3.82/1.85	YES
3.82/1.86	proof of /export/starexec/sandbox/benchmark/theBenchmark.pl
3.82/1.86	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.82/1.86	
3.82/1.86	
3.82/1.86	Left Termination of the query pattern
3.82/1.86	
3.82/1.86	con(g)
3.82/1.86	
3.82/1.86	w.r.t. the given Prolog program could successfully be proven:
3.82/1.86	
3.82/1.86	(0) Prolog
3.82/1.86	(1) PrologToPiTRSProof [SOUND, 0 ms]
3.82/1.86	(2) PiTRS
3.82/1.86	(3) DependencyPairsProof [EQUIVALENT, 3 ms]
3.82/1.86	(4) PiDP
3.82/1.86	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
3.82/1.86	(6) PiDP
3.82/1.86	(7) PiDPToQDPProof [EQUIVALENT, 1 ms]
3.82/1.86	(8) QDP
3.82/1.86	(9) QDPSizeChangeProof [EQUIVALENT, 0 ms]
3.82/1.86	(10) YES
3.82/1.86	
3.82/1.86	
3.82/1.86	----------------------------------------
3.82/1.86	
3.82/1.86	(0)
3.82/1.86	Obligation:
3.82/1.86	Clauses:
3.82/1.86	
3.82/1.86	dis(or(B1, B2)) :- ','(con(B1), dis(B2)).
3.82/1.86	dis(B) :- con(B).
3.82/1.86	con(and(B1, B2)) :- ','(dis(B1), con(B2)).
3.82/1.86	con(B) :- bool(B).
3.82/1.86	bool(0).
3.82/1.86	bool(1).
3.82/1.86	
3.82/1.86	
3.82/1.86	Query: con(g)
3.82/1.86	----------------------------------------
3.82/1.86	
3.82/1.86	(1) PrologToPiTRSProof (SOUND)
3.82/1.86	We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
3.82/1.86	
3.82/1.86	con_in_1: (b)
3.82/1.86	
3.82/1.86	dis_in_1: (b)
3.82/1.86	
3.82/1.86	Transforming Prolog into the following Term Rewriting System:
3.82/1.86	
3.82/1.86	Pi-finite rewrite system:
3.82/1.86	The TRS R consists of the following rules:
3.82/1.86	
3.82/1.86	   con_in_g(and(B1, B2)) -> U4_g(B1, B2, dis_in_g(B1))
3.82/1.86	   dis_in_g(or(B1, B2)) -> U1_g(B1, B2, con_in_g(B1))
3.82/1.86	   con_in_g(B) -> U6_g(B, bool_in_g(B))
3.82/1.86	   bool_in_g(0) -> bool_out_g(0)
3.82/1.86	   bool_in_g(1) -> bool_out_g(1)
3.82/1.86	   U6_g(B, bool_out_g(B)) -> con_out_g(B)
3.82/1.86	   U1_g(B1, B2, con_out_g(B1)) -> U2_g(B1, B2, dis_in_g(B2))
3.82/1.86	   dis_in_g(B) -> U3_g(B, con_in_g(B))
3.82/1.86	   U3_g(B, con_out_g(B)) -> dis_out_g(B)
3.82/1.86	   U2_g(B1, B2, dis_out_g(B2)) -> dis_out_g(or(B1, B2))
3.82/1.86	   U4_g(B1, B2, dis_out_g(B1)) -> U5_g(B1, B2, con_in_g(B2))
3.82/1.86	   U5_g(B1, B2, con_out_g(B2)) -> con_out_g(and(B1, B2))
3.82/1.86	
3.82/1.86	Pi is empty.
3.82/1.86	
3.82/1.86	
3.82/1.86	
3.82/1.86	Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
3.82/1.86	
3.82/1.86	
3.82/1.86	
3.82/1.86	----------------------------------------
3.82/1.86	
3.82/1.86	(2)
3.82/1.86	Obligation:
3.82/1.86	Pi-finite rewrite system:
3.82/1.86	The TRS R consists of the following rules:
3.82/1.86	
3.82/1.86	   con_in_g(and(B1, B2)) -> U4_g(B1, B2, dis_in_g(B1))
3.82/1.86	   dis_in_g(or(B1, B2)) -> U1_g(B1, B2, con_in_g(B1))
3.82/1.86	   con_in_g(B) -> U6_g(B, bool_in_g(B))
3.82/1.86	   bool_in_g(0) -> bool_out_g(0)
3.82/1.86	   bool_in_g(1) -> bool_out_g(1)
3.82/1.86	   U6_g(B, bool_out_g(B)) -> con_out_g(B)
3.82/1.86	   U1_g(B1, B2, con_out_g(B1)) -> U2_g(B1, B2, dis_in_g(B2))
3.82/1.86	   dis_in_g(B) -> U3_g(B, con_in_g(B))
3.82/1.86	   U3_g(B, con_out_g(B)) -> dis_out_g(B)
3.82/1.86	   U2_g(B1, B2, dis_out_g(B2)) -> dis_out_g(or(B1, B2))
3.82/1.86	   U4_g(B1, B2, dis_out_g(B1)) -> U5_g(B1, B2, con_in_g(B2))
3.82/1.86	   U5_g(B1, B2, con_out_g(B2)) -> con_out_g(and(B1, B2))
3.82/1.86	
3.82/1.86	Pi is empty.
3.82/1.86	
3.82/1.86	----------------------------------------
3.82/1.86	
3.82/1.86	(3) DependencyPairsProof (EQUIVALENT)
3.82/1.86	Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
3.82/1.86	Pi DP problem:
3.82/1.86	The TRS P consists of the following rules:
3.82/1.86	
3.82/1.86	   CON_IN_G(and(B1, B2)) -> U4_G(B1, B2, dis_in_g(B1))
3.82/1.86	   CON_IN_G(and(B1, B2)) -> DIS_IN_G(B1)
3.82/1.86	   DIS_IN_G(or(B1, B2)) -> U1_G(B1, B2, con_in_g(B1))
3.82/1.86	   DIS_IN_G(or(B1, B2)) -> CON_IN_G(B1)
3.82/1.86	   CON_IN_G(B) -> U6_G(B, bool_in_g(B))
3.82/1.86	   CON_IN_G(B) -> BOOL_IN_G(B)
3.82/1.86	   U1_G(B1, B2, con_out_g(B1)) -> U2_G(B1, B2, dis_in_g(B2))
3.82/1.86	   U1_G(B1, B2, con_out_g(B1)) -> DIS_IN_G(B2)
3.82/1.86	   DIS_IN_G(B) -> U3_G(B, con_in_g(B))
3.82/1.86	   DIS_IN_G(B) -> CON_IN_G(B)
3.82/1.86	   U4_G(B1, B2, dis_out_g(B1)) -> U5_G(B1, B2, con_in_g(B2))
3.82/1.86	   U4_G(B1, B2, dis_out_g(B1)) -> CON_IN_G(B2)
3.82/1.86	
3.82/1.86	The TRS R consists of the following rules:
3.82/1.86	
3.82/1.86	   con_in_g(and(B1, B2)) -> U4_g(B1, B2, dis_in_g(B1))
3.82/1.86	   dis_in_g(or(B1, B2)) -> U1_g(B1, B2, con_in_g(B1))
3.82/1.86	   con_in_g(B) -> U6_g(B, bool_in_g(B))
3.82/1.86	   bool_in_g(0) -> bool_out_g(0)
3.82/1.86	   bool_in_g(1) -> bool_out_g(1)
3.82/1.86	   U6_g(B, bool_out_g(B)) -> con_out_g(B)
3.82/1.86	   U1_g(B1, B2, con_out_g(B1)) -> U2_g(B1, B2, dis_in_g(B2))
3.82/1.86	   dis_in_g(B) -> U3_g(B, con_in_g(B))
3.82/1.86	   U3_g(B, con_out_g(B)) -> dis_out_g(B)
3.82/1.86	   U2_g(B1, B2, dis_out_g(B2)) -> dis_out_g(or(B1, B2))
3.82/1.86	   U4_g(B1, B2, dis_out_g(B1)) -> U5_g(B1, B2, con_in_g(B2))
3.82/1.86	   U5_g(B1, B2, con_out_g(B2)) -> con_out_g(and(B1, B2))
3.82/1.86	
3.82/1.86	Pi is empty.
3.82/1.86	We have to consider all (P,R,Pi)-chains
3.82/1.86	----------------------------------------
3.82/1.86	
3.82/1.86	(4)
3.82/1.86	Obligation:
3.82/1.86	Pi DP problem:
3.82/1.86	The TRS P consists of the following rules:
3.82/1.86	
3.82/1.86	   CON_IN_G(and(B1, B2)) -> U4_G(B1, B2, dis_in_g(B1))
3.82/1.86	   CON_IN_G(and(B1, B2)) -> DIS_IN_G(B1)
3.82/1.86	   DIS_IN_G(or(B1, B2)) -> U1_G(B1, B2, con_in_g(B1))
3.82/1.86	   DIS_IN_G(or(B1, B2)) -> CON_IN_G(B1)
3.82/1.86	   CON_IN_G(B) -> U6_G(B, bool_in_g(B))
3.82/1.86	   CON_IN_G(B) -> BOOL_IN_G(B)
3.82/1.86	   U1_G(B1, B2, con_out_g(B1)) -> U2_G(B1, B2, dis_in_g(B2))
3.82/1.86	   U1_G(B1, B2, con_out_g(B1)) -> DIS_IN_G(B2)
3.82/1.86	   DIS_IN_G(B) -> U3_G(B, con_in_g(B))
3.82/1.86	   DIS_IN_G(B) -> CON_IN_G(B)
3.82/1.86	   U4_G(B1, B2, dis_out_g(B1)) -> U5_G(B1, B2, con_in_g(B2))
3.82/1.86	   U4_G(B1, B2, dis_out_g(B1)) -> CON_IN_G(B2)
3.82/1.86	
3.82/1.86	The TRS R consists of the following rules:
3.82/1.86	
3.82/1.86	   con_in_g(and(B1, B2)) -> U4_g(B1, B2, dis_in_g(B1))
3.82/1.86	   dis_in_g(or(B1, B2)) -> U1_g(B1, B2, con_in_g(B1))
3.82/1.86	   con_in_g(B) -> U6_g(B, bool_in_g(B))
3.82/1.86	   bool_in_g(0) -> bool_out_g(0)
3.82/1.86	   bool_in_g(1) -> bool_out_g(1)
3.82/1.86	   U6_g(B, bool_out_g(B)) -> con_out_g(B)
3.82/1.86	   U1_g(B1, B2, con_out_g(B1)) -> U2_g(B1, B2, dis_in_g(B2))
3.82/1.86	   dis_in_g(B) -> U3_g(B, con_in_g(B))
3.82/1.86	   U3_g(B, con_out_g(B)) -> dis_out_g(B)
3.82/1.86	   U2_g(B1, B2, dis_out_g(B2)) -> dis_out_g(or(B1, B2))
3.82/1.86	   U4_g(B1, B2, dis_out_g(B1)) -> U5_g(B1, B2, con_in_g(B2))
3.82/1.86	   U5_g(B1, B2, con_out_g(B2)) -> con_out_g(and(B1, B2))
3.82/1.86	
3.82/1.86	Pi is empty.
3.82/1.86	We have to consider all (P,R,Pi)-chains
3.82/1.86	----------------------------------------
3.82/1.86	
3.82/1.86	(5) DependencyGraphProof (EQUIVALENT)
3.82/1.86	The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 5 less nodes.
3.82/1.86	----------------------------------------
3.82/1.86	
3.82/1.86	(6)
3.82/1.86	Obligation:
3.82/1.86	Pi DP problem:
3.82/1.86	The TRS P consists of the following rules:
3.82/1.86	
3.82/1.86	   U4_G(B1, B2, dis_out_g(B1)) -> CON_IN_G(B2)
3.82/1.86	   CON_IN_G(and(B1, B2)) -> U4_G(B1, B2, dis_in_g(B1))
3.82/1.86	   CON_IN_G(and(B1, B2)) -> DIS_IN_G(B1)
3.82/1.86	   DIS_IN_G(or(B1, B2)) -> U1_G(B1, B2, con_in_g(B1))
3.82/1.86	   U1_G(B1, B2, con_out_g(B1)) -> DIS_IN_G(B2)
3.82/1.86	   DIS_IN_G(or(B1, B2)) -> CON_IN_G(B1)
3.82/1.86	   DIS_IN_G(B) -> CON_IN_G(B)
3.82/1.86	
3.82/1.86	The TRS R consists of the following rules:
3.82/1.86	
3.82/1.86	   con_in_g(and(B1, B2)) -> U4_g(B1, B2, dis_in_g(B1))
3.82/1.86	   dis_in_g(or(B1, B2)) -> U1_g(B1, B2, con_in_g(B1))
3.82/1.86	   con_in_g(B) -> U6_g(B, bool_in_g(B))
3.82/1.86	   bool_in_g(0) -> bool_out_g(0)
3.82/1.86	   bool_in_g(1) -> bool_out_g(1)
3.82/1.86	   U6_g(B, bool_out_g(B)) -> con_out_g(B)
3.82/1.86	   U1_g(B1, B2, con_out_g(B1)) -> U2_g(B1, B2, dis_in_g(B2))
3.82/1.86	   dis_in_g(B) -> U3_g(B, con_in_g(B))
3.82/1.86	   U3_g(B, con_out_g(B)) -> dis_out_g(B)
3.82/1.86	   U2_g(B1, B2, dis_out_g(B2)) -> dis_out_g(or(B1, B2))
3.82/1.86	   U4_g(B1, B2, dis_out_g(B1)) -> U5_g(B1, B2, con_in_g(B2))
3.82/1.86	   U5_g(B1, B2, con_out_g(B2)) -> con_out_g(and(B1, B2))
3.82/1.86	
3.82/1.86	Pi is empty.
3.82/1.86	We have to consider all (P,R,Pi)-chains
3.82/1.86	----------------------------------------
3.82/1.86	
3.82/1.86	(7) PiDPToQDPProof (EQUIVALENT)
3.82/1.86	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
3.82/1.86	----------------------------------------
3.82/1.86	
3.82/1.86	(8)
3.82/1.86	Obligation:
3.82/1.86	Q DP problem:
3.82/1.86	The TRS P consists of the following rules:
3.82/1.86	
3.82/1.86	   U4_G(B1, B2, dis_out_g(B1)) -> CON_IN_G(B2)
3.82/1.86	   CON_IN_G(and(B1, B2)) -> U4_G(B1, B2, dis_in_g(B1))
3.82/1.86	   CON_IN_G(and(B1, B2)) -> DIS_IN_G(B1)
3.82/1.86	   DIS_IN_G(or(B1, B2)) -> U1_G(B1, B2, con_in_g(B1))
3.82/1.86	   U1_G(B1, B2, con_out_g(B1)) -> DIS_IN_G(B2)
3.82/1.86	   DIS_IN_G(or(B1, B2)) -> CON_IN_G(B1)
3.82/1.86	   DIS_IN_G(B) -> CON_IN_G(B)
3.82/1.86	
3.82/1.86	The TRS R consists of the following rules:
3.82/1.86	
3.82/1.86	   con_in_g(and(B1, B2)) -> U4_g(B1, B2, dis_in_g(B1))
3.82/1.86	   dis_in_g(or(B1, B2)) -> U1_g(B1, B2, con_in_g(B1))
3.82/1.86	   con_in_g(B) -> U6_g(B, bool_in_g(B))
3.82/1.86	   bool_in_g(0) -> bool_out_g(0)
3.82/1.86	   bool_in_g(1) -> bool_out_g(1)
3.82/1.86	   U6_g(B, bool_out_g(B)) -> con_out_g(B)
3.82/1.86	   U1_g(B1, B2, con_out_g(B1)) -> U2_g(B1, B2, dis_in_g(B2))
3.82/1.86	   dis_in_g(B) -> U3_g(B, con_in_g(B))
3.82/1.86	   U3_g(B, con_out_g(B)) -> dis_out_g(B)
3.82/1.86	   U2_g(B1, B2, dis_out_g(B2)) -> dis_out_g(or(B1, B2))
3.82/1.86	   U4_g(B1, B2, dis_out_g(B1)) -> U5_g(B1, B2, con_in_g(B2))
3.82/1.86	   U5_g(B1, B2, con_out_g(B2)) -> con_out_g(and(B1, B2))
3.82/1.86	
3.82/1.86	The set Q consists of the following terms:
3.82/1.86	
3.82/1.86	   con_in_g(x0)
3.82/1.86	   dis_in_g(x0)
3.82/1.86	   bool_in_g(x0)
3.82/1.86	   U6_g(x0, x1)
3.82/1.86	   U1_g(x0, x1, x2)
3.82/1.86	   U3_g(x0, x1)
3.82/1.86	   U2_g(x0, x1, x2)
3.82/1.86	   U4_g(x0, x1, x2)
3.82/1.86	   U5_g(x0, x1, x2)
3.82/1.86	
3.82/1.86	We have to consider all (P,Q,R)-chains.
3.82/1.86	----------------------------------------
3.82/1.86	
3.82/1.86	(9) QDPSizeChangeProof (EQUIVALENT)
3.82/1.86	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. 
3.82/1.86	
3.82/1.86	From the DPs we obtained the following set of size-change graphs:
3.82/1.86	*CON_IN_G(and(B1, B2)) -> U4_G(B1, B2, dis_in_g(B1))
3.82/1.86	The graph contains the following edges 1 > 1, 1 > 2
3.82/1.86	
3.82/1.86	
3.82/1.86	*CON_IN_G(and(B1, B2)) -> DIS_IN_G(B1)
3.82/1.86	The graph contains the following edges 1 > 1
3.82/1.86	
3.82/1.86	
3.82/1.86	*U4_G(B1, B2, dis_out_g(B1)) -> CON_IN_G(B2)
3.82/1.86	The graph contains the following edges 2 >= 1
3.82/1.86	
3.82/1.86	
3.82/1.86	*DIS_IN_G(or(B1, B2)) -> U1_G(B1, B2, con_in_g(B1))
3.82/1.86	The graph contains the following edges 1 > 1, 1 > 2
3.82/1.86	
3.82/1.86	
3.82/1.86	*U1_G(B1, B2, con_out_g(B1)) -> DIS_IN_G(B2)
3.82/1.86	The graph contains the following edges 2 >= 1
3.82/1.86	
3.82/1.86	
3.82/1.86	*DIS_IN_G(or(B1, B2)) -> CON_IN_G(B1)
3.82/1.86	The graph contains the following edges 1 > 1
3.82/1.86	
3.82/1.86	
3.82/1.86	*DIS_IN_G(B) -> CON_IN_G(B)
3.82/1.86	The graph contains the following edges 1 >= 1
3.82/1.86	
3.82/1.86	
3.82/1.86	----------------------------------------
3.82/1.86	
3.82/1.86	(10)
3.82/1.86	YES
4.08/1.88	EOF