Left Termination of the query pattern f_in_3(g, g, a) w.r.t. the given Prolog program could successfully be proven:



Prolog
  ↳ PrologToPiTRSProof

Clauses:

f(A, [], RES) :- g(A, [], RES).
f(.(A, As), .(B, Bs), RES) :- f(.(B, .(A, As)), Bs, RES).
g([], RES, RES).
g(.(C, Cs), D, RES) :- g(Cs, .(C, D), RES).

Queries:

f(g,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:

f_in(.(A, As), .(B, Bs), RES) → U2(A, As, B, Bs, RES, f_in(.(B, .(A, As)), Bs, RES))
f_in(A, [], RES) → U1(A, RES, g_in(A, [], RES))
g_in(.(C, Cs), D, RES) → U3(C, Cs, D, RES, g_in(Cs, .(C, D), RES))
g_in([], RES, RES) → g_out([], RES, RES)
U3(C, Cs, D, RES, g_out(Cs, .(C, D), RES)) → g_out(.(C, Cs), D, RES)
U1(A, RES, g_out(A, [], RES)) → f_out(A, [], RES)
U2(A, As, B, Bs, RES, f_out(.(B, .(A, As)), Bs, RES)) → f_out(.(A, As), .(B, Bs), RES)

The argument filtering Pi contains the following mapping:
f_in(x1, x2, x3)  =  f_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x6)
[]  =  []
U1(x1, x2, x3)  =  U1(x3)
g_in(x1, x2, x3)  =  g_in(x1, x2)
U3(x1, x2, x3, x4, x5)  =  U3(x5)
g_out(x1, x2, x3)  =  g_out(x3)
f_out(x1, x2, x3)  =  f_out(x3)

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:

f_in(.(A, As), .(B, Bs), RES) → U2(A, As, B, Bs, RES, f_in(.(B, .(A, As)), Bs, RES))
f_in(A, [], RES) → U1(A, RES, g_in(A, [], RES))
g_in(.(C, Cs), D, RES) → U3(C, Cs, D, RES, g_in(Cs, .(C, D), RES))
g_in([], RES, RES) → g_out([], RES, RES)
U3(C, Cs, D, RES, g_out(Cs, .(C, D), RES)) → g_out(.(C, Cs), D, RES)
U1(A, RES, g_out(A, [], RES)) → f_out(A, [], RES)
U2(A, As, B, Bs, RES, f_out(.(B, .(A, As)), Bs, RES)) → f_out(.(A, As), .(B, Bs), RES)

The argument filtering Pi contains the following mapping:
f_in(x1, x2, x3)  =  f_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x6)
[]  =  []
U1(x1, x2, x3)  =  U1(x3)
g_in(x1, x2, x3)  =  g_in(x1, x2)
U3(x1, x2, x3, x4, x5)  =  U3(x5)
g_out(x1, x2, x3)  =  g_out(x3)
f_out(x1, x2, x3)  =  f_out(x3)


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

F_IN(.(A, As), .(B, Bs), RES) → U21(A, As, B, Bs, RES, f_in(.(B, .(A, As)), Bs, RES))
F_IN(.(A, As), .(B, Bs), RES) → F_IN(.(B, .(A, As)), Bs, RES)
F_IN(A, [], RES) → U11(A, RES, g_in(A, [], RES))
F_IN(A, [], RES) → G_IN(A, [], RES)
G_IN(.(C, Cs), D, RES) → U31(C, Cs, D, RES, g_in(Cs, .(C, D), RES))
G_IN(.(C, Cs), D, RES) → G_IN(Cs, .(C, D), RES)

The TRS R consists of the following rules:

f_in(.(A, As), .(B, Bs), RES) → U2(A, As, B, Bs, RES, f_in(.(B, .(A, As)), Bs, RES))
f_in(A, [], RES) → U1(A, RES, g_in(A, [], RES))
g_in(.(C, Cs), D, RES) → U3(C, Cs, D, RES, g_in(Cs, .(C, D), RES))
g_in([], RES, RES) → g_out([], RES, RES)
U3(C, Cs, D, RES, g_out(Cs, .(C, D), RES)) → g_out(.(C, Cs), D, RES)
U1(A, RES, g_out(A, [], RES)) → f_out(A, [], RES)
U2(A, As, B, Bs, RES, f_out(.(B, .(A, As)), Bs, RES)) → f_out(.(A, As), .(B, Bs), RES)

The argument filtering Pi contains the following mapping:
f_in(x1, x2, x3)  =  f_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x6)
[]  =  []
U1(x1, x2, x3)  =  U1(x3)
g_in(x1, x2, x3)  =  g_in(x1, x2)
U3(x1, x2, x3, x4, x5)  =  U3(x5)
g_out(x1, x2, x3)  =  g_out(x3)
f_out(x1, x2, x3)  =  f_out(x3)
G_IN(x1, x2, x3)  =  G_IN(x1, x2)
F_IN(x1, x2, x3)  =  F_IN(x1, x2)
U31(x1, x2, x3, x4, x5)  =  U31(x5)
U21(x1, x2, x3, x4, x5, x6)  =  U21(x6)
U11(x1, x2, x3)  =  U11(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:

F_IN(.(A, As), .(B, Bs), RES) → U21(A, As, B, Bs, RES, f_in(.(B, .(A, As)), Bs, RES))
F_IN(.(A, As), .(B, Bs), RES) → F_IN(.(B, .(A, As)), Bs, RES)
F_IN(A, [], RES) → U11(A, RES, g_in(A, [], RES))
F_IN(A, [], RES) → G_IN(A, [], RES)
G_IN(.(C, Cs), D, RES) → U31(C, Cs, D, RES, g_in(Cs, .(C, D), RES))
G_IN(.(C, Cs), D, RES) → G_IN(Cs, .(C, D), RES)

The TRS R consists of the following rules:

f_in(.(A, As), .(B, Bs), RES) → U2(A, As, B, Bs, RES, f_in(.(B, .(A, As)), Bs, RES))
f_in(A, [], RES) → U1(A, RES, g_in(A, [], RES))
g_in(.(C, Cs), D, RES) → U3(C, Cs, D, RES, g_in(Cs, .(C, D), RES))
g_in([], RES, RES) → g_out([], RES, RES)
U3(C, Cs, D, RES, g_out(Cs, .(C, D), RES)) → g_out(.(C, Cs), D, RES)
U1(A, RES, g_out(A, [], RES)) → f_out(A, [], RES)
U2(A, As, B, Bs, RES, f_out(.(B, .(A, As)), Bs, RES)) → f_out(.(A, As), .(B, Bs), RES)

The argument filtering Pi contains the following mapping:
f_in(x1, x2, x3)  =  f_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x6)
[]  =  []
U1(x1, x2, x3)  =  U1(x3)
g_in(x1, x2, x3)  =  g_in(x1, x2)
U3(x1, x2, x3, x4, x5)  =  U3(x5)
g_out(x1, x2, x3)  =  g_out(x3)
f_out(x1, x2, x3)  =  f_out(x3)
G_IN(x1, x2, x3)  =  G_IN(x1, x2)
F_IN(x1, x2, x3)  =  F_IN(x1, x2)
U31(x1, x2, x3, x4, x5)  =  U31(x5)
U21(x1, x2, x3, x4, x5, x6)  =  U21(x6)
U11(x1, x2, x3)  =  U11(x3)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 4 less nodes.

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

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

G_IN(.(C, Cs), D, RES) → G_IN(Cs, .(C, D), RES)

The TRS R consists of the following rules:

f_in(.(A, As), .(B, Bs), RES) → U2(A, As, B, Bs, RES, f_in(.(B, .(A, As)), Bs, RES))
f_in(A, [], RES) → U1(A, RES, g_in(A, [], RES))
g_in(.(C, Cs), D, RES) → U3(C, Cs, D, RES, g_in(Cs, .(C, D), RES))
g_in([], RES, RES) → g_out([], RES, RES)
U3(C, Cs, D, RES, g_out(Cs, .(C, D), RES)) → g_out(.(C, Cs), D, RES)
U1(A, RES, g_out(A, [], RES)) → f_out(A, [], RES)
U2(A, As, B, Bs, RES, f_out(.(B, .(A, As)), Bs, RES)) → f_out(.(A, As), .(B, Bs), RES)

The argument filtering Pi contains the following mapping:
f_in(x1, x2, x3)  =  f_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x6)
[]  =  []
U1(x1, x2, x3)  =  U1(x3)
g_in(x1, x2, x3)  =  g_in(x1, x2)
U3(x1, x2, x3, x4, x5)  =  U3(x5)
g_out(x1, x2, x3)  =  g_out(x3)
f_out(x1, x2, x3)  =  f_out(x3)
G_IN(x1, x2, x3)  =  G_IN(x1, x2)

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

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

G_IN(.(C, Cs), D, RES) → G_IN(Cs, .(C, D), RES)

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

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

G_IN(.(C, Cs), D) → G_IN(Cs, .(C, D))

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:



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

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

F_IN(.(A, As), .(B, Bs), RES) → F_IN(.(B, .(A, As)), Bs, RES)

The TRS R consists of the following rules:

f_in(.(A, As), .(B, Bs), RES) → U2(A, As, B, Bs, RES, f_in(.(B, .(A, As)), Bs, RES))
f_in(A, [], RES) → U1(A, RES, g_in(A, [], RES))
g_in(.(C, Cs), D, RES) → U3(C, Cs, D, RES, g_in(Cs, .(C, D), RES))
g_in([], RES, RES) → g_out([], RES, RES)
U3(C, Cs, D, RES, g_out(Cs, .(C, D), RES)) → g_out(.(C, Cs), D, RES)
U1(A, RES, g_out(A, [], RES)) → f_out(A, [], RES)
U2(A, As, B, Bs, RES, f_out(.(B, .(A, As)), Bs, RES)) → f_out(.(A, As), .(B, Bs), RES)

The argument filtering Pi contains the following mapping:
f_in(x1, x2, x3)  =  f_in(x1, x2)
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4, x5, x6)  =  U2(x6)
[]  =  []
U1(x1, x2, x3)  =  U1(x3)
g_in(x1, x2, x3)  =  g_in(x1, x2)
U3(x1, x2, x3, x4, x5)  =  U3(x5)
g_out(x1, x2, x3)  =  g_out(x3)
f_out(x1, x2, x3)  =  f_out(x3)
F_IN(x1, x2, x3)  =  F_IN(x1, x2)

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

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

F_IN(.(A, As), .(B, Bs), RES) → F_IN(.(B, .(A, As)), Bs, RES)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
F_IN(x1, x2, x3)  =  F_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
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
QDP
                        ↳ QDPSizeChangeProof

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

F_IN(.(A, As), .(B, Bs)) → F_IN(.(B, .(A, As)), Bs)

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: