Left Termination of the query pattern member_in_2(a, g) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

member(X, .(Y, Xs)) :- member(X, Xs).
member(X, .(X, Xs)).

Queries:

member(a,g).

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

member_in_ag(X, .(Y, Xs)) → U1_ag(X, Y, Xs, member_in_ag(X, Xs))
member_in_ag(X, .(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(X, Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))

The argument filtering Pi contains the following mapping:
member_in_ag(x1, x2)  =  member_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x4)
member_out_ag(x1, x2)  =  member_out_ag(x1)

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:

member_in_ag(X, .(Y, Xs)) → U1_ag(X, Y, Xs, member_in_ag(X, Xs))
member_in_ag(X, .(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(X, Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))

The argument filtering Pi contains the following mapping:
member_in_ag(x1, x2)  =  member_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x4)
member_out_ag(x1, x2)  =  member_out_ag(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:

MEMBER_IN_AG(X, .(Y, Xs)) → U1_AG(X, Y, Xs, member_in_ag(X, Xs))
MEMBER_IN_AG(X, .(Y, Xs)) → MEMBER_IN_AG(X, Xs)

The TRS R consists of the following rules:

member_in_ag(X, .(Y, Xs)) → U1_ag(X, Y, Xs, member_in_ag(X, Xs))
member_in_ag(X, .(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(X, Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))

The argument filtering Pi contains the following mapping:
member_in_ag(x1, x2)  =  member_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x4)
member_out_ag(x1, x2)  =  member_out_ag(x1)
MEMBER_IN_AG(x1, x2)  =  MEMBER_IN_AG(x2)
U1_AG(x1, x2, x3, x4)  =  U1_AG(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:

MEMBER_IN_AG(X, .(Y, Xs)) → U1_AG(X, Y, Xs, member_in_ag(X, Xs))
MEMBER_IN_AG(X, .(Y, Xs)) → MEMBER_IN_AG(X, Xs)

The TRS R consists of the following rules:

member_in_ag(X, .(Y, Xs)) → U1_ag(X, Y, Xs, member_in_ag(X, Xs))
member_in_ag(X, .(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(X, Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))

The argument filtering Pi contains the following mapping:
member_in_ag(x1, x2)  =  member_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x4)
member_out_ag(x1, x2)  =  member_out_ag(x1)
MEMBER_IN_AG(x1, x2)  =  MEMBER_IN_AG(x2)
U1_AG(x1, x2, x3, x4)  =  U1_AG(x4)

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

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

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

MEMBER_IN_AG(X, .(Y, Xs)) → MEMBER_IN_AG(X, Xs)

The TRS R consists of the following rules:

member_in_ag(X, .(Y, Xs)) → U1_ag(X, Y, Xs, member_in_ag(X, Xs))
member_in_ag(X, .(X, Xs)) → member_out_ag(X, .(X, Xs))
U1_ag(X, Y, Xs, member_out_ag(X, Xs)) → member_out_ag(X, .(Y, Xs))

The argument filtering Pi contains the following mapping:
member_in_ag(x1, x2)  =  member_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U1_ag(x1, x2, x3, x4)  =  U1_ag(x4)
member_out_ag(x1, x2)  =  member_out_ag(x1)
MEMBER_IN_AG(x1, x2)  =  MEMBER_IN_AG(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:

MEMBER_IN_AG(X, .(Y, Xs)) → MEMBER_IN_AG(X, Xs)

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

MEMBER_IN_AG(.(Y, Xs)) → MEMBER_IN_AG(Xs)

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: