Left Termination of the query pattern goal_in_1(g) w.r.t. the given Prolog program could not be shown:



Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

Clauses:

append([], XS, XS).
append(.(X, XS), YS, .(X, ZS)) :- append(XS, YS, ZS).
s2l(s(X), .(Y, Xs)) :- s2l(X, Xs).
s2l(0, []).
goal(X) :- ','(s2l(X, XS), append(XS, YS, ZS)).

Queries:

goal(g).

We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
goal_in: (b)
s2l_in: (b,f)
append_in: (f,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

goal_in_g(X) → U3_g(X, s2l_in_ga(X, XS))
s2l_in_ga(s(X), .(Y, Xs)) → U2_ga(X, Y, Xs, s2l_in_ga(X, Xs))
s2l_in_ga(0, []) → s2l_out_ga(0, [])
U2_ga(X, Y, Xs, s2l_out_ga(X, Xs)) → s2l_out_ga(s(X), .(Y, Xs))
U3_g(X, s2l_out_ga(X, XS)) → U4_g(X, append_in_aaa(XS, YS, ZS))
append_in_aaa([], XS, XS) → append_out_aaa([], XS, XS)
append_in_aaa(.(X, XS), YS, .(X, ZS)) → U1_aaa(X, XS, YS, ZS, append_in_aaa(XS, YS, ZS))
U1_aaa(X, XS, YS, ZS, append_out_aaa(XS, YS, ZS)) → append_out_aaa(.(X, XS), YS, .(X, ZS))
U4_g(X, append_out_aaa(XS, YS, ZS)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U3_g(x1, x2)  =  U3_g(x2)
s2l_in_ga(x1, x2)  =  s2l_in_ga(x1)
s(x1)  =  s(x1)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x4)
0  =  0
s2l_out_ga(x1, x2)  =  s2l_out_ga
U4_g(x1, x2)  =  U4_g(x2)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
goal_out_g(x1)  =  goal_out_g

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
PiTRS
      ↳ DependencyPairsProof
  ↳ PrologToPiTRSProof

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

goal_in_g(X) → U3_g(X, s2l_in_ga(X, XS))
s2l_in_ga(s(X), .(Y, Xs)) → U2_ga(X, Y, Xs, s2l_in_ga(X, Xs))
s2l_in_ga(0, []) → s2l_out_ga(0, [])
U2_ga(X, Y, Xs, s2l_out_ga(X, Xs)) → s2l_out_ga(s(X), .(Y, Xs))
U3_g(X, s2l_out_ga(X, XS)) → U4_g(X, append_in_aaa(XS, YS, ZS))
append_in_aaa([], XS, XS) → append_out_aaa([], XS, XS)
append_in_aaa(.(X, XS), YS, .(X, ZS)) → U1_aaa(X, XS, YS, ZS, append_in_aaa(XS, YS, ZS))
U1_aaa(X, XS, YS, ZS, append_out_aaa(XS, YS, ZS)) → append_out_aaa(.(X, XS), YS, .(X, ZS))
U4_g(X, append_out_aaa(XS, YS, ZS)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U3_g(x1, x2)  =  U3_g(x2)
s2l_in_ga(x1, x2)  =  s2l_in_ga(x1)
s(x1)  =  s(x1)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x4)
0  =  0
s2l_out_ga(x1, x2)  =  s2l_out_ga
U4_g(x1, x2)  =  U4_g(x2)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
goal_out_g(x1)  =  goal_out_g


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

GOAL_IN_G(X) → U3_G(X, s2l_in_ga(X, XS))
GOAL_IN_G(X) → S2L_IN_GA(X, XS)
S2L_IN_GA(s(X), .(Y, Xs)) → U2_GA(X, Y, Xs, s2l_in_ga(X, Xs))
S2L_IN_GA(s(X), .(Y, Xs)) → S2L_IN_GA(X, Xs)
U3_G(X, s2l_out_ga(X, XS)) → U4_G(X, append_in_aaa(XS, YS, ZS))
U3_G(X, s2l_out_ga(X, XS)) → APPEND_IN_AAA(XS, YS, ZS)
APPEND_IN_AAA(.(X, XS), YS, .(X, ZS)) → U1_AAA(X, XS, YS, ZS, append_in_aaa(XS, YS, ZS))
APPEND_IN_AAA(.(X, XS), YS, .(X, ZS)) → APPEND_IN_AAA(XS, YS, ZS)

The TRS R consists of the following rules:

goal_in_g(X) → U3_g(X, s2l_in_ga(X, XS))
s2l_in_ga(s(X), .(Y, Xs)) → U2_ga(X, Y, Xs, s2l_in_ga(X, Xs))
s2l_in_ga(0, []) → s2l_out_ga(0, [])
U2_ga(X, Y, Xs, s2l_out_ga(X, Xs)) → s2l_out_ga(s(X), .(Y, Xs))
U3_g(X, s2l_out_ga(X, XS)) → U4_g(X, append_in_aaa(XS, YS, ZS))
append_in_aaa([], XS, XS) → append_out_aaa([], XS, XS)
append_in_aaa(.(X, XS), YS, .(X, ZS)) → U1_aaa(X, XS, YS, ZS, append_in_aaa(XS, YS, ZS))
U1_aaa(X, XS, YS, ZS, append_out_aaa(XS, YS, ZS)) → append_out_aaa(.(X, XS), YS, .(X, ZS))
U4_g(X, append_out_aaa(XS, YS, ZS)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U3_g(x1, x2)  =  U3_g(x2)
s2l_in_ga(x1, x2)  =  s2l_in_ga(x1)
s(x1)  =  s(x1)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x4)
0  =  0
s2l_out_ga(x1, x2)  =  s2l_out_ga
U4_g(x1, x2)  =  U4_g(x2)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
goal_out_g(x1)  =  goal_out_g
U4_G(x1, x2)  =  U4_G(x2)
GOAL_IN_G(x1)  =  GOAL_IN_G(x1)
S2L_IN_GA(x1, x2)  =  S2L_IN_GA(x1)
U2_GA(x1, x2, x3, x4)  =  U2_GA(x4)
U1_AAA(x1, x2, x3, x4, x5)  =  U1_AAA(x5)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA
U3_G(x1, x2)  =  U3_G(x2)

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

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
PiDP
          ↳ DependencyGraphProof
  ↳ PrologToPiTRSProof

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

GOAL_IN_G(X) → U3_G(X, s2l_in_ga(X, XS))
GOAL_IN_G(X) → S2L_IN_GA(X, XS)
S2L_IN_GA(s(X), .(Y, Xs)) → U2_GA(X, Y, Xs, s2l_in_ga(X, Xs))
S2L_IN_GA(s(X), .(Y, Xs)) → S2L_IN_GA(X, Xs)
U3_G(X, s2l_out_ga(X, XS)) → U4_G(X, append_in_aaa(XS, YS, ZS))
U3_G(X, s2l_out_ga(X, XS)) → APPEND_IN_AAA(XS, YS, ZS)
APPEND_IN_AAA(.(X, XS), YS, .(X, ZS)) → U1_AAA(X, XS, YS, ZS, append_in_aaa(XS, YS, ZS))
APPEND_IN_AAA(.(X, XS), YS, .(X, ZS)) → APPEND_IN_AAA(XS, YS, ZS)

The TRS R consists of the following rules:

goal_in_g(X) → U3_g(X, s2l_in_ga(X, XS))
s2l_in_ga(s(X), .(Y, Xs)) → U2_ga(X, Y, Xs, s2l_in_ga(X, Xs))
s2l_in_ga(0, []) → s2l_out_ga(0, [])
U2_ga(X, Y, Xs, s2l_out_ga(X, Xs)) → s2l_out_ga(s(X), .(Y, Xs))
U3_g(X, s2l_out_ga(X, XS)) → U4_g(X, append_in_aaa(XS, YS, ZS))
append_in_aaa([], XS, XS) → append_out_aaa([], XS, XS)
append_in_aaa(.(X, XS), YS, .(X, ZS)) → U1_aaa(X, XS, YS, ZS, append_in_aaa(XS, YS, ZS))
U1_aaa(X, XS, YS, ZS, append_out_aaa(XS, YS, ZS)) → append_out_aaa(.(X, XS), YS, .(X, ZS))
U4_g(X, append_out_aaa(XS, YS, ZS)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U3_g(x1, x2)  =  U3_g(x2)
s2l_in_ga(x1, x2)  =  s2l_in_ga(x1)
s(x1)  =  s(x1)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x4)
0  =  0
s2l_out_ga(x1, x2)  =  s2l_out_ga
U4_g(x1, x2)  =  U4_g(x2)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
goal_out_g(x1)  =  goal_out_g
U4_G(x1, x2)  =  U4_G(x2)
GOAL_IN_G(x1)  =  GOAL_IN_G(x1)
S2L_IN_GA(x1, x2)  =  S2L_IN_GA(x1)
U2_GA(x1, x2, x3, x4)  =  U2_GA(x4)
U1_AAA(x1, x2, x3, x4, x5)  =  U1_AAA(x5)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA
U3_G(x1, x2)  =  U3_G(x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 6 less nodes.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
PiDP
                ↳ UsableRulesProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

APPEND_IN_AAA(.(X, XS), YS, .(X, ZS)) → APPEND_IN_AAA(XS, YS, ZS)

The TRS R consists of the following rules:

goal_in_g(X) → U3_g(X, s2l_in_ga(X, XS))
s2l_in_ga(s(X), .(Y, Xs)) → U2_ga(X, Y, Xs, s2l_in_ga(X, Xs))
s2l_in_ga(0, []) → s2l_out_ga(0, [])
U2_ga(X, Y, Xs, s2l_out_ga(X, Xs)) → s2l_out_ga(s(X), .(Y, Xs))
U3_g(X, s2l_out_ga(X, XS)) → U4_g(X, append_in_aaa(XS, YS, ZS))
append_in_aaa([], XS, XS) → append_out_aaa([], XS, XS)
append_in_aaa(.(X, XS), YS, .(X, ZS)) → U1_aaa(X, XS, YS, ZS, append_in_aaa(XS, YS, ZS))
U1_aaa(X, XS, YS, ZS, append_out_aaa(XS, YS, ZS)) → append_out_aaa(.(X, XS), YS, .(X, ZS))
U4_g(X, append_out_aaa(XS, YS, ZS)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U3_g(x1, x2)  =  U3_g(x2)
s2l_in_ga(x1, x2)  =  s2l_in_ga(x1)
s(x1)  =  s(x1)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x4)
0  =  0
s2l_out_ga(x1, x2)  =  s2l_out_ga
U4_g(x1, x2)  =  U4_g(x2)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
goal_out_g(x1)  =  goal_out_g
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA

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
                ↳ UsableRulesProof
PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

APPEND_IN_AAA(.(X, XS), YS, .(X, ZS)) → APPEND_IN_AAA(XS, YS, ZS)

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

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
                        ↳ NonTerminationProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

APPEND_IN_AAAAPPEND_IN_AAA

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

APPEND_IN_AAAAPPEND_IN_AAA

The TRS R consists of the following rules:none


s = APPEND_IN_AAA evaluates to t =APPEND_IN_AAA

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:




Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from APPEND_IN_AAA to APPEND_IN_AAA.





↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
PiDP
                ↳ UsableRulesProof
  ↳ PrologToPiTRSProof

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

S2L_IN_GA(s(X), .(Y, Xs)) → S2L_IN_GA(X, Xs)

The TRS R consists of the following rules:

goal_in_g(X) → U3_g(X, s2l_in_ga(X, XS))
s2l_in_ga(s(X), .(Y, Xs)) → U2_ga(X, Y, Xs, s2l_in_ga(X, Xs))
s2l_in_ga(0, []) → s2l_out_ga(0, [])
U2_ga(X, Y, Xs, s2l_out_ga(X, Xs)) → s2l_out_ga(s(X), .(Y, Xs))
U3_g(X, s2l_out_ga(X, XS)) → U4_g(X, append_in_aaa(XS, YS, ZS))
append_in_aaa([], XS, XS) → append_out_aaa([], XS, XS)
append_in_aaa(.(X, XS), YS, .(X, ZS)) → U1_aaa(X, XS, YS, ZS, append_in_aaa(XS, YS, ZS))
U1_aaa(X, XS, YS, ZS, append_out_aaa(XS, YS, ZS)) → append_out_aaa(.(X, XS), YS, .(X, ZS))
U4_g(X, append_out_aaa(XS, YS, ZS)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U3_g(x1, x2)  =  U3_g(x2)
s2l_in_ga(x1, x2)  =  s2l_in_ga(x1)
s(x1)  =  s(x1)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x4)
0  =  0
s2l_out_ga(x1, x2)  =  s2l_out_ga
U4_g(x1, x2)  =  U4_g(x2)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
goal_out_g(x1)  =  goal_out_g
S2L_IN_GA(x1, x2)  =  S2L_IN_GA(x1)

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
                ↳ UsableRulesProof
PiDP
                    ↳ PiDPToQDPProof
  ↳ PrologToPiTRSProof

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

S2L_IN_GA(s(X), .(Y, Xs)) → S2L_IN_GA(X, Xs)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
S2L_IN_GA(x1, x2)  =  S2L_IN_GA(x1)

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
  ↳ PrologToPiTRSProof

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

S2L_IN_GA(s(X)) → S2L_IN_GA(X)

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:


We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
goal_in: (b)
s2l_in: (b,f)
append_in: (f,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

goal_in_g(X) → U3_g(X, s2l_in_ga(X, XS))
s2l_in_ga(s(X), .(Y, Xs)) → U2_ga(X, Y, Xs, s2l_in_ga(X, Xs))
s2l_in_ga(0, []) → s2l_out_ga(0, [])
U2_ga(X, Y, Xs, s2l_out_ga(X, Xs)) → s2l_out_ga(s(X), .(Y, Xs))
U3_g(X, s2l_out_ga(X, XS)) → U4_g(X, append_in_aaa(XS, YS, ZS))
append_in_aaa([], XS, XS) → append_out_aaa([], XS, XS)
append_in_aaa(.(X, XS), YS, .(X, ZS)) → U1_aaa(X, XS, YS, ZS, append_in_aaa(XS, YS, ZS))
U1_aaa(X, XS, YS, ZS, append_out_aaa(XS, YS, ZS)) → append_out_aaa(.(X, XS), YS, .(X, ZS))
U4_g(X, append_out_aaa(XS, YS, ZS)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)
s2l_in_ga(x1, x2)  =  s2l_in_ga(x1)
s(x1)  =  s(x1)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x1, x4)
0  =  0
s2l_out_ga(x1, x2)  =  s2l_out_ga(x1)
U4_g(x1, x2)  =  U4_g(x1, x2)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
goal_out_g(x1)  =  goal_out_g(x1)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
PiTRS
      ↳ DependencyPairsProof

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

goal_in_g(X) → U3_g(X, s2l_in_ga(X, XS))
s2l_in_ga(s(X), .(Y, Xs)) → U2_ga(X, Y, Xs, s2l_in_ga(X, Xs))
s2l_in_ga(0, []) → s2l_out_ga(0, [])
U2_ga(X, Y, Xs, s2l_out_ga(X, Xs)) → s2l_out_ga(s(X), .(Y, Xs))
U3_g(X, s2l_out_ga(X, XS)) → U4_g(X, append_in_aaa(XS, YS, ZS))
append_in_aaa([], XS, XS) → append_out_aaa([], XS, XS)
append_in_aaa(.(X, XS), YS, .(X, ZS)) → U1_aaa(X, XS, YS, ZS, append_in_aaa(XS, YS, ZS))
U1_aaa(X, XS, YS, ZS, append_out_aaa(XS, YS, ZS)) → append_out_aaa(.(X, XS), YS, .(X, ZS))
U4_g(X, append_out_aaa(XS, YS, ZS)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)
s2l_in_ga(x1, x2)  =  s2l_in_ga(x1)
s(x1)  =  s(x1)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x1, x4)
0  =  0
s2l_out_ga(x1, x2)  =  s2l_out_ga(x1)
U4_g(x1, x2)  =  U4_g(x1, x2)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
goal_out_g(x1)  =  goal_out_g(x1)


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

GOAL_IN_G(X) → U3_G(X, s2l_in_ga(X, XS))
GOAL_IN_G(X) → S2L_IN_GA(X, XS)
S2L_IN_GA(s(X), .(Y, Xs)) → U2_GA(X, Y, Xs, s2l_in_ga(X, Xs))
S2L_IN_GA(s(X), .(Y, Xs)) → S2L_IN_GA(X, Xs)
U3_G(X, s2l_out_ga(X, XS)) → U4_G(X, append_in_aaa(XS, YS, ZS))
U3_G(X, s2l_out_ga(X, XS)) → APPEND_IN_AAA(XS, YS, ZS)
APPEND_IN_AAA(.(X, XS), YS, .(X, ZS)) → U1_AAA(X, XS, YS, ZS, append_in_aaa(XS, YS, ZS))
APPEND_IN_AAA(.(X, XS), YS, .(X, ZS)) → APPEND_IN_AAA(XS, YS, ZS)

The TRS R consists of the following rules:

goal_in_g(X) → U3_g(X, s2l_in_ga(X, XS))
s2l_in_ga(s(X), .(Y, Xs)) → U2_ga(X, Y, Xs, s2l_in_ga(X, Xs))
s2l_in_ga(0, []) → s2l_out_ga(0, [])
U2_ga(X, Y, Xs, s2l_out_ga(X, Xs)) → s2l_out_ga(s(X), .(Y, Xs))
U3_g(X, s2l_out_ga(X, XS)) → U4_g(X, append_in_aaa(XS, YS, ZS))
append_in_aaa([], XS, XS) → append_out_aaa([], XS, XS)
append_in_aaa(.(X, XS), YS, .(X, ZS)) → U1_aaa(X, XS, YS, ZS, append_in_aaa(XS, YS, ZS))
U1_aaa(X, XS, YS, ZS, append_out_aaa(XS, YS, ZS)) → append_out_aaa(.(X, XS), YS, .(X, ZS))
U4_g(X, append_out_aaa(XS, YS, ZS)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)
s2l_in_ga(x1, x2)  =  s2l_in_ga(x1)
s(x1)  =  s(x1)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x1, x4)
0  =  0
s2l_out_ga(x1, x2)  =  s2l_out_ga(x1)
U4_g(x1, x2)  =  U4_g(x1, x2)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
goal_out_g(x1)  =  goal_out_g(x1)
U4_G(x1, x2)  =  U4_G(x1, x2)
GOAL_IN_G(x1)  =  GOAL_IN_G(x1)
S2L_IN_GA(x1, x2)  =  S2L_IN_GA(x1)
U2_GA(x1, x2, x3, x4)  =  U2_GA(x1, x4)
U1_AAA(x1, x2, x3, x4, x5)  =  U1_AAA(x5)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA
U3_G(x1, x2)  =  U3_G(x1, x2)

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

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
PiDP
          ↳ DependencyGraphProof

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

GOAL_IN_G(X) → U3_G(X, s2l_in_ga(X, XS))
GOAL_IN_G(X) → S2L_IN_GA(X, XS)
S2L_IN_GA(s(X), .(Y, Xs)) → U2_GA(X, Y, Xs, s2l_in_ga(X, Xs))
S2L_IN_GA(s(X), .(Y, Xs)) → S2L_IN_GA(X, Xs)
U3_G(X, s2l_out_ga(X, XS)) → U4_G(X, append_in_aaa(XS, YS, ZS))
U3_G(X, s2l_out_ga(X, XS)) → APPEND_IN_AAA(XS, YS, ZS)
APPEND_IN_AAA(.(X, XS), YS, .(X, ZS)) → U1_AAA(X, XS, YS, ZS, append_in_aaa(XS, YS, ZS))
APPEND_IN_AAA(.(X, XS), YS, .(X, ZS)) → APPEND_IN_AAA(XS, YS, ZS)

The TRS R consists of the following rules:

goal_in_g(X) → U3_g(X, s2l_in_ga(X, XS))
s2l_in_ga(s(X), .(Y, Xs)) → U2_ga(X, Y, Xs, s2l_in_ga(X, Xs))
s2l_in_ga(0, []) → s2l_out_ga(0, [])
U2_ga(X, Y, Xs, s2l_out_ga(X, Xs)) → s2l_out_ga(s(X), .(Y, Xs))
U3_g(X, s2l_out_ga(X, XS)) → U4_g(X, append_in_aaa(XS, YS, ZS))
append_in_aaa([], XS, XS) → append_out_aaa([], XS, XS)
append_in_aaa(.(X, XS), YS, .(X, ZS)) → U1_aaa(X, XS, YS, ZS, append_in_aaa(XS, YS, ZS))
U1_aaa(X, XS, YS, ZS, append_out_aaa(XS, YS, ZS)) → append_out_aaa(.(X, XS), YS, .(X, ZS))
U4_g(X, append_out_aaa(XS, YS, ZS)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)
s2l_in_ga(x1, x2)  =  s2l_in_ga(x1)
s(x1)  =  s(x1)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x1, x4)
0  =  0
s2l_out_ga(x1, x2)  =  s2l_out_ga(x1)
U4_g(x1, x2)  =  U4_g(x1, x2)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
goal_out_g(x1)  =  goal_out_g(x1)
U4_G(x1, x2)  =  U4_G(x1, x2)
GOAL_IN_G(x1)  =  GOAL_IN_G(x1)
S2L_IN_GA(x1, x2)  =  S2L_IN_GA(x1)
U2_GA(x1, x2, x3, x4)  =  U2_GA(x1, x4)
U1_AAA(x1, x2, x3, x4, x5)  =  U1_AAA(x5)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA
U3_G(x1, x2)  =  U3_G(x1, x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 6 less nodes.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
PiDP
                ↳ UsableRulesProof
              ↳ PiDP

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

APPEND_IN_AAA(.(X, XS), YS, .(X, ZS)) → APPEND_IN_AAA(XS, YS, ZS)

The TRS R consists of the following rules:

goal_in_g(X) → U3_g(X, s2l_in_ga(X, XS))
s2l_in_ga(s(X), .(Y, Xs)) → U2_ga(X, Y, Xs, s2l_in_ga(X, Xs))
s2l_in_ga(0, []) → s2l_out_ga(0, [])
U2_ga(X, Y, Xs, s2l_out_ga(X, Xs)) → s2l_out_ga(s(X), .(Y, Xs))
U3_g(X, s2l_out_ga(X, XS)) → U4_g(X, append_in_aaa(XS, YS, ZS))
append_in_aaa([], XS, XS) → append_out_aaa([], XS, XS)
append_in_aaa(.(X, XS), YS, .(X, ZS)) → U1_aaa(X, XS, YS, ZS, append_in_aaa(XS, YS, ZS))
U1_aaa(X, XS, YS, ZS, append_out_aaa(XS, YS, ZS)) → append_out_aaa(.(X, XS), YS, .(X, ZS))
U4_g(X, append_out_aaa(XS, YS, ZS)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)
s2l_in_ga(x1, x2)  =  s2l_in_ga(x1)
s(x1)  =  s(x1)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x1, x4)
0  =  0
s2l_out_ga(x1, x2)  =  s2l_out_ga(x1)
U4_g(x1, x2)  =  U4_g(x1, x2)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
goal_out_g(x1)  =  goal_out_g(x1)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA

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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP

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

APPEND_IN_AAA(.(X, XS), YS, .(X, ZS)) → APPEND_IN_AAA(XS, YS, ZS)

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

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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
QDP
                        ↳ NonTerminationProof
              ↳ PiDP

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

APPEND_IN_AAAAPPEND_IN_AAA

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

APPEND_IN_AAAAPPEND_IN_AAA

The TRS R consists of the following rules:none


s = APPEND_IN_AAA evaluates to t =APPEND_IN_AAA

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:




Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from APPEND_IN_AAA to APPEND_IN_AAA.





↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
PiDP
                ↳ UsableRulesProof

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

S2L_IN_GA(s(X), .(Y, Xs)) → S2L_IN_GA(X, Xs)

The TRS R consists of the following rules:

goal_in_g(X) → U3_g(X, s2l_in_ga(X, XS))
s2l_in_ga(s(X), .(Y, Xs)) → U2_ga(X, Y, Xs, s2l_in_ga(X, Xs))
s2l_in_ga(0, []) → s2l_out_ga(0, [])
U2_ga(X, Y, Xs, s2l_out_ga(X, Xs)) → s2l_out_ga(s(X), .(Y, Xs))
U3_g(X, s2l_out_ga(X, XS)) → U4_g(X, append_in_aaa(XS, YS, ZS))
append_in_aaa([], XS, XS) → append_out_aaa([], XS, XS)
append_in_aaa(.(X, XS), YS, .(X, ZS)) → U1_aaa(X, XS, YS, ZS, append_in_aaa(XS, YS, ZS))
U1_aaa(X, XS, YS, ZS, append_out_aaa(XS, YS, ZS)) → append_out_aaa(.(X, XS), YS, .(X, ZS))
U4_g(X, append_out_aaa(XS, YS, ZS)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U3_g(x1, x2)  =  U3_g(x1, x2)
s2l_in_ga(x1, x2)  =  s2l_in_ga(x1)
s(x1)  =  s(x1)
U2_ga(x1, x2, x3, x4)  =  U2_ga(x1, x4)
0  =  0
s2l_out_ga(x1, x2)  =  s2l_out_ga(x1)
U4_g(x1, x2)  =  U4_g(x1, x2)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa
U1_aaa(x1, x2, x3, x4, x5)  =  U1_aaa(x5)
goal_out_g(x1)  =  goal_out_g(x1)
S2L_IN_GA(x1, x2)  =  S2L_IN_GA(x1)

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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
PiDP
                    ↳ PiDPToQDPProof

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

S2L_IN_GA(s(X), .(Y, Xs)) → S2L_IN_GA(X, Xs)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
S2L_IN_GA(x1, x2)  =  S2L_IN_GA(x1)

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
  ↳ 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_IN_GA(s(X)) → S2L_IN_GA(X)

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: