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(RES, [], RES).
f([], .(Head, Tail), RES) :- f(.(Head, Tail), Tail, RES).
f(.(Head, Tail), Y, RES) :- f(Y, Tail, 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(.(Head, Tail), Y, RES) → U2(Head, Tail, Y, RES, f_in(Y, Tail, RES))
f_in([], .(Head, Tail), RES) → U1(Head, Tail, RES, f_in(.(Head, Tail), Tail, RES))
f_in(RES, [], RES) → f_out(RES, [], RES)
U1(Head, Tail, RES, f_out(.(Head, Tail), Tail, RES)) → f_out([], .(Head, Tail), RES)
U2(Head, Tail, Y, RES, f_out(Y, Tail, RES)) → f_out(.(Head, Tail), Y, 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)  =  U2(x5)
[]  =  []
U1(x1, x2, x3, x4)  =  U1(x4)
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(.(Head, Tail), Y, RES) → U2(Head, Tail, Y, RES, f_in(Y, Tail, RES))
f_in([], .(Head, Tail), RES) → U1(Head, Tail, RES, f_in(.(Head, Tail), Tail, RES))
f_in(RES, [], RES) → f_out(RES, [], RES)
U1(Head, Tail, RES, f_out(.(Head, Tail), Tail, RES)) → f_out([], .(Head, Tail), RES)
U2(Head, Tail, Y, RES, f_out(Y, Tail, RES)) → f_out(.(Head, Tail), Y, 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)  =  U2(x5)
[]  =  []
U1(x1, x2, x3, x4)  =  U1(x4)
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(.(Head, Tail), Y, RES) → U21(Head, Tail, Y, RES, f_in(Y, Tail, RES))
F_IN(.(Head, Tail), Y, RES) → F_IN(Y, Tail, RES)
F_IN([], .(Head, Tail), RES) → U11(Head, Tail, RES, f_in(.(Head, Tail), Tail, RES))
F_IN([], .(Head, Tail), RES) → F_IN(.(Head, Tail), Tail, RES)

The TRS R consists of the following rules:

f_in(.(Head, Tail), Y, RES) → U2(Head, Tail, Y, RES, f_in(Y, Tail, RES))
f_in([], .(Head, Tail), RES) → U1(Head, Tail, RES, f_in(.(Head, Tail), Tail, RES))
f_in(RES, [], RES) → f_out(RES, [], RES)
U1(Head, Tail, RES, f_out(.(Head, Tail), Tail, RES)) → f_out([], .(Head, Tail), RES)
U2(Head, Tail, Y, RES, f_out(Y, Tail, RES)) → f_out(.(Head, Tail), Y, 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)  =  U2(x5)
[]  =  []
U1(x1, x2, x3, x4)  =  U1(x4)
f_out(x1, x2, x3)  =  f_out(x3)
F_IN(x1, x2, x3)  =  F_IN(x1, x2)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3, x4)  =  U11(x4)

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(.(Head, Tail), Y, RES) → U21(Head, Tail, Y, RES, f_in(Y, Tail, RES))
F_IN(.(Head, Tail), Y, RES) → F_IN(Y, Tail, RES)
F_IN([], .(Head, Tail), RES) → U11(Head, Tail, RES, f_in(.(Head, Tail), Tail, RES))
F_IN([], .(Head, Tail), RES) → F_IN(.(Head, Tail), Tail, RES)

The TRS R consists of the following rules:

f_in(.(Head, Tail), Y, RES) → U2(Head, Tail, Y, RES, f_in(Y, Tail, RES))
f_in([], .(Head, Tail), RES) → U1(Head, Tail, RES, f_in(.(Head, Tail), Tail, RES))
f_in(RES, [], RES) → f_out(RES, [], RES)
U1(Head, Tail, RES, f_out(.(Head, Tail), Tail, RES)) → f_out([], .(Head, Tail), RES)
U2(Head, Tail, Y, RES, f_out(Y, Tail, RES)) → f_out(.(Head, Tail), Y, 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)  =  U2(x5)
[]  =  []
U1(x1, x2, x3, x4)  =  U1(x4)
f_out(x1, x2, x3)  =  f_out(x3)
F_IN(x1, x2, x3)  =  F_IN(x1, x2)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3, x4)  =  U11(x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 2 less nodes.

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

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

F_IN([], .(Head, Tail), RES) → F_IN(.(Head, Tail), Tail, RES)
F_IN(.(Head, Tail), Y, RES) → F_IN(Y, Tail, RES)

The TRS R consists of the following rules:

f_in(.(Head, Tail), Y, RES) → U2(Head, Tail, Y, RES, f_in(Y, Tail, RES))
f_in([], .(Head, Tail), RES) → U1(Head, Tail, RES, f_in(.(Head, Tail), Tail, RES))
f_in(RES, [], RES) → f_out(RES, [], RES)
U1(Head, Tail, RES, f_out(.(Head, Tail), Tail, RES)) → f_out([], .(Head, Tail), RES)
U2(Head, Tail, Y, RES, f_out(Y, Tail, RES)) → f_out(.(Head, Tail), Y, 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)  =  U2(x5)
[]  =  []
U1(x1, x2, x3, x4)  =  U1(x4)
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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof

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

F_IN([], .(Head, Tail), RES) → F_IN(.(Head, Tail), Tail, RES)
F_IN(.(Head, Tail), Y, RES) → F_IN(Y, Tail, 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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ QDPSizeChangeProof

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

F_IN(.(Head, Tail), Y) → F_IN(Y, Tail)
F_IN([], .(Head, Tail)) → F_IN(.(Head, Tail), Tail)

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: