Left Termination proof of /tmp/tmpArR6Wv/bappend.pl

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

↳ PROLOG

  ↳ PrologToPiTRSProof

append₃({}₀, XS, XS).
append₃(.₂(X, XS), YS, .₂(X, ZS)) :- append₃(XS, YS, ZS).
s2l₂(s₁(X), .₂(Y, Xs)) :- s2l₂(X, Xs).
s2l₂(0₀, {}₀).
goal₁(X) :- s2l₂(X, XS), append₃(XS, YS, ZS).

With regard to the inferred argument filtering the predicates were used in the following modes:
goal₁: (b)
s2l₂: (b,f)
append₃: (b,f,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

goal_1_in_g₁(X) -> if_goal_1_in_1_g₂(X, s2l_2_in_ga₂(X, XS))
s2l_2_in_ga₂(s_1₁(X), ._2₂(Y, Xs)) -> if_s2l_2_in_1_ga₄(X, Y, Xs, s2l_2_in_ga₂(X, Xs))
s2l_2_in_ga₂(0_0, []_0) -> s2l_2_out_ga₂(0_0, []_0)
if_s2l_2_in_1_ga₄(X, Y, Xs, s2l_2_out_ga₂(X, Xs)) -> s2l_2_out_ga₂(s_1₁(X), ._2₂(Y, Xs))
if_goal_1_in_1_g₂(X, s2l_2_out_ga₂(X, XS)) -> if_goal_1_in_2_g₃(X, XS, append_3_in_gaa₃(XS, YS, ZS))
append_3_in_gaa₃([]_0, XS, XS) -> append_3_out_gaa₃([]_0, XS, XS)
append_3_in_gaa₃(._2₂(X, XS), YS, ._2₂(X, ZS)) -> if_append_3_in_1_gaa₅(X, XS, YS, ZS, append_3_in_gaa₃(XS, YS, ZS))
if_append_3_in_1_gaa₅(X, XS, YS, ZS, append_3_out_gaa₃(XS, YS, ZS)) -> append_3_out_gaa₃(._2₂(X, XS), YS, ._2₂(X, ZS))
if_goal_1_in_2_g₃(X, XS, append_3_out_gaa₃(XS, YS, ZS)) -> goal_1_out_g₁(X)

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:

goal_1_in_g₁(X) -> if_goal_1_in_1_g₂(X, s2l_2_in_ga₂(X, XS))
s2l_2_in_ga₂(s_1₁(X), ._2₂(Y, Xs)) -> if_s2l_2_in_1_ga₄(X, Y, Xs, s2l_2_in_ga₂(X, Xs))
s2l_2_in_ga₂(0_0, []_0) -> s2l_2_out_ga₂(0_0, []_0)
if_s2l_2_in_1_ga₄(X, Y, Xs, s2l_2_out_ga₂(X, Xs)) -> s2l_2_out_ga₂(s_1₁(X), ._2₂(Y, Xs))
if_goal_1_in_1_g₂(X, s2l_2_out_ga₂(X, XS)) -> if_goal_1_in_2_g₃(X, XS, append_3_in_gaa₃(XS, YS, ZS))
append_3_in_gaa₃([]_0, XS, XS) -> append_3_out_gaa₃([]_0, XS, XS)
append_3_in_gaa₃(._2₂(X, XS), YS, ._2₂(X, ZS)) -> if_append_3_in_1_gaa₅(X, XS, YS, ZS, append_3_in_gaa₃(XS, YS, ZS))
if_append_3_in_1_gaa₅(X, XS, YS, ZS, append_3_out_gaa₃(XS, YS, ZS)) -> append_3_out_gaa₃(._2₂(X, XS), YS, ._2₂(X, ZS))
if_goal_1_in_2_g₃(X, XS, append_3_out_gaa₃(XS, YS, ZS)) -> goal_1_out_g₁(X)

GOAL_1_IN_G₁(X) -> IF_GOAL_1_IN_1_G₂(X, s2l_2_in_ga₂(X, XS))
GOAL_1_IN_G₁(X) -> S2L_2_IN_GA₂(X, XS)
S2L_2_IN_GA₂(s_1₁(X), ._2₂(Y, Xs)) -> IF_S2L_2_IN_1_GA₄(X, Y, Xs, s2l_2_in_ga₂(X, Xs))
S2L_2_IN_GA₂(s_1₁(X), ._2₂(Y, Xs)) -> S2L_2_IN_GA₂(X, Xs)
IF_GOAL_1_IN_1_G₂(X, s2l_2_out_ga₂(X, XS)) -> IF_GOAL_1_IN_2_G₃(X, XS, append_3_in_gaa₃(XS, YS, ZS))
IF_GOAL_1_IN_1_G₂(X, s2l_2_out_ga₂(X, XS)) -> APPEND_3_IN_GAA₃(XS, YS, ZS)
APPEND_3_IN_GAA₃(._2₂(X, XS), YS, ._2₂(X, ZS)) -> IF_APPEND_3_IN_1_GAA₅(X, XS, YS, ZS, append_3_in_gaa₃(XS, YS, ZS))
APPEND_3_IN_GAA₃(._2₂(X, XS), YS, ._2₂(X, ZS)) -> APPEND_3_IN_GAA₃(XS, YS, ZS)

The TRS R consists of the following rules:

goal_1_in_g₁(X) -> if_goal_1_in_1_g₂(X, s2l_2_in_ga₂(X, XS))
s2l_2_in_ga₂(s_1₁(X), ._2₂(Y, Xs)) -> if_s2l_2_in_1_ga₄(X, Y, Xs, s2l_2_in_ga₂(X, Xs))
s2l_2_in_ga₂(0_0, []_0) -> s2l_2_out_ga₂(0_0, []_0)
if_s2l_2_in_1_ga₄(X, Y, Xs, s2l_2_out_ga₂(X, Xs)) -> s2l_2_out_ga₂(s_1₁(X), ._2₂(Y, Xs))
if_goal_1_in_1_g₂(X, s2l_2_out_ga₂(X, XS)) -> if_goal_1_in_2_g₃(X, XS, append_3_in_gaa₃(XS, YS, ZS))
append_3_in_gaa₃([]_0, XS, XS) -> append_3_out_gaa₃([]_0, XS, XS)
append_3_in_gaa₃(._2₂(X, XS), YS, ._2₂(X, ZS)) -> if_append_3_in_1_gaa₅(X, XS, YS, ZS, append_3_in_gaa₃(XS, YS, ZS))
if_append_3_in_1_gaa₅(X, XS, YS, ZS, append_3_out_gaa₃(XS, YS, ZS)) -> append_3_out_gaa₃(._2₂(X, XS), YS, ._2₂(X, ZS))
if_goal_1_in_2_g₃(X, XS, append_3_out_gaa₃(XS, YS, ZS)) -> goal_1_out_g₁(X)

The argument filtering Pi contains the following mapping:
goal_1_in_g₁(x1) = goal_1_in_g₁(x1)
[]_0 = []_0
._2₂(x1, x2) = ._2₁(x2)
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
if_goal_1_in_1_g₂(x1, x2) = if_goal_1_in_1_g₁(x2)
s2l_2_in_ga₂(x1, x2) = s2l_2_in_ga₁(x1)
if_s2l_2_in_1_ga₄(x1, x2, x3, x4) = if_s2l_2_in_1_ga₁(x4)
s2l_2_out_ga₂(x1, x2) = s2l_2_out_ga₁(x2)
if_goal_1_in_2_g₃(x1, x2, x3) = if_goal_1_in_2_g₁(x3)
append_3_in_gaa₃(x1, x2, x3) = append_3_in_gaa₁(x1)
append_3_out_gaa₃(x1, x2, x3) = append_3_out_gaa
if_append_3_in_1_gaa₅(x1, x2, x3, x4, x5) = if_append_3_in_1_gaa₁(x5)
goal_1_out_g₁(x1) = goal_1_out_g
S2L_2_IN_GA₂(x1, x2) = S2L_2_IN_GA₁(x1)
GOAL_1_IN_G₁(x1) = GOAL_1_IN_G₁(x1)
APPEND_3_IN_GAA₃(x1, x2, x3) = APPEND_3_IN_GAA₁(x1)
IF_S2L_2_IN_1_GA₄(x1, x2, x3, x4) = IF_S2L_2_IN_1_GA₁(x4)
IF_APPEND_3_IN_1_GAA₅(x1, x2, x3, x4, x5) = IF_APPEND_3_IN_1_GAA₁(x5)
IF_GOAL_1_IN_1_G₂(x1, x2) = IF_GOAL_1_IN_1_G₁(x2)
IF_GOAL_1_IN_2_G₃(x1, x2, x3) = IF_GOAL_1_IN_2_G₁(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:

GOAL_1_IN_G₁(X) -> IF_GOAL_1_IN_1_G₂(X, s2l_2_in_ga₂(X, XS))
GOAL_1_IN_G₁(X) -> S2L_2_IN_GA₂(X, XS)
S2L_2_IN_GA₂(s_1₁(X), ._2₂(Y, Xs)) -> IF_S2L_2_IN_1_GA₄(X, Y, Xs, s2l_2_in_ga₂(X, Xs))
S2L_2_IN_GA₂(s_1₁(X), ._2₂(Y, Xs)) -> S2L_2_IN_GA₂(X, Xs)
IF_GOAL_1_IN_1_G₂(X, s2l_2_out_ga₂(X, XS)) -> IF_GOAL_1_IN_2_G₃(X, XS, append_3_in_gaa₃(XS, YS, ZS))
IF_GOAL_1_IN_1_G₂(X, s2l_2_out_ga₂(X, XS)) -> APPEND_3_IN_GAA₃(XS, YS, ZS)
APPEND_3_IN_GAA₃(._2₂(X, XS), YS, ._2₂(X, ZS)) -> IF_APPEND_3_IN_1_GAA₅(X, XS, YS, ZS, append_3_in_gaa₃(XS, YS, ZS))
APPEND_3_IN_GAA₃(._2₂(X, XS), YS, ._2₂(X, ZS)) -> APPEND_3_IN_GAA₃(XS, YS, ZS)

The TRS R consists of the following rules:

goal_1_in_g₁(X) -> if_goal_1_in_1_g₂(X, s2l_2_in_ga₂(X, XS))
s2l_2_in_ga₂(s_1₁(X), ._2₂(Y, Xs)) -> if_s2l_2_in_1_ga₄(X, Y, Xs, s2l_2_in_ga₂(X, Xs))
s2l_2_in_ga₂(0_0, []_0) -> s2l_2_out_ga₂(0_0, []_0)
if_s2l_2_in_1_ga₄(X, Y, Xs, s2l_2_out_ga₂(X, Xs)) -> s2l_2_out_ga₂(s_1₁(X), ._2₂(Y, Xs))
if_goal_1_in_1_g₂(X, s2l_2_out_ga₂(X, XS)) -> if_goal_1_in_2_g₃(X, XS, append_3_in_gaa₃(XS, YS, ZS))
append_3_in_gaa₃([]_0, XS, XS) -> append_3_out_gaa₃([]_0, XS, XS)
append_3_in_gaa₃(._2₂(X, XS), YS, ._2₂(X, ZS)) -> if_append_3_in_1_gaa₅(X, XS, YS, ZS, append_3_in_gaa₃(XS, YS, ZS))
if_append_3_in_1_gaa₅(X, XS, YS, ZS, append_3_out_gaa₃(XS, YS, ZS)) -> append_3_out_gaa₃(._2₂(X, XS), YS, ._2₂(X, ZS))
if_goal_1_in_2_g₃(X, XS, append_3_out_gaa₃(XS, YS, ZS)) -> goal_1_out_g₁(X)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

APPEND_3_IN_GAA₃(._2₂(X, XS), YS, ._2₂(X, ZS)) -> APPEND_3_IN_GAA₃(XS, YS, ZS)

The TRS R consists of the following rules:

goal_1_in_g₁(X) -> if_goal_1_in_1_g₂(X, s2l_2_in_ga₂(X, XS))
s2l_2_in_ga₂(s_1₁(X), ._2₂(Y, Xs)) -> if_s2l_2_in_1_ga₄(X, Y, Xs, s2l_2_in_ga₂(X, Xs))
s2l_2_in_ga₂(0_0, []_0) -> s2l_2_out_ga₂(0_0, []_0)
if_s2l_2_in_1_ga₄(X, Y, Xs, s2l_2_out_ga₂(X, Xs)) -> s2l_2_out_ga₂(s_1₁(X), ._2₂(Y, Xs))
if_goal_1_in_1_g₂(X, s2l_2_out_ga₂(X, XS)) -> if_goal_1_in_2_g₃(X, XS, append_3_in_gaa₃(XS, YS, ZS))
append_3_in_gaa₃([]_0, XS, XS) -> append_3_out_gaa₃([]_0, XS, XS)
append_3_in_gaa₃(._2₂(X, XS), YS, ._2₂(X, ZS)) -> if_append_3_in_1_gaa₅(X, XS, YS, ZS, append_3_in_gaa₃(XS, YS, ZS))
if_append_3_in_1_gaa₅(X, XS, YS, ZS, append_3_out_gaa₃(XS, YS, ZS)) -> append_3_out_gaa₃(._2₂(X, XS), YS, ._2₂(X, ZS))
if_goal_1_in_2_g₃(X, XS, append_3_out_gaa₃(XS, YS, ZS)) -> goal_1_out_g₁(X)

The argument filtering Pi contains the following mapping:
goal_1_in_g₁(x1) = goal_1_in_g₁(x1)
[]_0 = []_0
._2₂(x1, x2) = ._2₁(x2)
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
if_goal_1_in_1_g₂(x1, x2) = if_goal_1_in_1_g₁(x2)
s2l_2_in_ga₂(x1, x2) = s2l_2_in_ga₁(x1)
if_s2l_2_in_1_ga₄(x1, x2, x3, x4) = if_s2l_2_in_1_ga₁(x4)
s2l_2_out_ga₂(x1, x2) = s2l_2_out_ga₁(x2)
if_goal_1_in_2_g₃(x1, x2, x3) = if_goal_1_in_2_g₁(x3)
append_3_in_gaa₃(x1, x2, x3) = append_3_in_gaa₁(x1)
append_3_out_gaa₃(x1, x2, x3) = append_3_out_gaa
if_append_3_in_1_gaa₅(x1, x2, x3, x4, x5) = if_append_3_in_1_gaa₁(x5)
goal_1_out_g₁(x1) = goal_1_out_g
APPEND_3_IN_GAA₃(x1, x2, x3) = APPEND_3_IN_GAA₁(x1)

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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

APPEND_3_IN_GAA₃(._2₂(X, XS), YS, ._2₂(X, ZS)) -> APPEND_3_IN_GAA₃(XS, YS, ZS)

R is empty.
The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₁(x2)
APPEND_3_IN_GAA₃(x1, x2, x3) = APPEND_3_IN_GAA₁(x1)

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

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

APPEND_3_IN_GAA₁(._2₁(XS)) -> APPEND_3_IN_GAA₁(XS)

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_GAA₁}.
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_GAA₁(._2₁(XS)) -> APPEND_3_IN_GAA₁(XS)
The graph contains the following edges 1 > 1

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

S2L_2_IN_GA₂(s_1₁(X), ._2₂(Y, Xs)) -> S2L_2_IN_GA₂(X, Xs)

The TRS R consists of the following rules:

goal_1_in_g₁(X) -> if_goal_1_in_1_g₂(X, s2l_2_in_ga₂(X, XS))
s2l_2_in_ga₂(s_1₁(X), ._2₂(Y, Xs)) -> if_s2l_2_in_1_ga₄(X, Y, Xs, s2l_2_in_ga₂(X, Xs))
s2l_2_in_ga₂(0_0, []_0) -> s2l_2_out_ga₂(0_0, []_0)
if_s2l_2_in_1_ga₄(X, Y, Xs, s2l_2_out_ga₂(X, Xs)) -> s2l_2_out_ga₂(s_1₁(X), ._2₂(Y, Xs))
if_goal_1_in_1_g₂(X, s2l_2_out_ga₂(X, XS)) -> if_goal_1_in_2_g₃(X, XS, append_3_in_gaa₃(XS, YS, ZS))
append_3_in_gaa₃([]_0, XS, XS) -> append_3_out_gaa₃([]_0, XS, XS)
append_3_in_gaa₃(._2₂(X, XS), YS, ._2₂(X, ZS)) -> if_append_3_in_1_gaa₅(X, XS, YS, ZS, append_3_in_gaa₃(XS, YS, ZS))
if_append_3_in_1_gaa₅(X, XS, YS, ZS, append_3_out_gaa₃(XS, YS, ZS)) -> append_3_out_gaa₃(._2₂(X, XS), YS, ._2₂(X, ZS))
if_goal_1_in_2_g₃(X, XS, append_3_out_gaa₃(XS, YS, ZS)) -> goal_1_out_g₁(X)

The argument filtering Pi contains the following mapping:
goal_1_in_g₁(x1) = goal_1_in_g₁(x1)
[]_0 = []_0
._2₂(x1, x2) = ._2₁(x2)
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
if_goal_1_in_1_g₂(x1, x2) = if_goal_1_in_1_g₁(x2)
s2l_2_in_ga₂(x1, x2) = s2l_2_in_ga₁(x1)
if_s2l_2_in_1_ga₄(x1, x2, x3, x4) = if_s2l_2_in_1_ga₁(x4)
s2l_2_out_ga₂(x1, x2) = s2l_2_out_ga₁(x2)
if_goal_1_in_2_g₃(x1, x2, x3) = if_goal_1_in_2_g₁(x3)
append_3_in_gaa₃(x1, x2, x3) = append_3_in_gaa₁(x1)
append_3_out_gaa₃(x1, x2, x3) = append_3_out_gaa
if_append_3_in_1_gaa₅(x1, x2, x3, x4, x5) = if_append_3_in_1_gaa₁(x5)
goal_1_out_g₁(x1) = goal_1_out_g
S2L_2_IN_GA₂(x1, x2) = S2L_2_IN_GA₁(x1)

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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

S2L_2_IN_GA₂(s_1₁(X), ._2₂(Y, Xs)) -> S2L_2_IN_GA₂(X, Xs)

R is empty.
The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₁(x2)
s_1₁(x1) = s_1₁(x1)
S2L_2_IN_GA₂(x1, x2) = S2L_2_IN_GA₁(x1)

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

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

S2L_2_IN_GA₁(s_1₁(X)) -> S2L_2_IN_GA₁(X)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {S2L_2_IN_GA₁}.
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:

S2L_2_IN_GA₁(s_1₁(X)) -> S2L_2_IN_GA₁(X)
The graph contains the following edges 1 > 1