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

0 Prolog
↳1 PrologToPiTRSProof (⇒, 0 ms)
↳2 PiTRS
↳3 DependencyPairsProof (⇔, 13 ms)
↳4 PiDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 PiDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 PiDP
        ↳10 PiDPToQDPProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔, 0 ms)
        ↳13 YES
    ↳14 PiDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 PiDP
        ↳17 PiDPToQDPProof (⇒, 0 ms)
        ↳18 QDP
        ↳19 QDPSizeChangeProof (⇔, 0 ms)
        ↳20 YES

(0) Obligation:

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)).

Query: overlap(g,g)

(1) PrologToPiTRSProof (SOUND transformation)

We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
overlap_in: (b,b)
member2_in: (f,b)
member1_in: (b,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

overlap_in_gg(Xs, Ys) → U1_gg(Xs, Ys, member2_in_ag(X, Xs))
member2_in_ag(X, .(Y, Xs)) → U5_ag(X, Y, Xs, member2_in_ag(X, Xs))
member2_in_ag(X, .(X, Xs)) → member2_out_ag(X, .(X, Xs))
U5_ag(X, Y, Xs, member2_out_ag(X, Xs)) → member2_out_ag(X, .(Y, Xs))
U1_gg(Xs, Ys, member2_out_ag(X, Xs)) → U2_gg(Xs, Ys, member1_in_gg(X, Ys))
member1_in_gg(X, .(Y, Xs)) → U4_gg(X, Y, Xs, member1_in_gg(X, Xs))
member1_in_gg(X, .(X, Xs)) → member1_out_gg(X, .(X, Xs))
U4_gg(X, Y, Xs, member1_out_gg(X, Xs)) → member1_out_gg(X, .(Y, Xs))
U2_gg(Xs, Ys, member1_out_gg(X, Ys)) → overlap_out_gg(Xs, Ys)

The argument filtering Pi contains the following mapping:
overlap_in_gg(x1, x2)  =  overlap_in_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x2, x3)
member2_in_ag(x1, x2)  =  member2_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U5_ag(x1, x2, x3, x4)  =  U5_ag(x2, x3, x4)
member2_out_ag(x1, x2)  =  member2_out_ag(x1, x2)
U2_gg(x1, x2, x3)  =  U2_gg(x1, x2, x3)
member1_in_gg(x1, x2)  =  member1_in_gg(x1, x2)
U4_gg(x1, x2, x3, x4)  =  U4_gg(x1, x2, x3, x4)
member1_out_gg(x1, x2)  =  member1_out_gg(x1, x2)
overlap_out_gg(x1, x2)  =  overlap_out_gg(x1, x2)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(2) Obligation:

Pi-finite rewrite system:
The TRS R consists of the following rules:

overlap_in_gg(Xs, Ys) → U1_gg(Xs, Ys, member2_in_ag(X, Xs))
member2_in_ag(X, .(Y, Xs)) → U5_ag(X, Y, Xs, member2_in_ag(X, Xs))
member2_in_ag(X, .(X, Xs)) → member2_out_ag(X, .(X, Xs))
U5_ag(X, Y, Xs, member2_out_ag(X, Xs)) → member2_out_ag(X, .(Y, Xs))
U1_gg(Xs, Ys, member2_out_ag(X, Xs)) → U2_gg(Xs, Ys, member1_in_gg(X, Ys))
member1_in_gg(X, .(Y, Xs)) → U4_gg(X, Y, Xs, member1_in_gg(X, Xs))
member1_in_gg(X, .(X, Xs)) → member1_out_gg(X, .(X, Xs))
U4_gg(X, Y, Xs, member1_out_gg(X, Xs)) → member1_out_gg(X, .(Y, Xs))
U2_gg(Xs, Ys, member1_out_gg(X, Ys)) → overlap_out_gg(Xs, Ys)

The argument filtering Pi contains the following mapping:
overlap_in_gg(x1, x2)  =  overlap_in_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x2, x3)
member2_in_ag(x1, x2)  =  member2_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U5_ag(x1, x2, x3, x4)  =  U5_ag(x2, x3, x4)
member2_out_ag(x1, x2)  =  member2_out_ag(x1, x2)
U2_gg(x1, x2, x3)  =  U2_gg(x1, x2, x3)
member1_in_gg(x1, x2)  =  member1_in_gg(x1, x2)
U4_gg(x1, x2, x3, x4)  =  U4_gg(x1, x2, x3, x4)
member1_out_gg(x1, x2)  =  member1_out_gg(x1, x2)
overlap_out_gg(x1, x2)  =  overlap_out_gg(x1, x2)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

OVERLAP_IN_GG(Xs, Ys) → U1_GG(Xs, Ys, member2_in_ag(X, Xs))
OVERLAP_IN_GG(Xs, Ys) → MEMBER2_IN_AG(X, Xs)
MEMBER2_IN_AG(X, .(Y, Xs)) → U5_AG(X, Y, Xs, member2_in_ag(X, Xs))
MEMBER2_IN_AG(X, .(Y, Xs)) → MEMBER2_IN_AG(X, Xs)
U1_GG(Xs, Ys, member2_out_ag(X, Xs)) → U2_GG(Xs, Ys, member1_in_gg(X, Ys))
U1_GG(Xs, Ys, member2_out_ag(X, Xs)) → MEMBER1_IN_GG(X, Ys)
MEMBER1_IN_GG(X, .(Y, Xs)) → U4_GG(X, Y, Xs, member1_in_gg(X, Xs))
MEMBER1_IN_GG(X, .(Y, Xs)) → MEMBER1_IN_GG(X, Xs)

The TRS R consists of the following rules:

overlap_in_gg(Xs, Ys) → U1_gg(Xs, Ys, member2_in_ag(X, Xs))
member2_in_ag(X, .(Y, Xs)) → U5_ag(X, Y, Xs, member2_in_ag(X, Xs))
member2_in_ag(X, .(X, Xs)) → member2_out_ag(X, .(X, Xs))
U5_ag(X, Y, Xs, member2_out_ag(X, Xs)) → member2_out_ag(X, .(Y, Xs))
U1_gg(Xs, Ys, member2_out_ag(X, Xs)) → U2_gg(Xs, Ys, member1_in_gg(X, Ys))
member1_in_gg(X, .(Y, Xs)) → U4_gg(X, Y, Xs, member1_in_gg(X, Xs))
member1_in_gg(X, .(X, Xs)) → member1_out_gg(X, .(X, Xs))
U4_gg(X, Y, Xs, member1_out_gg(X, Xs)) → member1_out_gg(X, .(Y, Xs))
U2_gg(Xs, Ys, member1_out_gg(X, Ys)) → overlap_out_gg(Xs, Ys)

The argument filtering Pi contains the following mapping:
overlap_in_gg(x1, x2)  =  overlap_in_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x2, x3)
member2_in_ag(x1, x2)  =  member2_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U5_ag(x1, x2, x3, x4)  =  U5_ag(x2, x3, x4)
member2_out_ag(x1, x2)  =  member2_out_ag(x1, x2)
U2_gg(x1, x2, x3)  =  U2_gg(x1, x2, x3)
member1_in_gg(x1, x2)  =  member1_in_gg(x1, x2)
U4_gg(x1, x2, x3, x4)  =  U4_gg(x1, x2, x3, x4)
member1_out_gg(x1, x2)  =  member1_out_gg(x1, x2)
overlap_out_gg(x1, x2)  =  overlap_out_gg(x1, x2)
OVERLAP_IN_GG(x1, x2)  =  OVERLAP_IN_GG(x1, x2)
U1_GG(x1, x2, x3)  =  U1_GG(x1, x2, x3)
MEMBER2_IN_AG(x1, x2)  =  MEMBER2_IN_AG(x2)
U5_AG(x1, x2, x3, x4)  =  U5_AG(x2, x3, x4)
U2_GG(x1, x2, x3)  =  U2_GG(x1, x2, x3)
MEMBER1_IN_GG(x1, x2)  =  MEMBER1_IN_GG(x1, x2)
U4_GG(x1, x2, x3, x4)  =  U4_GG(x1, x2, x3, x4)

We have to consider all (P,R,Pi)-chains

(4) Obligation:

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

OVERLAP_IN_GG(Xs, Ys) → U1_GG(Xs, Ys, member2_in_ag(X, Xs))
OVERLAP_IN_GG(Xs, Ys) → MEMBER2_IN_AG(X, Xs)
MEMBER2_IN_AG(X, .(Y, Xs)) → U5_AG(X, Y, Xs, member2_in_ag(X, Xs))
MEMBER2_IN_AG(X, .(Y, Xs)) → MEMBER2_IN_AG(X, Xs)
U1_GG(Xs, Ys, member2_out_ag(X, Xs)) → U2_GG(Xs, Ys, member1_in_gg(X, Ys))
U1_GG(Xs, Ys, member2_out_ag(X, Xs)) → MEMBER1_IN_GG(X, Ys)
MEMBER1_IN_GG(X, .(Y, Xs)) → U4_GG(X, Y, Xs, member1_in_gg(X, Xs))
MEMBER1_IN_GG(X, .(Y, Xs)) → MEMBER1_IN_GG(X, Xs)

The TRS R consists of the following rules:

overlap_in_gg(Xs, Ys) → U1_gg(Xs, Ys, member2_in_ag(X, Xs))
member2_in_ag(X, .(Y, Xs)) → U5_ag(X, Y, Xs, member2_in_ag(X, Xs))
member2_in_ag(X, .(X, Xs)) → member2_out_ag(X, .(X, Xs))
U5_ag(X, Y, Xs, member2_out_ag(X, Xs)) → member2_out_ag(X, .(Y, Xs))
U1_gg(Xs, Ys, member2_out_ag(X, Xs)) → U2_gg(Xs, Ys, member1_in_gg(X, Ys))
member1_in_gg(X, .(Y, Xs)) → U4_gg(X, Y, Xs, member1_in_gg(X, Xs))
member1_in_gg(X, .(X, Xs)) → member1_out_gg(X, .(X, Xs))
U4_gg(X, Y, Xs, member1_out_gg(X, Xs)) → member1_out_gg(X, .(Y, Xs))
U2_gg(Xs, Ys, member1_out_gg(X, Ys)) → overlap_out_gg(Xs, Ys)

The argument filtering Pi contains the following mapping:
overlap_in_gg(x1, x2)  =  overlap_in_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x2, x3)
member2_in_ag(x1, x2)  =  member2_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U5_ag(x1, x2, x3, x4)  =  U5_ag(x2, x3, x4)
member2_out_ag(x1, x2)  =  member2_out_ag(x1, x2)
U2_gg(x1, x2, x3)  =  U2_gg(x1, x2, x3)
member1_in_gg(x1, x2)  =  member1_in_gg(x1, x2)
U4_gg(x1, x2, x3, x4)  =  U4_gg(x1, x2, x3, x4)
member1_out_gg(x1, x2)  =  member1_out_gg(x1, x2)
overlap_out_gg(x1, x2)  =  overlap_out_gg(x1, x2)
OVERLAP_IN_GG(x1, x2)  =  OVERLAP_IN_GG(x1, x2)
U1_GG(x1, x2, x3)  =  U1_GG(x1, x2, x3)
MEMBER2_IN_AG(x1, x2)  =  MEMBER2_IN_AG(x2)
U5_AG(x1, x2, x3, x4)  =  U5_AG(x2, x3, x4)
U2_GG(x1, x2, x3)  =  U2_GG(x1, x2, x3)
MEMBER1_IN_GG(x1, x2)  =  MEMBER1_IN_GG(x1, x2)
U4_GG(x1, x2, x3, x4)  =  U4_GG(x1, x2, x3, x4)

We have to consider all (P,R,Pi)-chains

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 6 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

MEMBER1_IN_GG(X, .(Y, Xs)) → MEMBER1_IN_GG(X, Xs)

The TRS R consists of the following rules:

overlap_in_gg(Xs, Ys) → U1_gg(Xs, Ys, member2_in_ag(X, Xs))
member2_in_ag(X, .(Y, Xs)) → U5_ag(X, Y, Xs, member2_in_ag(X, Xs))
member2_in_ag(X, .(X, Xs)) → member2_out_ag(X, .(X, Xs))
U5_ag(X, Y, Xs, member2_out_ag(X, Xs)) → member2_out_ag(X, .(Y, Xs))
U1_gg(Xs, Ys, member2_out_ag(X, Xs)) → U2_gg(Xs, Ys, member1_in_gg(X, Ys))
member1_in_gg(X, .(Y, Xs)) → U4_gg(X, Y, Xs, member1_in_gg(X, Xs))
member1_in_gg(X, .(X, Xs)) → member1_out_gg(X, .(X, Xs))
U4_gg(X, Y, Xs, member1_out_gg(X, Xs)) → member1_out_gg(X, .(Y, Xs))
U2_gg(Xs, Ys, member1_out_gg(X, Ys)) → overlap_out_gg(Xs, Ys)

The argument filtering Pi contains the following mapping:
overlap_in_gg(x1, x2)  =  overlap_in_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x2, x3)
member2_in_ag(x1, x2)  =  member2_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U5_ag(x1, x2, x3, x4)  =  U5_ag(x2, x3, x4)
member2_out_ag(x1, x2)  =  member2_out_ag(x1, x2)
U2_gg(x1, x2, x3)  =  U2_gg(x1, x2, x3)
member1_in_gg(x1, x2)  =  member1_in_gg(x1, x2)
U4_gg(x1, x2, x3, x4)  =  U4_gg(x1, x2, x3, x4)
member1_out_gg(x1, x2)  =  member1_out_gg(x1, x2)
overlap_out_gg(x1, x2)  =  overlap_out_gg(x1, x2)
MEMBER1_IN_GG(x1, x2)  =  MEMBER1_IN_GG(x1, x2)

We have to consider all (P,R,Pi)-chains

(8) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(9) Obligation:

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

MEMBER1_IN_GG(X, .(Y, Xs)) → MEMBER1_IN_GG(X, Xs)

R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains

(10) PiDPToQDPProof (EQUIVALENT transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(11) Obligation:

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

MEMBER1_IN_GG(X, .(Y, Xs)) → MEMBER1_IN_GG(X, Xs)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • MEMBER1_IN_GG(X, .(Y, Xs)) → MEMBER1_IN_GG(X, Xs)
    The graph contains the following edges 1 >= 1, 2 > 2

(13) YES

(14) Obligation:

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

MEMBER2_IN_AG(X, .(Y, Xs)) → MEMBER2_IN_AG(X, Xs)

The TRS R consists of the following rules:

overlap_in_gg(Xs, Ys) → U1_gg(Xs, Ys, member2_in_ag(X, Xs))
member2_in_ag(X, .(Y, Xs)) → U5_ag(X, Y, Xs, member2_in_ag(X, Xs))
member2_in_ag(X, .(X, Xs)) → member2_out_ag(X, .(X, Xs))
U5_ag(X, Y, Xs, member2_out_ag(X, Xs)) → member2_out_ag(X, .(Y, Xs))
U1_gg(Xs, Ys, member2_out_ag(X, Xs)) → U2_gg(Xs, Ys, member1_in_gg(X, Ys))
member1_in_gg(X, .(Y, Xs)) → U4_gg(X, Y, Xs, member1_in_gg(X, Xs))
member1_in_gg(X, .(X, Xs)) → member1_out_gg(X, .(X, Xs))
U4_gg(X, Y, Xs, member1_out_gg(X, Xs)) → member1_out_gg(X, .(Y, Xs))
U2_gg(Xs, Ys, member1_out_gg(X, Ys)) → overlap_out_gg(Xs, Ys)

The argument filtering Pi contains the following mapping:
overlap_in_gg(x1, x2)  =  overlap_in_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x2, x3)
member2_in_ag(x1, x2)  =  member2_in_ag(x2)
.(x1, x2)  =  .(x1, x2)
U5_ag(x1, x2, x3, x4)  =  U5_ag(x2, x3, x4)
member2_out_ag(x1, x2)  =  member2_out_ag(x1, x2)
U2_gg(x1, x2, x3)  =  U2_gg(x1, x2, x3)
member1_in_gg(x1, x2)  =  member1_in_gg(x1, x2)
U4_gg(x1, x2, x3, x4)  =  U4_gg(x1, x2, x3, x4)
member1_out_gg(x1, x2)  =  member1_out_gg(x1, x2)
overlap_out_gg(x1, x2)  =  overlap_out_gg(x1, x2)
MEMBER2_IN_AG(x1, x2)  =  MEMBER2_IN_AG(x2)

We have to consider all (P,R,Pi)-chains

(15) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(16) Obligation:

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

MEMBER2_IN_AG(X, .(Y, Xs)) → MEMBER2_IN_AG(X, Xs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
MEMBER2_IN_AG(x1, x2)  =  MEMBER2_IN_AG(x2)

We have to consider all (P,R,Pi)-chains

(17) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(18) Obligation:

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

MEMBER2_IN_AG(.(Y, Xs)) → MEMBER2_IN_AG(Xs)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(19) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • MEMBER2_IN_AG(.(Y, Xs)) → MEMBER2_IN_AG(Xs)
    The graph contains the following edges 1 > 1

(20) YES