Left Termination of the query pattern dis_in_1(g) w.r.t. the given Prolog program could successfully be proven:

0 Prolog
  ↳1 PrologToPiTRSProof (⇐)
    ↳2 PiTRS
    ↳3 DependencyPairsProof (⇔)
    ↳4 PiDP
    ↳5 DependencyGraphProof (⇔)
    ↳6 PiDP
    ↳7 PiDPToQDPProof (⇐)
    ↳8 QDP
  ↳9 PrologToPiTRSProof (⇐)
    ↳10 PiTRS
    ↳11 DependencyPairsProof (⇔)
    ↳12 PiDP
    ↳13 DependencyGraphProof (⇔)
    ↳14 PiDP
    ↳15 PiDPToQDPProof (⇔)
    ↳16 QDP
    ↳17 QDPSizeChangeProof (⇔)
    ↳18 TRUE

(0) Obligation:

Clauses:

dis(or(B1, B2)) :- ','(con(B1), dis(B2)).
dis(B) :- con(B).
con(and(B1, B2)) :- ','(dis(B1), con(B2)).
con(B) :- bool(B).
bool(0).
bool(1).

Queries:

dis(g).

(1) PrologToPiTRSProof (SOUND transformation)

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

dis_in_g(or(B1, B2)) → U1_g(B1, B2, con_in_g(B1))
con_in_g(and(B1, B2)) → U4_g(B1, B2, dis_in_g(B1))
dis_in_g(B) → U3_g(B, con_in_g(B))
con_in_g(B) → U6_g(B, bool_in_g(B))
bool_in_g(0) → bool_out_g(0)
bool_in_g(1) → bool_out_g(1)
U6_g(B, bool_out_g(B)) → con_out_g(B)
U3_g(B, con_out_g(B)) → dis_out_g(B)
U4_g(B1, B2, dis_out_g(B1)) → U5_g(B1, B2, con_in_g(B2))
U5_g(B1, B2, con_out_g(B2)) → con_out_g(and(B1, B2))
U1_g(B1, B2, con_out_g(B1)) → U2_g(B1, B2, dis_in_g(B2))
U2_g(B1, B2, dis_out_g(B2)) → dis_out_g(or(B1, B2))

The argument filtering Pi contains the following mapping:
dis_in_g(x1)  =  dis_in_g(x1)
or(x1, x2)  =  or(x1, x2)
U1_g(x1, x2, x3)  =  U1_g(x2, x3)
con_in_g(x1)  =  con_in_g(x1)
and(x1, x2)  =  and(x1, x2)
U4_g(x1, x2, x3)  =  U4_g(x2, x3)
U3_g(x1, x2)  =  U3_g(x2)
U6_g(x1, x2)  =  U6_g(x2)
bool_in_g(x1)  =  bool_in_g(x1)
0  =  0
bool_out_g(x1)  =  bool_out_g
1  =  1
con_out_g(x1)  =  con_out_g
dis_out_g(x1)  =  dis_out_g
U5_g(x1, x2, x3)  =  U5_g(x3)
U2_g(x1, x2, x3)  =  U2_g(x3)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(2) Obligation:

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

dis_in_g(or(B1, B2)) → U1_g(B1, B2, con_in_g(B1))
con_in_g(and(B1, B2)) → U4_g(B1, B2, dis_in_g(B1))
dis_in_g(B) → U3_g(B, con_in_g(B))
con_in_g(B) → U6_g(B, bool_in_g(B))
bool_in_g(0) → bool_out_g(0)
bool_in_g(1) → bool_out_g(1)
U6_g(B, bool_out_g(B)) → con_out_g(B)
U3_g(B, con_out_g(B)) → dis_out_g(B)
U4_g(B1, B2, dis_out_g(B1)) → U5_g(B1, B2, con_in_g(B2))
U5_g(B1, B2, con_out_g(B2)) → con_out_g(and(B1, B2))
U1_g(B1, B2, con_out_g(B1)) → U2_g(B1, B2, dis_in_g(B2))
U2_g(B1, B2, dis_out_g(B2)) → dis_out_g(or(B1, B2))

The argument filtering Pi contains the following mapping:
dis_in_g(x1)  =  dis_in_g(x1)
or(x1, x2)  =  or(x1, x2)
U1_g(x1, x2, x3)  =  U1_g(x2, x3)
con_in_g(x1)  =  con_in_g(x1)
and(x1, x2)  =  and(x1, x2)
U4_g(x1, x2, x3)  =  U4_g(x2, x3)
U3_g(x1, x2)  =  U3_g(x2)
U6_g(x1, x2)  =  U6_g(x2)
bool_in_g(x1)  =  bool_in_g(x1)
0  =  0
bool_out_g(x1)  =  bool_out_g
1  =  1
con_out_g(x1)  =  con_out_g
dis_out_g(x1)  =  dis_out_g
U5_g(x1, x2, x3)  =  U5_g(x3)
U2_g(x1, x2, x3)  =  U2_g(x3)

(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:

DIS_IN_G(or(B1, B2)) → U1_G(B1, B2, con_in_g(B1))
DIS_IN_G(or(B1, B2)) → CON_IN_G(B1)
CON_IN_G(and(B1, B2)) → U4_G(B1, B2, dis_in_g(B1))
CON_IN_G(and(B1, B2)) → DIS_IN_G(B1)
DIS_IN_G(B) → U3_G(B, con_in_g(B))
DIS_IN_G(B) → CON_IN_G(B)
CON_IN_G(B) → U6_G(B, bool_in_g(B))
CON_IN_G(B) → BOOL_IN_G(B)
U4_G(B1, B2, dis_out_g(B1)) → U5_G(B1, B2, con_in_g(B2))
U4_G(B1, B2, dis_out_g(B1)) → CON_IN_G(B2)
U1_G(B1, B2, con_out_g(B1)) → U2_G(B1, B2, dis_in_g(B2))
U1_G(B1, B2, con_out_g(B1)) → DIS_IN_G(B2)

The TRS R consists of the following rules:

dis_in_g(or(B1, B2)) → U1_g(B1, B2, con_in_g(B1))
con_in_g(and(B1, B2)) → U4_g(B1, B2, dis_in_g(B1))
dis_in_g(B) → U3_g(B, con_in_g(B))
con_in_g(B) → U6_g(B, bool_in_g(B))
bool_in_g(0) → bool_out_g(0)
bool_in_g(1) → bool_out_g(1)
U6_g(B, bool_out_g(B)) → con_out_g(B)
U3_g(B, con_out_g(B)) → dis_out_g(B)
U4_g(B1, B2, dis_out_g(B1)) → U5_g(B1, B2, con_in_g(B2))
U5_g(B1, B2, con_out_g(B2)) → con_out_g(and(B1, B2))
U1_g(B1, B2, con_out_g(B1)) → U2_g(B1, B2, dis_in_g(B2))
U2_g(B1, B2, dis_out_g(B2)) → dis_out_g(or(B1, B2))

The argument filtering Pi contains the following mapping:
dis_in_g(x1)  =  dis_in_g(x1)
or(x1, x2)  =  or(x1, x2)
U1_g(x1, x2, x3)  =  U1_g(x2, x3)
con_in_g(x1)  =  con_in_g(x1)
and(x1, x2)  =  and(x1, x2)
U4_g(x1, x2, x3)  =  U4_g(x2, x3)
U3_g(x1, x2)  =  U3_g(x2)
U6_g(x1, x2)  =  U6_g(x2)
bool_in_g(x1)  =  bool_in_g(x1)
0  =  0
bool_out_g(x1)  =  bool_out_g
1  =  1
con_out_g(x1)  =  con_out_g
dis_out_g(x1)  =  dis_out_g
U5_g(x1, x2, x3)  =  U5_g(x3)
U2_g(x1, x2, x3)  =  U2_g(x3)
DIS_IN_G(x1)  =  DIS_IN_G(x1)
U1_G(x1, x2, x3)  =  U1_G(x2, x3)
CON_IN_G(x1)  =  CON_IN_G(x1)
U4_G(x1, x2, x3)  =  U4_G(x2, x3)
U3_G(x1, x2)  =  U3_G(x2)
U6_G(x1, x2)  =  U6_G(x2)
BOOL_IN_G(x1)  =  BOOL_IN_G(x1)
U5_G(x1, x2, x3)  =  U5_G(x3)
U2_G(x1, x2, x3)  =  U2_G(x3)

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

(4) Obligation:

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

DIS_IN_G(or(B1, B2)) → U1_G(B1, B2, con_in_g(B1))
DIS_IN_G(or(B1, B2)) → CON_IN_G(B1)
CON_IN_G(and(B1, B2)) → U4_G(B1, B2, dis_in_g(B1))
CON_IN_G(and(B1, B2)) → DIS_IN_G(B1)
DIS_IN_G(B) → U3_G(B, con_in_g(B))
DIS_IN_G(B) → CON_IN_G(B)
CON_IN_G(B) → U6_G(B, bool_in_g(B))
CON_IN_G(B) → BOOL_IN_G(B)
U4_G(B1, B2, dis_out_g(B1)) → U5_G(B1, B2, con_in_g(B2))
U4_G(B1, B2, dis_out_g(B1)) → CON_IN_G(B2)
U1_G(B1, B2, con_out_g(B1)) → U2_G(B1, B2, dis_in_g(B2))
U1_G(B1, B2, con_out_g(B1)) → DIS_IN_G(B2)

The TRS R consists of the following rules:

dis_in_g(or(B1, B2)) → U1_g(B1, B2, con_in_g(B1))
con_in_g(and(B1, B2)) → U4_g(B1, B2, dis_in_g(B1))
dis_in_g(B) → U3_g(B, con_in_g(B))
con_in_g(B) → U6_g(B, bool_in_g(B))
bool_in_g(0) → bool_out_g(0)
bool_in_g(1) → bool_out_g(1)
U6_g(B, bool_out_g(B)) → con_out_g(B)
U3_g(B, con_out_g(B)) → dis_out_g(B)
U4_g(B1, B2, dis_out_g(B1)) → U5_g(B1, B2, con_in_g(B2))
U5_g(B1, B2, con_out_g(B2)) → con_out_g(and(B1, B2))
U1_g(B1, B2, con_out_g(B1)) → U2_g(B1, B2, dis_in_g(B2))
U2_g(B1, B2, dis_out_g(B2)) → dis_out_g(or(B1, B2))

The argument filtering Pi contains the following mapping:
dis_in_g(x1)  =  dis_in_g(x1)
or(x1, x2)  =  or(x1, x2)
U1_g(x1, x2, x3)  =  U1_g(x2, x3)
con_in_g(x1)  =  con_in_g(x1)
and(x1, x2)  =  and(x1, x2)
U4_g(x1, x2, x3)  =  U4_g(x2, x3)
U3_g(x1, x2)  =  U3_g(x2)
U6_g(x1, x2)  =  U6_g(x2)
bool_in_g(x1)  =  bool_in_g(x1)
0  =  0
bool_out_g(x1)  =  bool_out_g
1  =  1
con_out_g(x1)  =  con_out_g
dis_out_g(x1)  =  dis_out_g
U5_g(x1, x2, x3)  =  U5_g(x3)
U2_g(x1, x2, x3)  =  U2_g(x3)
DIS_IN_G(x1)  =  DIS_IN_G(x1)
U1_G(x1, x2, x3)  =  U1_G(x2, x3)
CON_IN_G(x1)  =  CON_IN_G(x1)
U4_G(x1, x2, x3)  =  U4_G(x2, x3)
U3_G(x1, x2)  =  U3_G(x2)
U6_G(x1, x2)  =  U6_G(x2)
BOOL_IN_G(x1)  =  BOOL_IN_G(x1)
U5_G(x1, x2, x3)  =  U5_G(x3)
U2_G(x1, x2, x3)  =  U2_G(x3)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 5 less nodes.

(6) Obligation:

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

U1_G(B1, B2, con_out_g(B1)) → DIS_IN_G(B2)
DIS_IN_G(or(B1, B2)) → U1_G(B1, B2, con_in_g(B1))
DIS_IN_G(or(B1, B2)) → CON_IN_G(B1)
CON_IN_G(and(B1, B2)) → U4_G(B1, B2, dis_in_g(B1))
U4_G(B1, B2, dis_out_g(B1)) → CON_IN_G(B2)
CON_IN_G(and(B1, B2)) → DIS_IN_G(B1)
DIS_IN_G(B) → CON_IN_G(B)

The TRS R consists of the following rules:

dis_in_g(or(B1, B2)) → U1_g(B1, B2, con_in_g(B1))
con_in_g(and(B1, B2)) → U4_g(B1, B2, dis_in_g(B1))
dis_in_g(B) → U3_g(B, con_in_g(B))
con_in_g(B) → U6_g(B, bool_in_g(B))
bool_in_g(0) → bool_out_g(0)
bool_in_g(1) → bool_out_g(1)
U6_g(B, bool_out_g(B)) → con_out_g(B)
U3_g(B, con_out_g(B)) → dis_out_g(B)
U4_g(B1, B2, dis_out_g(B1)) → U5_g(B1, B2, con_in_g(B2))
U5_g(B1, B2, con_out_g(B2)) → con_out_g(and(B1, B2))
U1_g(B1, B2, con_out_g(B1)) → U2_g(B1, B2, dis_in_g(B2))
U2_g(B1, B2, dis_out_g(B2)) → dis_out_g(or(B1, B2))

The argument filtering Pi contains the following mapping:
dis_in_g(x1)  =  dis_in_g(x1)
or(x1, x2)  =  or(x1, x2)
U1_g(x1, x2, x3)  =  U1_g(x2, x3)
con_in_g(x1)  =  con_in_g(x1)
and(x1, x2)  =  and(x1, x2)
U4_g(x1, x2, x3)  =  U4_g(x2, x3)
U3_g(x1, x2)  =  U3_g(x2)
U6_g(x1, x2)  =  U6_g(x2)
bool_in_g(x1)  =  bool_in_g(x1)
0  =  0
bool_out_g(x1)  =  bool_out_g
1  =  1
con_out_g(x1)  =  con_out_g
dis_out_g(x1)  =  dis_out_g
U5_g(x1, x2, x3)  =  U5_g(x3)
U2_g(x1, x2, x3)  =  U2_g(x3)
DIS_IN_G(x1)  =  DIS_IN_G(x1)
U1_G(x1, x2, x3)  =  U1_G(x2, x3)
CON_IN_G(x1)  =  CON_IN_G(x1)
U4_G(x1, x2, x3)  =  U4_G(x2, x3)

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

(7) PiDPToQDPProof (SOUND transformation)

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

(8) Obligation:

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

U1_G(B2, con_out_g) → DIS_IN_G(B2)
DIS_IN_G(or(B1, B2)) → U1_G(B2, con_in_g(B1))
DIS_IN_G(or(B1, B2)) → CON_IN_G(B1)
CON_IN_G(and(B1, B2)) → U4_G(B2, dis_in_g(B1))
U4_G(B2, dis_out_g) → CON_IN_G(B2)
CON_IN_G(and(B1, B2)) → DIS_IN_G(B1)
DIS_IN_G(B) → CON_IN_G(B)

The TRS R consists of the following rules:

dis_in_g(or(B1, B2)) → U1_g(B2, con_in_g(B1))
con_in_g(and(B1, B2)) → U4_g(B2, dis_in_g(B1))
dis_in_g(B) → U3_g(con_in_g(B))
con_in_g(B) → U6_g(bool_in_g(B))
bool_in_g(0) → bool_out_g
bool_in_g(1) → bool_out_g
U6_g(bool_out_g) → con_out_g
U3_g(con_out_g) → dis_out_g
U4_g(B2, dis_out_g) → U5_g(con_in_g(B2))
U5_g(con_out_g) → con_out_g
U1_g(B2, con_out_g) → U2_g(dis_in_g(B2))
U2_g(dis_out_g) → dis_out_g

The set Q consists of the following terms:

dis_in_g(x0)
con_in_g(x0)
bool_in_g(x0)
U6_g(x0)
U3_g(x0)
U4_g(x0, x1)
U5_g(x0)
U1_g(x0, x1)
U2_g(x0)

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

(9) PrologToPiTRSProof (SOUND transformation)

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

dis_in_g(or(B1, B2)) → U1_g(B1, B2, con_in_g(B1))
con_in_g(and(B1, B2)) → U4_g(B1, B2, dis_in_g(B1))
dis_in_g(B) → U3_g(B, con_in_g(B))
con_in_g(B) → U6_g(B, bool_in_g(B))
bool_in_g(0) → bool_out_g(0)
bool_in_g(1) → bool_out_g(1)
U6_g(B, bool_out_g(B)) → con_out_g(B)
U3_g(B, con_out_g(B)) → dis_out_g(B)
U4_g(B1, B2, dis_out_g(B1)) → U5_g(B1, B2, con_in_g(B2))
U5_g(B1, B2, con_out_g(B2)) → con_out_g(and(B1, B2))
U1_g(B1, B2, con_out_g(B1)) → U2_g(B1, B2, dis_in_g(B2))
U2_g(B1, B2, dis_out_g(B2)) → dis_out_g(or(B1, B2))

Pi is empty.

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(10) Obligation:

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

dis_in_g(or(B1, B2)) → U1_g(B1, B2, con_in_g(B1))
con_in_g(and(B1, B2)) → U4_g(B1, B2, dis_in_g(B1))
dis_in_g(B) → U3_g(B, con_in_g(B))
con_in_g(B) → U6_g(B, bool_in_g(B))
bool_in_g(0) → bool_out_g(0)
bool_in_g(1) → bool_out_g(1)
U6_g(B, bool_out_g(B)) → con_out_g(B)
U3_g(B, con_out_g(B)) → dis_out_g(B)
U4_g(B1, B2, dis_out_g(B1)) → U5_g(B1, B2, con_in_g(B2))
U5_g(B1, B2, con_out_g(B2)) → con_out_g(and(B1, B2))
U1_g(B1, B2, con_out_g(B1)) → U2_g(B1, B2, dis_in_g(B2))
U2_g(B1, B2, dis_out_g(B2)) → dis_out_g(or(B1, B2))

Pi is empty.

(11) 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:

DIS_IN_G(or(B1, B2)) → U1_G(B1, B2, con_in_g(B1))
DIS_IN_G(or(B1, B2)) → CON_IN_G(B1)
CON_IN_G(and(B1, B2)) → U4_G(B1, B2, dis_in_g(B1))
CON_IN_G(and(B1, B2)) → DIS_IN_G(B1)
DIS_IN_G(B) → U3_G(B, con_in_g(B))
DIS_IN_G(B) → CON_IN_G(B)
CON_IN_G(B) → U6_G(B, bool_in_g(B))
CON_IN_G(B) → BOOL_IN_G(B)
U4_G(B1, B2, dis_out_g(B1)) → U5_G(B1, B2, con_in_g(B2))
U4_G(B1, B2, dis_out_g(B1)) → CON_IN_G(B2)
U1_G(B1, B2, con_out_g(B1)) → U2_G(B1, B2, dis_in_g(B2))
U1_G(B1, B2, con_out_g(B1)) → DIS_IN_G(B2)

The TRS R consists of the following rules:

dis_in_g(or(B1, B2)) → U1_g(B1, B2, con_in_g(B1))
con_in_g(and(B1, B2)) → U4_g(B1, B2, dis_in_g(B1))
dis_in_g(B) → U3_g(B, con_in_g(B))
con_in_g(B) → U6_g(B, bool_in_g(B))
bool_in_g(0) → bool_out_g(0)
bool_in_g(1) → bool_out_g(1)
U6_g(B, bool_out_g(B)) → con_out_g(B)
U3_g(B, con_out_g(B)) → dis_out_g(B)
U4_g(B1, B2, dis_out_g(B1)) → U5_g(B1, B2, con_in_g(B2))
U5_g(B1, B2, con_out_g(B2)) → con_out_g(and(B1, B2))
U1_g(B1, B2, con_out_g(B1)) → U2_g(B1, B2, dis_in_g(B2))
U2_g(B1, B2, dis_out_g(B2)) → dis_out_g(or(B1, B2))

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

(12) Obligation:

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

DIS_IN_G(or(B1, B2)) → U1_G(B1, B2, con_in_g(B1))
DIS_IN_G(or(B1, B2)) → CON_IN_G(B1)
CON_IN_G(and(B1, B2)) → U4_G(B1, B2, dis_in_g(B1))
CON_IN_G(and(B1, B2)) → DIS_IN_G(B1)
DIS_IN_G(B) → U3_G(B, con_in_g(B))
DIS_IN_G(B) → CON_IN_G(B)
CON_IN_G(B) → U6_G(B, bool_in_g(B))
CON_IN_G(B) → BOOL_IN_G(B)
U4_G(B1, B2, dis_out_g(B1)) → U5_G(B1, B2, con_in_g(B2))
U4_G(B1, B2, dis_out_g(B1)) → CON_IN_G(B2)
U1_G(B1, B2, con_out_g(B1)) → U2_G(B1, B2, dis_in_g(B2))
U1_G(B1, B2, con_out_g(B1)) → DIS_IN_G(B2)

The TRS R consists of the following rules:

dis_in_g(or(B1, B2)) → U1_g(B1, B2, con_in_g(B1))
con_in_g(and(B1, B2)) → U4_g(B1, B2, dis_in_g(B1))
dis_in_g(B) → U3_g(B, con_in_g(B))
con_in_g(B) → U6_g(B, bool_in_g(B))
bool_in_g(0) → bool_out_g(0)
bool_in_g(1) → bool_out_g(1)
U6_g(B, bool_out_g(B)) → con_out_g(B)
U3_g(B, con_out_g(B)) → dis_out_g(B)
U4_g(B1, B2, dis_out_g(B1)) → U5_g(B1, B2, con_in_g(B2))
U5_g(B1, B2, con_out_g(B2)) → con_out_g(and(B1, B2))
U1_g(B1, B2, con_out_g(B1)) → U2_g(B1, B2, dis_in_g(B2))
U2_g(B1, B2, dis_out_g(B2)) → dis_out_g(or(B1, B2))

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

(13) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 5 less nodes.

(14) Obligation:

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

U1_G(B1, B2, con_out_g(B1)) → DIS_IN_G(B2)
DIS_IN_G(or(B1, B2)) → U1_G(B1, B2, con_in_g(B1))
DIS_IN_G(or(B1, B2)) → CON_IN_G(B1)
CON_IN_G(and(B1, B2)) → U4_G(B1, B2, dis_in_g(B1))
U4_G(B1, B2, dis_out_g(B1)) → CON_IN_G(B2)
CON_IN_G(and(B1, B2)) → DIS_IN_G(B1)
DIS_IN_G(B) → CON_IN_G(B)

The TRS R consists of the following rules:

dis_in_g(or(B1, B2)) → U1_g(B1, B2, con_in_g(B1))
con_in_g(and(B1, B2)) → U4_g(B1, B2, dis_in_g(B1))
dis_in_g(B) → U3_g(B, con_in_g(B))
con_in_g(B) → U6_g(B, bool_in_g(B))
bool_in_g(0) → bool_out_g(0)
bool_in_g(1) → bool_out_g(1)
U6_g(B, bool_out_g(B)) → con_out_g(B)
U3_g(B, con_out_g(B)) → dis_out_g(B)
U4_g(B1, B2, dis_out_g(B1)) → U5_g(B1, B2, con_in_g(B2))
U5_g(B1, B2, con_out_g(B2)) → con_out_g(and(B1, B2))
U1_g(B1, B2, con_out_g(B1)) → U2_g(B1, B2, dis_in_g(B2))
U2_g(B1, B2, dis_out_g(B2)) → dis_out_g(or(B1, B2))

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

(15) PiDPToQDPProof (EQUIVALENT transformation)

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

(16) Obligation:

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

U1_G(B1, B2, con_out_g(B1)) → DIS_IN_G(B2)
DIS_IN_G(or(B1, B2)) → U1_G(B1, B2, con_in_g(B1))
DIS_IN_G(or(B1, B2)) → CON_IN_G(B1)
CON_IN_G(and(B1, B2)) → U4_G(B1, B2, dis_in_g(B1))
U4_G(B1, B2, dis_out_g(B1)) → CON_IN_G(B2)
CON_IN_G(and(B1, B2)) → DIS_IN_G(B1)
DIS_IN_G(B) → CON_IN_G(B)

The TRS R consists of the following rules:

dis_in_g(or(B1, B2)) → U1_g(B1, B2, con_in_g(B1))
con_in_g(and(B1, B2)) → U4_g(B1, B2, dis_in_g(B1))
dis_in_g(B) → U3_g(B, con_in_g(B))
con_in_g(B) → U6_g(B, bool_in_g(B))
bool_in_g(0) → bool_out_g(0)
bool_in_g(1) → bool_out_g(1)
U6_g(B, bool_out_g(B)) → con_out_g(B)
U3_g(B, con_out_g(B)) → dis_out_g(B)
U4_g(B1, B2, dis_out_g(B1)) → U5_g(B1, B2, con_in_g(B2))
U5_g(B1, B2, con_out_g(B2)) → con_out_g(and(B1, B2))
U1_g(B1, B2, con_out_g(B1)) → U2_g(B1, B2, dis_in_g(B2))
U2_g(B1, B2, dis_out_g(B2)) → dis_out_g(or(B1, B2))

The set Q consists of the following terms:

dis_in_g(x0)
con_in_g(x0)
bool_in_g(x0)
U6_g(x0, x1)
U3_g(x0, x1)
U4_g(x0, x1, x2)
U5_g(x0, x1, x2)
U1_g(x0, x1, x2)
U2_g(x0, x1, x2)

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

(17) 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:

  • DIS_IN_G(or(B1, B2)) → U1_G(B1, B2, con_in_g(B1))
    The graph contains the following edges 1 > 1, 1 > 2

  • U1_G(B1, B2, con_out_g(B1)) → DIS_IN_G(B2)
    The graph contains the following edges 2 >= 1

  • CON_IN_G(and(B1, B2)) → DIS_IN_G(B1)
    The graph contains the following edges 1 > 1

  • CON_IN_G(and(B1, B2)) → U4_G(B1, B2, dis_in_g(B1))
    The graph contains the following edges 1 > 1, 1 > 2

  • U4_G(B1, B2, dis_out_g(B1)) → CON_IN_G(B2)
    The graph contains the following edges 2 >= 1

  • DIS_IN_G(or(B1, B2)) → CON_IN_G(B1)
    The graph contains the following edges 1 > 1

  • DIS_IN_G(B) → CON_IN_G(B)
    The graph contains the following edges 1 >= 1

(18) TRUE