(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).
Query: con(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:
con_in: (b)
dis_in: (b)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
con_in_g(and(B1, B2)) → U4_g(B1, B2, dis_in_g(B1))
dis_in_g(or(B1, B2)) → U1_g(B1, B2, con_in_g(B1))
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)
U1_g(B1, B2, con_out_g(B1)) → U2_g(B1, B2, dis_in_g(B2))
dis_in_g(B) → U3_g(B, con_in_g(B))
U3_g(B, con_out_g(B)) → dis_out_g(B)
U2_g(B1, B2, dis_out_g(B2)) → dis_out_g(or(B1, B2))
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))
The argument filtering Pi contains the following mapping:
con_in_g(
x1) =
con_in_g(
x1)
and(
x1,
x2) =
and(
x1,
x2)
U4_g(
x1,
x2,
x3) =
U4_g(
x2,
x3)
dis_in_g(
x1) =
dis_in_g(
x1)
or(
x1,
x2) =
or(
x1,
x2)
U1_g(
x1,
x2,
x3) =
U1_g(
x2,
x3)
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
U2_g(
x1,
x2,
x3) =
U2_g(
x3)
U3_g(
x1,
x2) =
U3_g(
x2)
dis_out_g(
x1) =
dis_out_g
U5_g(
x1,
x2,
x3) =
U5_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:
con_in_g(and(B1, B2)) → U4_g(B1, B2, dis_in_g(B1))
dis_in_g(or(B1, B2)) → U1_g(B1, B2, con_in_g(B1))
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)
U1_g(B1, B2, con_out_g(B1)) → U2_g(B1, B2, dis_in_g(B2))
dis_in_g(B) → U3_g(B, con_in_g(B))
U3_g(B, con_out_g(B)) → dis_out_g(B)
U2_g(B1, B2, dis_out_g(B2)) → dis_out_g(or(B1, B2))
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))
The argument filtering Pi contains the following mapping:
con_in_g(
x1) =
con_in_g(
x1)
and(
x1,
x2) =
and(
x1,
x2)
U4_g(
x1,
x2,
x3) =
U4_g(
x2,
x3)
dis_in_g(
x1) =
dis_in_g(
x1)
or(
x1,
x2) =
or(
x1,
x2)
U1_g(
x1,
x2,
x3) =
U1_g(
x2,
x3)
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
U2_g(
x1,
x2,
x3) =
U2_g(
x3)
U3_g(
x1,
x2) =
U3_g(
x2)
dis_out_g(
x1) =
dis_out_g
U5_g(
x1,
x2,
x3) =
U5_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:
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(or(B1, B2)) → U1_G(B1, B2, con_in_g(B1))
DIS_IN_G(or(B1, B2)) → CON_IN_G(B1)
CON_IN_G(B) → U6_G(B, bool_in_g(B))
CON_IN_G(B) → BOOL_IN_G(B)
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)
DIS_IN_G(B) → U3_G(B, con_in_g(B))
DIS_IN_G(B) → CON_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)
The TRS R consists of the following rules:
con_in_g(and(B1, B2)) → U4_g(B1, B2, dis_in_g(B1))
dis_in_g(or(B1, B2)) → U1_g(B1, B2, con_in_g(B1))
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)
U1_g(B1, B2, con_out_g(B1)) → U2_g(B1, B2, dis_in_g(B2))
dis_in_g(B) → U3_g(B, con_in_g(B))
U3_g(B, con_out_g(B)) → dis_out_g(B)
U2_g(B1, B2, dis_out_g(B2)) → dis_out_g(or(B1, B2))
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))
The argument filtering Pi contains the following mapping:
con_in_g(
x1) =
con_in_g(
x1)
and(
x1,
x2) =
and(
x1,
x2)
U4_g(
x1,
x2,
x3) =
U4_g(
x2,
x3)
dis_in_g(
x1) =
dis_in_g(
x1)
or(
x1,
x2) =
or(
x1,
x2)
U1_g(
x1,
x2,
x3) =
U1_g(
x2,
x3)
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
U2_g(
x1,
x2,
x3) =
U2_g(
x3)
U3_g(
x1,
x2) =
U3_g(
x2)
dis_out_g(
x1) =
dis_out_g
U5_g(
x1,
x2,
x3) =
U5_g(
x3)
CON_IN_G(
x1) =
CON_IN_G(
x1)
U4_G(
x1,
x2,
x3) =
U4_G(
x2,
x3)
DIS_IN_G(
x1) =
DIS_IN_G(
x1)
U1_G(
x1,
x2,
x3) =
U1_G(
x2,
x3)
U6_G(
x1,
x2) =
U6_G(
x2)
BOOL_IN_G(
x1) =
BOOL_IN_G(
x1)
U2_G(
x1,
x2,
x3) =
U2_G(
x3)
U3_G(
x1,
x2) =
U3_G(
x2)
U5_G(
x1,
x2,
x3) =
U5_G(
x3)
We have to consider all (P,R,Pi)-chains
(4) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
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(or(B1, B2)) → U1_G(B1, B2, con_in_g(B1))
DIS_IN_G(or(B1, B2)) → CON_IN_G(B1)
CON_IN_G(B) → U6_G(B, bool_in_g(B))
CON_IN_G(B) → BOOL_IN_G(B)
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)
DIS_IN_G(B) → U3_G(B, con_in_g(B))
DIS_IN_G(B) → CON_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)
The TRS R consists of the following rules:
con_in_g(and(B1, B2)) → U4_g(B1, B2, dis_in_g(B1))
dis_in_g(or(B1, B2)) → U1_g(B1, B2, con_in_g(B1))
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)
U1_g(B1, B2, con_out_g(B1)) → U2_g(B1, B2, dis_in_g(B2))
dis_in_g(B) → U3_g(B, con_in_g(B))
U3_g(B, con_out_g(B)) → dis_out_g(B)
U2_g(B1, B2, dis_out_g(B2)) → dis_out_g(or(B1, B2))
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))
The argument filtering Pi contains the following mapping:
con_in_g(
x1) =
con_in_g(
x1)
and(
x1,
x2) =
and(
x1,
x2)
U4_g(
x1,
x2,
x3) =
U4_g(
x2,
x3)
dis_in_g(
x1) =
dis_in_g(
x1)
or(
x1,
x2) =
or(
x1,
x2)
U1_g(
x1,
x2,
x3) =
U1_g(
x2,
x3)
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
U2_g(
x1,
x2,
x3) =
U2_g(
x3)
U3_g(
x1,
x2) =
U3_g(
x2)
dis_out_g(
x1) =
dis_out_g
U5_g(
x1,
x2,
x3) =
U5_g(
x3)
CON_IN_G(
x1) =
CON_IN_G(
x1)
U4_G(
x1,
x2,
x3) =
U4_G(
x2,
x3)
DIS_IN_G(
x1) =
DIS_IN_G(
x1)
U1_G(
x1,
x2,
x3) =
U1_G(
x2,
x3)
U6_G(
x1,
x2) =
U6_G(
x2)
BOOL_IN_G(
x1) =
BOOL_IN_G(
x1)
U2_G(
x1,
x2,
x3) =
U2_G(
x3)
U3_G(
x1,
x2) =
U3_G(
x2)
U5_G(
x1,
x2,
x3) =
U5_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:
U4_G(B1, B2, dis_out_g(B1)) → CON_IN_G(B2)
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(or(B1, B2)) → U1_G(B1, B2, con_in_g(B1))
U1_G(B1, B2, con_out_g(B1)) → DIS_IN_G(B2)
DIS_IN_G(or(B1, B2)) → CON_IN_G(B1)
DIS_IN_G(B) → CON_IN_G(B)
The TRS R consists of the following rules:
con_in_g(and(B1, B2)) → U4_g(B1, B2, dis_in_g(B1))
dis_in_g(or(B1, B2)) → U1_g(B1, B2, con_in_g(B1))
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)
U1_g(B1, B2, con_out_g(B1)) → U2_g(B1, B2, dis_in_g(B2))
dis_in_g(B) → U3_g(B, con_in_g(B))
U3_g(B, con_out_g(B)) → dis_out_g(B)
U2_g(B1, B2, dis_out_g(B2)) → dis_out_g(or(B1, B2))
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))
The argument filtering Pi contains the following mapping:
con_in_g(
x1) =
con_in_g(
x1)
and(
x1,
x2) =
and(
x1,
x2)
U4_g(
x1,
x2,
x3) =
U4_g(
x2,
x3)
dis_in_g(
x1) =
dis_in_g(
x1)
or(
x1,
x2) =
or(
x1,
x2)
U1_g(
x1,
x2,
x3) =
U1_g(
x2,
x3)
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
U2_g(
x1,
x2,
x3) =
U2_g(
x3)
U3_g(
x1,
x2) =
U3_g(
x2)
dis_out_g(
x1) =
dis_out_g
U5_g(
x1,
x2,
x3) =
U5_g(
x3)
CON_IN_G(
x1) =
CON_IN_G(
x1)
U4_G(
x1,
x2,
x3) =
U4_G(
x2,
x3)
DIS_IN_G(
x1) =
DIS_IN_G(
x1)
U1_G(
x1,
x2,
x3) =
U1_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:
U4_G(B2, dis_out_g) → CON_IN_G(B2)
CON_IN_G(and(B1, B2)) → U4_G(B2, dis_in_g(B1))
CON_IN_G(and(B1, B2)) → DIS_IN_G(B1)
DIS_IN_G(or(B1, B2)) → U1_G(B2, con_in_g(B1))
U1_G(B2, con_out_g) → DIS_IN_G(B2)
DIS_IN_G(or(B1, B2)) → CON_IN_G(B1)
DIS_IN_G(B) → CON_IN_G(B)
The TRS R consists of the following rules:
con_in_g(and(B1, B2)) → U4_g(B2, dis_in_g(B1))
dis_in_g(or(B1, B2)) → U1_g(B2, con_in_g(B1))
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
U1_g(B2, con_out_g) → U2_g(dis_in_g(B2))
dis_in_g(B) → U3_g(con_in_g(B))
U3_g(con_out_g) → dis_out_g
U2_g(dis_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
The set Q consists of the following terms:
con_in_g(x0)
dis_in_g(x0)
bool_in_g(x0)
U6_g(x0)
U1_g(x0, x1)
U3_g(x0)
U2_g(x0)
U4_g(x0, x1)
U5_g(x0)
We have to consider all (P,Q,R)-chains.
(9) 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:
- CON_IN_G(and(B1, B2)) → U4_G(B2, dis_in_g(B1))
The graph contains the following edges 1 > 1
- CON_IN_G(and(B1, B2)) → DIS_IN_G(B1)
The graph contains the following edges 1 > 1
- U4_G(B2, dis_out_g) → CON_IN_G(B2)
The graph contains the following edges 1 >= 1
- DIS_IN_G(or(B1, B2)) → U1_G(B2, con_in_g(B1))
The graph contains the following edges 1 > 1
- U1_G(B2, con_out_g) → DIS_IN_G(B2)
The graph contains the following edges 1 >= 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
(10) YES