2₂(A, X), Y, Z))
if_merge_3_in_4_gga₆(A, X, B, Y, Z, merge_3_out_gga₃(._2₂(A, X), Y, Z)) -> merge_3_out_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(B, Z))
if_merge_3_in_2_gga₆(A, X, B, Y, Z, merge_3_out_gga₃(X, ._2₂(B, Y), Z)) -> merge_3_out_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(A, Z))

MERGE_3_IN_GGA₃(._2₂(A, X), ._2₂(B, Y), ._2₂(A, Z)) -> IF_MERGE_3_IN_1_GGA₆(A, X, B, Y, Z, le_2_in_gg₂(A, B))
MERGE_3_IN_GGA₃(._2₂(A, X), ._2₂(B, Y), ._2₂(A, Z)) -> LE_2_IN_GG₂(A, B)
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_MERGE_3_IN_1_GGA₆(A, X, B, Y, Z, le_2_out_gg₂(A, B)) -> IF_MERGE_3_IN_2_GGA₆(A, X, B, Y, Z, merge_3_in_gga₃(X, ._2₂(B, Y), Z))
IF_MERGE_3_IN_1_GGA₆(A, X, B, Y, Z, le_2_out_gg₂(A, B)) -> MERGE_3_IN_GGA₃(X, ._2₂(B, Y), Z)
MERGE_3_IN_GGA₃(._2₂(A, X), ._2₂(B, Y), ._2₂(B, Z)) -> IF_MERGE_3_IN_3_GGA₆(A, X, B, Y, Z, gt_2_in_gg₂(A, B))
MERGE_3_IN_GGA₃(._2₂(A, X), ._2₂(B, Y), ._2₂(B, Z)) -> GT_2_IN_GG₂(A, B)
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_MERGE_3_IN_3_GGA₆(A, X, B, Y, Z, gt_2_out_gg₂(A, B)) -> IF_MERGE_3_IN_4_GGA₆(A, X, B, Y, Z, merge_3_in_gga₃(._2₂(A, X), Y, Z))
IF_MERGE_3_IN_3_GGA₆(A, X, B, Y, Z, gt_2_out_gg₂(A, B)) -> MERGE_3_IN_GGA₃(._2₂(A, X), Y, Z)

The TRS R consists of the following rules:

merge_3_in_gga₃(X, []_0, X) -> merge_3_out_gga₃(X, []_0, X)
merge_3_in_gga₃([]_0, X, X) -> merge_3_out_gga₃([]_0, X, X)
merge_3_in_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(A, Z)) -> if_merge_3_in_1_gga₆(A, X, B, Y, Z, le_2_in_gg₂(A, B))
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(zero_0, s_1₁(Y)) -> le_2_out_gg₂(zero_0, s_1₁(Y))
le_2_in_gg₂(zero_0, zero_0) -> le_2_out_gg₂(zero_0, zero_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_merge_3_in_1_gga₆(A, X, B, Y, Z, le_2_out_gg₂(A, B)) -> if_merge_3_in_2_gga₆(A, X, B, Y, Z, merge_3_in_gga₃(X, ._2₂(B, Y), Z))
merge_3_in_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(B, Z)) -> if_merge_3_in_3_gga₆(A, X, B, Y, Z, gt_2_in_gg₂(A, B))
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), zero_0) -> gt_2_out_gg₂(s_1₁(X), zero_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_merge_3_in_3_gga₆(A, X, B, Y, Z, gt_2_out_gg₂(A, B)) -> if_merge_3_in_4_gga₆(A, X, B, Y, Z, merge_3_in_gga₃(._2₂(A, X), Y, Z))
if_merge_3_in_4_gga₆(A, X, B, Y, Z, merge_3_out_gga₃(._2₂(A, X), Y, Z)) -> merge_3_out_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(B, Z))
if_merge_3_in_2_gga₆(A, X, B, Y, Z, merge_3_out_gga₃(X, ._2₂(B, Y), Z)) -> merge_3_out_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(A, Z))

The argument filtering Pi contains the following mapping:
merge_3_in_gga₃(x1, x2, x3) = merge_3_in_gga₂(x1, x2)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
zero_0 = zero_0
merge_3_out_gga₃(x1, x2, x3) = merge_3_out_gga₁(x3)
if_merge_3_in_1_gga₆(x1, x2, x3, x4, x5, x6) = if_merge_3_in_1_gga₅(x1, x2, x3, x4, 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_merge_3_in_2_gga₆(x1, x2, x3, x4, x5, x6) = if_merge_3_in_2_gga₂(x1, x6)
if_merge_3_in_3_gga₆(x1, x2, x3, x4, x5, x6) = if_merge_3_in_3_gga₅(x1, x2, x3, x4, 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_merge_3_in_4_gga₆(x1, x2, x3, x4, x5, x6) = if_merge_3_in_4_gga₂(x3, x6)
GT_2_IN_GG₂(x1, x2) = GT_2_IN_GG₂(x1, x2)
IF_MERGE_3_IN_2_GGA₆(x1, x2, x3, x4, x5, x6) = IF_MERGE_3_IN_2_GGA₂(x1, x6)
MERGE_3_IN_GGA₃(x1, x2, x3) = MERGE_3_IN_GGA₂(x1, x2)
IF_MERGE_3_IN_1_GGA₆(x1, x2, x3, x4, x5, x6) = IF_MERGE_3_IN_1_GGA₅(x1, x2, x3, x4, x6)
IF_GT_2_IN_1_GG₃(x1, x2, x3) = IF_GT_2_IN_1_GG₁(x3)
IF_MERGE_3_IN_3_GGA₆(x1, x2, x3, x4, x5, x6) = IF_MERGE_3_IN_3_GGA₅(x1, x2, x3, x4, x6)
IF_MERGE_3_IN_4_GGA₆(x1, x2, x3, x4, x5, x6) = IF_MERGE_3_IN_4_GGA₂(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)

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:

MERGE_3_IN_GGA₃(._2₂(A, X), ._2₂(B, Y), ._2₂(A, Z)) -> IF_MERGE_3_IN_1_GGA₆(A, X, B, Y, Z, le_2_in_gg₂(A, B))
MERGE_3_IN_GGA₃(._2₂(A, X), ._2₂(B, Y), ._2₂(A, Z)) -> LE_2_IN_GG₂(A, B)
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_MERGE_3_IN_1_GGA₆(A, X, B, Y, Z, le_2_out_gg₂(A, B)) -> IF_MERGE_3_IN_2_GGA₆(A, X, B, Y, Z, merge_3_in_gga₃(X, ._2₂(B, Y), Z))
IF_MERGE_3_IN_1_GGA₆(A, X, B, Y, Z, le_2_out_gg₂(A, B)) -> MERGE_3_IN_GGA₃(X, ._2₂(B, Y), Z)
MERGE_3_IN_GGA₃(._2₂(A, X), ._2₂(B, Y), ._2₂(B, Z)) -> IF_MERGE_3_IN_3_GGA₆(A, X, B, Y, Z, gt_2_in_gg₂(A, B))
MERGE_3_IN_GGA₃(._2₂(A, X), ._2₂(B, Y), ._2₂(B, Z)) -> GT_2_IN_GG₂(A, B)
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_MERGE_3_IN_3_GGA₆(A, X, B, Y, Z, gt_2_out_gg₂(A, B)) -> IF_MERGE_3_IN_4_GGA₆(A, X, B, Y, Z, merge_3_in_gga₃(._2₂(A, X), Y, Z))
IF_MERGE_3_IN_3_GGA₆(A, X, B, Y, Z, gt_2_out_gg₂(A, B)) -> MERGE_3_IN_GGA₃(._2₂(A, X), Y, Z)

The TRS R consists of the following rules:

merge_3_in_gga₃(X, []_0, X) -> merge_3_out_gga₃(X, []_0, X)
merge_3_in_gga₃([]_0, X, X) -> merge_3_out_gga₃([]_0, X, X)
merge_3_in_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(A, Z)) -> if_merge_3_in_1_gga₆(A, X, B, Y, Z, le_2_in_gg₂(A, B))
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(zero_0, s_1₁(Y)) -> le_2_out_gg₂(zero_0, s_1₁(Y))
le_2_in_gg₂(zero_0, zero_0) -> le_2_out_gg₂(zero_0, zero_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_merge_3_in_1_gga₆(A, X, B, Y, Z, le_2_out_gg₂(A, B)) -> if_merge_3_in_2_gga₆(A, X, B, Y, Z, merge_3_in_gga₃(X, ._2₂(B, Y), Z))
merge_3_in_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(B, Z)) -> if_merge_3_in_3_gga₆(A, X, B, Y, Z, gt_2_in_gg₂(A, B))
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), zero_0) -> gt_2_out_gg₂(s_1₁(X), zero_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_merge_3_in_3_gga₆(A, X, B, Y, Z, gt_2_out_gg₂(A, B)) -> if_merge_3_in_4_gga₆(A, X, B, Y, Z, merge_3_in_gga₃(._2₂(A, X), Y, Z))
if_merge_3_in_4_gga₆(A, X, B, Y, Z, merge_3_out_gga₃(._2₂(A, X), Y, Z)) -> merge_3_out_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(B, Z))
if_merge_3_in_2_gga₆(A, X, B, Y, Z, merge_3_out_gga₃(X, ._2₂(B, Y), Z)) -> merge_3_out_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(A, Z))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ 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:

merge_3_in_gga₃(X, []_0, X) -> merge_3_out_gga₃(X, []_0, X)
merge_3_in_gga₃([]_0, X, X) -> merge_3_out_gga₃([]_0, X, X)
merge_3_in_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(A, Z)) -> if_merge_3_in_1_gga₆(A, X, B, Y, Z, le_2_in_gg₂(A, B))
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(zero_0, s_1₁(Y)) -> le_2_out_gg₂(zero_0, s_1₁(Y))
le_2_in_gg₂(zero_0, zero_0) -> le_2_out_gg₂(zero_0, zero_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_merge_3_in_1_gga₆(A, X, B, Y, Z, le_2_out_gg₂(A, B)) -> if_merge_3_in_2_gga₆(A, X, B, Y, Z, merge_3_in_gga₃(X, ._2₂(B, Y), Z))
merge_3_in_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(B, Z)) -> if_merge_3_in_3_gga₆(A, X, B, Y, Z, gt_2_in_gg₂(A, B))
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), zero_0) -> gt_2_out_gg₂(s_1₁(X), zero_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_merge_3_in_3_gga₆(A, X, B, Y, Z, gt_2_out_gg₂(A, B)) -> if_merge_3_in_4_gga₆(A, X, B, Y, Z, merge_3_in_gga₃(._2₂(A, X), Y, Z))
if_merge_3_in_4_gga₆(A, X, B, Y, Z, merge_3_out_gga₃(._2₂(A, X), Y, Z)) -> merge_3_out_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(B, Z))
if_merge_3_in_2_gga₆(A, X, B, Y, Z, merge_3_out_gga₃(X, ._2₂(B, Y), Z)) -> merge_3_out_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(A, Z))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ 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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ 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

                ↳ UsableRulesProof

              ↳ 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:

merge_3_in_gga₃(X, []_0, X) -> merge_3_out_gga₃(X, []_0, X)
merge_3_in_gga₃([]_0, X, X) -> merge_3_out_gga₃([]_0, X, X)
merge_3_in_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(A, Z)) -> if_merge_3_in_1_gga₆(A, X, B, Y, Z, le_2_in_gg₂(A, B))
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(zero_0, s_1₁(Y)) -> le_2_out_gg₂(zero_0, s_1₁(Y))
le_2_in_gg₂(zero_0, zero_0) -> le_2_out_gg₂(zero_0, zero_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_merge_3_in_1_gga₆(A, X, B, Y, Z, le_2_out_gg₂(A, B)) -> if_merge_3_in_2_gga₆(A, X, B, Y, Z, merge_3_in_gga₃(X, ._2₂(B, Y), Z))
merge_3_in_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(B, Z)) -> if_merge_3_in_3_gga₆(A, X, B, Y, Z, gt_2_in_gg₂(A, B))
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), zero_0) -> gt_2_out_gg₂(s_1₁(X), zero_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_merge_3_in_3_gga₆(A, X, B, Y, Z, gt_2_out_gg₂(A, B)) -> if_merge_3_in_4_gga₆(A, X, B, Y, Z, merge_3_in_gga₃(._2₂(A, X), Y, Z))
if_merge_3_in_4_gga₆(A, X, B, Y, Z, merge_3_out_gga₃(._2₂(A, X), Y, Z)) -> merge_3_out_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(B, Z))
if_merge_3_in_2_gga₆(A, X, B, Y, Z, merge_3_out_gga₃(X, ._2₂(B, Y), Z)) -> merge_3_out_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(A, Z))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ 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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ 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

                ↳ UsableRulesProof

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

MERGE_3_IN_GGA₃(._2₂(A, X), ._2₂(B, Y), ._2₂(B, Z)) -> IF_MERGE_3_IN_3_GGA₆(A, X, B, Y, Z, gt_2_in_gg₂(A, B))
MERGE_3_IN_GGA₃(._2₂(A, X), ._2₂(B, Y), ._2₂(A, Z)) -> IF_MERGE_3_IN_1_GGA₆(A, X, B, Y, Z, le_2_in_gg₂(A, B))
IF_MERGE_3_IN_1_GGA₆(A, X, B, Y, Z, le_2_out_gg₂(A, B)) -> MERGE_3_IN_GGA₃(X, ._2₂(B, Y), Z)
IF_MERGE_3_IN_3_GGA₆(A, X, B, Y, Z, gt_2_out_gg₂(A, B)) -> MERGE_3_IN_GGA₃(._2₂(A, X), Y, Z)

The TRS R consists of the following rules:

merge_3_in_gga₃(X, []_0, X) -> merge_3_out_gga₃(X, []_0, X)
merge_3_in_gga₃([]_0, X, X) -> merge_3_out_gga₃([]_0, X, X)
merge_3_in_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(A, Z)) -> if_merge_3_in_1_gga₆(A, X, B, Y, Z, le_2_in_gg₂(A, B))
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(zero_0, s_1₁(Y)) -> le_2_out_gg₂(zero_0, s_1₁(Y))
le_2_in_gg₂(zero_0, zero_0) -> le_2_out_gg₂(zero_0, zero_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_merge_3_in_1_gga₆(A, X, B, Y, Z, le_2_out_gg₂(A, B)) -> if_merge_3_in_2_gga₆(A, X, B, Y, Z, merge_3_in_gga₃(X, ._2₂(B, Y), Z))
merge_3_in_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(B, Z)) -> if_merge_3_in_3_gga₆(A, X, B, Y, Z, gt_2_in_gg₂(A, B))
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), zero_0) -> gt_2_out_gg₂(s_1₁(X), zero_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_merge_3_in_3_gga₆(A, X, B, Y, Z, gt_2_out_gg₂(A, B)) -> if_merge_3_in_4_gga₆(A, X, B, Y, Z, merge_3_in_gga₃(._2₂(A, X), Y, Z))
if_merge_3_in_4_gga₆(A, X, B, Y, Z, merge_3_out_gga₃(._2₂(A, X), Y, Z)) -> merge_3_out_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(B, Z))
if_merge_3_in_2_gga₆(A, X, B, Y, Z, merge_3_out_gga₃(X, ._2₂(B, Y), Z)) -> merge_3_out_gga₃(._2₂(A, X), ._2₂(B, Y), ._2₂(A, Z))

The argument filtering Pi contains the following mapping:
merge_3_in_gga₃(x1, x2, x3) = merge_3_in_gga₂(x1, x2)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
zero_0 = zero_0
merge_3_out_gga₃(x1, x2, x3) = merge_3_out_gga₁(x3)
if_merge_3_in_1_gga₆(x1, x2, x3, x4, x5, x6) = if_merge_3_in_1_gga₅(x1, x2, x3, x4, 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_merge_3_in_2_gga₆(x1, x2, x3, x4, x5, x6) = if_merge_3_in_2_gga₂(x1, x6)
if_merge_3_in_3_gga₆(x1, x2, x3, x4, x5, x6) = if_merge_3_in_3_gga₅(x1, x2, x3, x4, 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_merge_3_in_4_gga₆(x1, x2, x3, x4, x5, x6) = if_merge_3_in_4_gga₂(x3, x6)
MERGE_3_IN_GGA₃(x1, x2, x3) = MERGE_3_IN_GGA₂(x1, x2)
IF_MERGE_3_IN_1_GGA₆(x1, x2, x3, x4, x5, x6) = IF_MERGE_3_IN_1_GGA₅(x1, x2, x3, x4, x6)
IF_MERGE_3_IN_3_GGA₆(x1, x2, x3, x4, x5, x6) = IF_MERGE_3_IN_3_GGA₅(x1, x2, x3, x4, x6)

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

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

MERGE_3_IN_GGA₃(._2₂(A, X), ._2₂(B, Y), ._2₂(B, Z)) -> IF_MERGE_3_IN_3_GGA₆(A, X, B, Y, Z, gt_2_in_gg₂(A, B))
MERGE_3_IN_GGA₃(._2₂(A, X), ._2₂(B, Y), ._2₂(A, Z)) -> IF_MERGE_3_IN_1_GGA₆(A, X, B, Y, Z, le_2_in_gg₂(A, B))
IF_MERGE_3_IN_1_GGA₆(A, X, B, Y, Z, le_2_out_gg₂(A, B)) -> MERGE_3_IN_GGA₃(X, ._2₂(B, Y), Z)
IF_MERGE_3_IN_3_GGA₆(A, X, B, Y, Z, gt_2_out_gg₂(A, B)) -> MERGE_3_IN_GGA₃(._2₂(A, X), Y, Z)

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), zero_0) -> gt_2_out_gg₂(s_1₁(X), zero_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(zero_0, s_1₁(Y)) -> le_2_out_gg₂(zero_0, s_1₁(Y))
le_2_in_gg₂(zero_0, zero_0) -> le_2_out_gg₂(zero_0, zero_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)
zero_0 = zero_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
MERGE_3_IN_GGA₃(x1, x2, x3) = MERGE_3_IN_GGA₂(x1, x2)
IF_MERGE_3_IN_1_GGA₆(x1, x2, x3, x4, x5, x6) = IF_MERGE_3_IN_1_GGA₅(x1, x2, x3, x4, x6)
IF_MERGE_3_IN_3_GGA₆(x1, x2, x3, x4, x5, x6) = IF_MERGE_3_IN_3_GGA₅(x1, x2, x3, x4, 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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPPoloProof

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

MERGE_3_IN_GGA₂(._2₂(A, X), ._2₂(B, Y)) -> IF_MERGE_3_IN_3_GGA₅(A, X, B, Y, gt_2_in_gg₂(A, B))
MERGE_3_IN_GGA₂(._2₂(A, X), ._2₂(B, Y)) -> IF_MERGE_3_IN_1_GGA₅(A, X, B, Y, le_2_in_gg₂(A, B))
IF_MERGE_3_IN_1_GGA₅(A, X, B, Y, le_2_out_gg) -> MERGE_3_IN_GGA₂(X, ._2₂(B, Y))
IF_MERGE_3_IN_3_GGA₅(A, X, B, Y, gt_2_out_gg) -> MERGE_3_IN_GGA₂(._2₂(A, 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), zero_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(zero_0, s_1₁(Y)) -> le_2_out_gg
le_2_in_gg₂(zero_0, zero_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_MERGE_3_IN_3_GGA₅, MERGE_3_IN_GGA₂, IF_MERGE_3_IN_1_GGA₅}.
By using a polynomial ordering, the following set of Dependency Pairs of this DP problem can be strictly oriented.

MERGE_3_IN_GGA₂(._2₂(A, X), ._2₂(B, Y)) -> IF_MERGE_3_IN_1_GGA₅(A, X, B, Y, le_2_in_gg₂(A, B))

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

MERGE_3_IN_GGA₂(._2₂(A, X), ._2₂(B, Y)) -> IF_MERGE_3_IN_3_GGA₅(A, X, B, Y, gt_2_in_gg₂(A, B))
IF_MERGE_3_IN_1_GGA₅(A, X, B, Y, le_2_out_gg) -> MERGE_3_IN_GGA₂(X, ._2₂(B, Y))
IF_MERGE_3_IN_3_GGA₅(A, X, B, Y, gt_2_out_gg) -> MERGE_3_IN_GGA₂(._2₂(A, X), Y)

With the implicit AFS there is no usable rule.

Used ordering: POLO with Polynomial interpretation:

POL(gt_2_in_gg₂(x₁, x₂)) = 0
POL(if_gt_2_in_1_gg₁(x₁)) = 0
POL(MERGE_3_IN_GGA₂(x₁, x₂)) = x₁
POL(._2₂(x₁, x₂)) = 1 + x₂
POL(zero_0) = 0
POL(IF_MERGE_3_IN_1_GGA₅(x₁, x₂, x₃, x₄, x₅)) = x₂
POL(le_2_out_gg) = 0
POL(le_2_in_gg₂(x₁, x₂)) = 0
POL(s_1₁(x₁)) = 0
POL(gt_2_out_gg) = 0
POL(IF_MERGE_3_IN_3_GGA₅(x₁, x₂, x₃, x₄, x₅)) = 1 + x₂
POL(if_le_2_in_1_gg₁(x₁)) = 0

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPPoloProof

                          ↳ QDP

                            ↳ DependencyGraphProof

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

MERGE_3_IN_GGA₂(._2₂(A, X), ._2₂(B, Y)) -> IF_MERGE_3_IN_3_GGA₅(A, X, B, Y, gt_2_in_gg₂(A, B))
IF_MERGE_3_IN_1_GGA₅(A, X, B, Y, le_2_out_gg) -> MERGE_3_IN_GGA₂(X, ._2₂(B, Y))
IF_MERGE_3_IN_3_GGA₅(A, X, B, Y, gt_2_out_gg) -> MERGE_3_IN_GGA₂(._2₂(A, 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), zero_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(zero_0, s_1₁(Y)) -> le_2_out_gg
le_2_in_gg₂(zero_0, zero_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_MERGE_3_IN_3_GGA₅, MERGE_3_IN_GGA₂, IF_MERGE_3_IN_1_GGA₅}.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPPoloProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

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

MERGE_3_IN_GGA₂(._2₂(A, X), ._2₂(B, Y)) -> IF_MERGE_3_IN_3_GGA₅(A, X, B, Y, gt_2_in_gg₂(A, B))
IF_MERGE_3_IN_3_GGA₅(A, X, B, Y, gt_2_out_gg) -> MERGE_3_IN_GGA₂(._2₂(A, 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), zero_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(zero_0, s_1₁(Y)) -> le_2_out_gg
le_2_in_gg₂(zero_0, zero_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_MERGE_3_IN_3_GGA₅, MERGE_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:

MERGE_3_IN_GGA₂(._2₂(A, X), ._2₂(B, Y)) -> IF_MERGE_3_IN_3_GGA₅(A, X, B, Y, gt_2_in_gg₂(A, B))
The graph contains the following edges 1 > 1, 1 > 2, 2 > 3, 2 > 4
IF_MERGE_3_IN_3_GGA₅(A, X, B, Y, gt_2_out_gg) -> MERGE_3_IN_GGA₂(._2₂(A, X), Y)
The graph contains the following edges 4 >= 2