Left Termination of the query pattern rev_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

rev(LS, RES) :- r1(LS, [], RES).
r1([], RES, RES).
r1(.(X, Xs), Accm, RES) :- r1(Xs, .(X, Accm), RES).

Queries:

rev(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:

rev_in(LS, RES) → U1(LS, RES, r1_in(LS, [], RES))
r1_in(.(X, Xs), Accm, RES) → U2(X, Xs, Accm, RES, r1_in(Xs, .(X, Accm), RES))
r1_in([], RES, RES) → r1_out([], RES, RES)
U2(X, Xs, Accm, RES, r1_out(Xs, .(X, Accm), RES)) → r1_out(.(X, Xs), Accm, RES)
U1(LS, RES, r1_out(LS, [], RES)) → rev_out(LS, RES)

The argument filtering Pi contains the following mapping:
rev_in(x1, x2)  =  rev_in(x1)
U1(x1, x2, x3)  =  U1(x3)
r1_in(x1, x2, x3)  =  r1_in(x1, x2)
[]  =  []
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
r1_out(x1, x2, x3)  =  r1_out(x3)
rev_out(x1, x2)  =  rev_out(x2)

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:

rev_in(LS, RES) → U1(LS, RES, r1_in(LS, [], RES))
r1_in(.(X, Xs), Accm, RES) → U2(X, Xs, Accm, RES, r1_in(Xs, .(X, Accm), RES))
r1_in([], RES, RES) → r1_out([], RES, RES)
U2(X, Xs, Accm, RES, r1_out(Xs, .(X, Accm), RES)) → r1_out(.(X, Xs), Accm, RES)
U1(LS, RES, r1_out(LS, [], RES)) → rev_out(LS, RES)

The argument filtering Pi contains the following mapping:
rev_in(x1, x2)  =  rev_in(x1)
U1(x1, x2, x3)  =  U1(x3)
r1_in(x1, x2, x3)  =  r1_in(x1, x2)
[]  =  []
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
r1_out(x1, x2, x3)  =  r1_out(x3)
rev_out(x1, x2)  =  rev_out(x2)


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

REV_IN(LS, RES) → U11(LS, RES, r1_in(LS, [], RES))
REV_IN(LS, RES) → R1_IN(LS, [], RES)
R1_IN(.(X, Xs), Accm, RES) → U21(X, Xs, Accm, RES, r1_in(Xs, .(X, Accm), RES))
R1_IN(.(X, Xs), Accm, RES) → R1_IN(Xs, .(X, Accm), RES)

The TRS R consists of the following rules:

rev_in(LS, RES) → U1(LS, RES, r1_in(LS, [], RES))
r1_in(.(X, Xs), Accm, RES) → U2(X, Xs, Accm, RES, r1_in(Xs, .(X, Accm), RES))
r1_in([], RES, RES) → r1_out([], RES, RES)
U2(X, Xs, Accm, RES, r1_out(Xs, .(X, Accm), RES)) → r1_out(.(X, Xs), Accm, RES)
U1(LS, RES, r1_out(LS, [], RES)) → rev_out(LS, RES)

The argument filtering Pi contains the following mapping:
rev_in(x1, x2)  =  rev_in(x1)
U1(x1, x2, x3)  =  U1(x3)
r1_in(x1, x2, x3)  =  r1_in(x1, x2)
[]  =  []
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
r1_out(x1, x2, x3)  =  r1_out(x3)
rev_out(x1, x2)  =  rev_out(x2)
REV_IN(x1, x2)  =  REV_IN(x1)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3)  =  U11(x3)
R1_IN(x1, x2, x3)  =  R1_IN(x1, x2)

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:

REV_IN(LS, RES) → U11(LS, RES, r1_in(LS, [], RES))
REV_IN(LS, RES) → R1_IN(LS, [], RES)
R1_IN(.(X, Xs), Accm, RES) → U21(X, Xs, Accm, RES, r1_in(Xs, .(X, Accm), RES))
R1_IN(.(X, Xs), Accm, RES) → R1_IN(Xs, .(X, Accm), RES)

The TRS R consists of the following rules:

rev_in(LS, RES) → U1(LS, RES, r1_in(LS, [], RES))
r1_in(.(X, Xs), Accm, RES) → U2(X, Xs, Accm, RES, r1_in(Xs, .(X, Accm), RES))
r1_in([], RES, RES) → r1_out([], RES, RES)
U2(X, Xs, Accm, RES, r1_out(Xs, .(X, Accm), RES)) → r1_out(.(X, Xs), Accm, RES)
U1(LS, RES, r1_out(LS, [], RES)) → rev_out(LS, RES)

The argument filtering Pi contains the following mapping:
rev_in(x1, x2)  =  rev_in(x1)
U1(x1, x2, x3)  =  U1(x3)
r1_in(x1, x2, x3)  =  r1_in(x1, x2)
[]  =  []
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
r1_out(x1, x2, x3)  =  r1_out(x3)
rev_out(x1, x2)  =  rev_out(x2)
REV_IN(x1, x2)  =  REV_IN(x1)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3)  =  U11(x3)
R1_IN(x1, x2, x3)  =  R1_IN(x1, x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 3 less nodes.

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

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

R1_IN(.(X, Xs), Accm, RES) → R1_IN(Xs, .(X, Accm), RES)

The TRS R consists of the following rules:

rev_in(LS, RES) → U1(LS, RES, r1_in(LS, [], RES))
r1_in(.(X, Xs), Accm, RES) → U2(X, Xs, Accm, RES, r1_in(Xs, .(X, Accm), RES))
r1_in([], RES, RES) → r1_out([], RES, RES)
U2(X, Xs, Accm, RES, r1_out(Xs, .(X, Accm), RES)) → r1_out(.(X, Xs), Accm, RES)
U1(LS, RES, r1_out(LS, [], RES)) → rev_out(LS, RES)

The argument filtering Pi contains the following mapping:
rev_in(x1, x2)  =  rev_in(x1)
U1(x1, x2, x3)  =  U1(x3)
r1_in(x1, x2, x3)  =  r1_in(x1, x2)
[]  =  []
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
r1_out(x1, x2, x3)  =  r1_out(x3)
rev_out(x1, x2)  =  rev_out(x2)
R1_IN(x1, x2, x3)  =  R1_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:

R1_IN(.(X, Xs), Accm, RES) → R1_IN(Xs, .(X, Accm), RES)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
R1_IN(x1, x2, x3)  =  R1_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:

R1_IN(.(X, Xs), Accm) → R1_IN(Xs, .(X, Accm))

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: