Left Termination of the query pattern overlap_in_2(g, g) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

overlap(Xs, Ys) :- ','(member2(X, Xs), member1(X, Ys)).
has_a_or_b(Xs) :- overlap(Xs, .(a, .(b, []))).
member1(X, .(Y, Xs)) :- member1(X, Xs).
member1(X, .(X, Xs)).
member2(X, .(Y, Xs)) :- member2(X, Xs).
member2(X, .(X, Xs)).

Queries:

overlap(g,g).

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:

overlap_in(Xs, Ys) → U1(Xs, Ys, member2_in(X, Xs))
member2_in(X, .(X, Xs)) → member2_out(X, .(X, Xs))
member2_in(X, .(Y, Xs)) → U5(X, Y, Xs, member2_in(X, Xs))
U5(X, Y, Xs, member2_out(X, Xs)) → member2_out(X, .(Y, Xs))
U1(Xs, Ys, member2_out(X, Xs)) → U2(Xs, Ys, member1_in(X, Ys))
member1_in(X, .(X, Xs)) → member1_out(X, .(X, Xs))
member1_in(X, .(Y, Xs)) → U4(X, Y, Xs, member1_in(X, Xs))
U4(X, Y, Xs, member1_out(X, Xs)) → member1_out(X, .(Y, Xs))
U2(Xs, Ys, member1_out(X, Ys)) → overlap_out(Xs, Ys)

The argument filtering Pi contains the following mapping:
overlap_in(x1, x2)  =  overlap_in(x1, x2)
U1(x1, x2, x3)  =  U1(x2, x3)
member2_in(x1, x2)  =  member2_in(x2)
.(x1, x2)  =  .(x1, x2)
member2_out(x1, x2)  =  member2_out(x1)
U5(x1, x2, x3, x4)  =  U5(x4)
U2(x1, x2, x3)  =  U2(x3)
member1_in(x1, x2)  =  member1_in(x1, x2)
member1_out(x1, x2)  =  member1_out
U4(x1, x2, x3, x4)  =  U4(x4)
overlap_out(x1, x2)  =  overlap_out

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:

overlap_in(Xs, Ys) → U1(Xs, Ys, member2_in(X, Xs))
member2_in(X, .(X, Xs)) → member2_out(X, .(X, Xs))
member2_in(X, .(Y, Xs)) → U5(X, Y, Xs, member2_in(X, Xs))
U5(X, Y, Xs, member2_out(X, Xs)) → member2_out(X, .(Y, Xs))
U1(Xs, Ys, member2_out(X, Xs)) → U2(Xs, Ys, member1_in(X, Ys))
member1_in(X, .(X, Xs)) → member1_out(X, .(X, Xs))
member1_in(X, .(Y, Xs)) → U4(X, Y, Xs, member1_in(X, Xs))
U4(X, Y, Xs, member1_out(X, Xs)) → member1_out(X, .(Y, Xs))
U2(Xs, Ys, member1_out(X, Ys)) → overlap_out(Xs, Ys)

The argument filtering Pi contains the following mapping:
overlap_in(x1, x2)  =  overlap_in(x1, x2)
U1(x1, x2, x3)  =  U1(x2, x3)
member2_in(x1, x2)  =  member2_in(x2)
.(x1, x2)  =  .(x1, x2)
member2_out(x1, x2)  =  member2_out(x1)
U5(x1, x2, x3, x4)  =  U5(x4)
U2(x1, x2, x3)  =  U2(x3)
member1_in(x1, x2)  =  member1_in(x1, x2)
member1_out(x1, x2)  =  member1_out
U4(x1, x2, x3, x4)  =  U4(x4)
overlap_out(x1, x2)  =  overlap_out


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

OVERLAP_IN(Xs, Ys) → U11(Xs, Ys, member2_in(X, Xs))
OVERLAP_IN(Xs, Ys) → MEMBER2_IN(X, Xs)
MEMBER2_IN(X, .(Y, Xs)) → U51(X, Y, Xs, member2_in(X, Xs))
MEMBER2_IN(X, .(Y, Xs)) → MEMBER2_IN(X, Xs)
U11(Xs, Ys, member2_out(X, Xs)) → U21(Xs, Ys, member1_in(X, Ys))
U11(Xs, Ys, member2_out(X, Xs)) → MEMBER1_IN(X, Ys)
MEMBER1_IN(X, .(Y, Xs)) → U41(X, Y, Xs, member1_in(X, Xs))
MEMBER1_IN(X, .(Y, Xs)) → MEMBER1_IN(X, Xs)

The TRS R consists of the following rules:

overlap_in(Xs, Ys) → U1(Xs, Ys, member2_in(X, Xs))
member2_in(X, .(X, Xs)) → member2_out(X, .(X, Xs))
member2_in(X, .(Y, Xs)) → U5(X, Y, Xs, member2_in(X, Xs))
U5(X, Y, Xs, member2_out(X, Xs)) → member2_out(X, .(Y, Xs))
U1(Xs, Ys, member2_out(X, Xs)) → U2(Xs, Ys, member1_in(X, Ys))
member1_in(X, .(X, Xs)) → member1_out(X, .(X, Xs))
member1_in(X, .(Y, Xs)) → U4(X, Y, Xs, member1_in(X, Xs))
U4(X, Y, Xs, member1_out(X, Xs)) → member1_out(X, .(Y, Xs))
U2(Xs, Ys, member1_out(X, Ys)) → overlap_out(Xs, Ys)

The argument filtering Pi contains the following mapping:
overlap_in(x1, x2)  =  overlap_in(x1, x2)
U1(x1, x2, x3)  =  U1(x2, x3)
member2_in(x1, x2)  =  member2_in(x2)
.(x1, x2)  =  .(x1, x2)
member2_out(x1, x2)  =  member2_out(x1)
U5(x1, x2, x3, x4)  =  U5(x4)
U2(x1, x2, x3)  =  U2(x3)
member1_in(x1, x2)  =  member1_in(x1, x2)
member1_out(x1, x2)  =  member1_out
U4(x1, x2, x3, x4)  =  U4(x4)
overlap_out(x1, x2)  =  overlap_out
U51(x1, x2, x3, x4)  =  U51(x4)
U41(x1, x2, x3, x4)  =  U41(x4)
MEMBER1_IN(x1, x2)  =  MEMBER1_IN(x1, x2)
MEMBER2_IN(x1, x2)  =  MEMBER2_IN(x2)
U21(x1, x2, x3)  =  U21(x3)
OVERLAP_IN(x1, x2)  =  OVERLAP_IN(x1, x2)
U11(x1, x2, x3)  =  U11(x2, 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:

OVERLAP_IN(Xs, Ys) → U11(Xs, Ys, member2_in(X, Xs))
OVERLAP_IN(Xs, Ys) → MEMBER2_IN(X, Xs)
MEMBER2_IN(X, .(Y, Xs)) → U51(X, Y, Xs, member2_in(X, Xs))
MEMBER2_IN(X, .(Y, Xs)) → MEMBER2_IN(X, Xs)
U11(Xs, Ys, member2_out(X, Xs)) → U21(Xs, Ys, member1_in(X, Ys))
U11(Xs, Ys, member2_out(X, Xs)) → MEMBER1_IN(X, Ys)
MEMBER1_IN(X, .(Y, Xs)) → U41(X, Y, Xs, member1_in(X, Xs))
MEMBER1_IN(X, .(Y, Xs)) → MEMBER1_IN(X, Xs)

The TRS R consists of the following rules:

overlap_in(Xs, Ys) → U1(Xs, Ys, member2_in(X, Xs))
member2_in(X, .(X, Xs)) → member2_out(X, .(X, Xs))
member2_in(X, .(Y, Xs)) → U5(X, Y, Xs, member2_in(X, Xs))
U5(X, Y, Xs, member2_out(X, Xs)) → member2_out(X, .(Y, Xs))
U1(Xs, Ys, member2_out(X, Xs)) → U2(Xs, Ys, member1_in(X, Ys))
member1_in(X, .(X, Xs)) → member1_out(X, .(X, Xs))
member1_in(X, .(Y, Xs)) → U4(X, Y, Xs, member1_in(X, Xs))
U4(X, Y, Xs, member1_out(X, Xs)) → member1_out(X, .(Y, Xs))
U2(Xs, Ys, member1_out(X, Ys)) → overlap_out(Xs, Ys)

The argument filtering Pi contains the following mapping:
overlap_in(x1, x2)  =  overlap_in(x1, x2)
U1(x1, x2, x3)  =  U1(x2, x3)
member2_in(x1, x2)  =  member2_in(x2)
.(x1, x2)  =  .(x1, x2)
member2_out(x1, x2)  =  member2_out(x1)
U5(x1, x2, x3, x4)  =  U5(x4)
U2(x1, x2, x3)  =  U2(x3)
member1_in(x1, x2)  =  member1_in(x1, x2)
member1_out(x1, x2)  =  member1_out
U4(x1, x2, x3, x4)  =  U4(x4)
overlap_out(x1, x2)  =  overlap_out
U51(x1, x2, x3, x4)  =  U51(x4)
U41(x1, x2, x3, x4)  =  U41(x4)
MEMBER1_IN(x1, x2)  =  MEMBER1_IN(x1, x2)
MEMBER2_IN(x1, x2)  =  MEMBER2_IN(x2)
U21(x1, x2, x3)  =  U21(x3)
OVERLAP_IN(x1, x2)  =  OVERLAP_IN(x1, x2)
U11(x1, x2, x3)  =  U11(x2, x3)

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

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

MEMBER1_IN(X, .(Y, Xs)) → MEMBER1_IN(X, Xs)

The TRS R consists of the following rules:

overlap_in(Xs, Ys) → U1(Xs, Ys, member2_in(X, Xs))
member2_in(X, .(X, Xs)) → member2_out(X, .(X, Xs))
member2_in(X, .(Y, Xs)) → U5(X, Y, Xs, member2_in(X, Xs))
U5(X, Y, Xs, member2_out(X, Xs)) → member2_out(X, .(Y, Xs))
U1(Xs, Ys, member2_out(X, Xs)) → U2(Xs, Ys, member1_in(X, Ys))
member1_in(X, .(X, Xs)) → member1_out(X, .(X, Xs))
member1_in(X, .(Y, Xs)) → U4(X, Y, Xs, member1_in(X, Xs))
U4(X, Y, Xs, member1_out(X, Xs)) → member1_out(X, .(Y, Xs))
U2(Xs, Ys, member1_out(X, Ys)) → overlap_out(Xs, Ys)

The argument filtering Pi contains the following mapping:
overlap_in(x1, x2)  =  overlap_in(x1, x2)
U1(x1, x2, x3)  =  U1(x2, x3)
member2_in(x1, x2)  =  member2_in(x2)
.(x1, x2)  =  .(x1, x2)
member2_out(x1, x2)  =  member2_out(x1)
U5(x1, x2, x3, x4)  =  U5(x4)
U2(x1, x2, x3)  =  U2(x3)
member1_in(x1, x2)  =  member1_in(x1, x2)
member1_out(x1, x2)  =  member1_out
U4(x1, x2, x3, x4)  =  U4(x4)
overlap_out(x1, x2)  =  overlap_out
MEMBER1_IN(x1, x2)  =  MEMBER1_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:

MEMBER1_IN(X, .(Y, Xs)) → MEMBER1_IN(X, Xs)

R is empty.
Pi is empty.
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:

MEMBER1_IN(X, .(Y, Xs)) → MEMBER1_IN(X, 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: 



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

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

MEMBER2_IN(X, .(Y, Xs)) → MEMBER2_IN(X, Xs)

The TRS R consists of the following rules:

overlap_in(Xs, Ys) → U1(Xs, Ys, member2_in(X, Xs))
member2_in(X, .(X, Xs)) → member2_out(X, .(X, Xs))
member2_in(X, .(Y, Xs)) → U5(X, Y, Xs, member2_in(X, Xs))
U5(X, Y, Xs, member2_out(X, Xs)) → member2_out(X, .(Y, Xs))
U1(Xs, Ys, member2_out(X, Xs)) → U2(Xs, Ys, member1_in(X, Ys))
member1_in(X, .(X, Xs)) → member1_out(X, .(X, Xs))
member1_in(X, .(Y, Xs)) → U4(X, Y, Xs, member1_in(X, Xs))
U4(X, Y, Xs, member1_out(X, Xs)) → member1_out(X, .(Y, Xs))
U2(Xs, Ys, member1_out(X, Ys)) → overlap_out(Xs, Ys)

The argument filtering Pi contains the following mapping:
overlap_in(x1, x2)  =  overlap_in(x1, x2)
U1(x1, x2, x3)  =  U1(x2, x3)
member2_in(x1, x2)  =  member2_in(x2)
.(x1, x2)  =  .(x1, x2)
member2_out(x1, x2)  =  member2_out(x1)
U5(x1, x2, x3, x4)  =  U5(x4)
U2(x1, x2, x3)  =  U2(x3)
member1_in(x1, x2)  =  member1_in(x1, x2)
member1_out(x1, x2)  =  member1_out
U4(x1, x2, x3, x4)  =  U4(x4)
overlap_out(x1, x2)  =  overlap_out
MEMBER2_IN(x1, x2)  =  MEMBER2_IN(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:

MEMBER2_IN(X, .(Y, Xs)) → MEMBER2_IN(X, Xs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
MEMBER2_IN(x1, x2)  =  MEMBER2_IN(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:

MEMBER2_IN(.(Y, Xs)) → MEMBER2_IN(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: