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)

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

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)

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


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)

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)

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
The approximation of the Dependency Graph contains 2 SCCs with 6 less nodes.

↳ 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:



↳ 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: