3.83/1.72	YES
3.83/1.74	proof of /export/starexec/sandbox2/benchmark/theBenchmark.pl
3.83/1.74	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.83/1.74	
3.83/1.74	
3.83/1.74	Left Termination of the query pattern
3.83/1.74	
3.83/1.74	dis(g)
3.83/1.74	
3.83/1.74	w.r.t. the given Prolog program could successfully be proven:
3.83/1.74	
3.83/1.74	(0) Prolog
3.83/1.74	(1) PrologToPiTRSProof [SOUND, 0 ms]
3.83/1.74	(2) PiTRS
3.83/1.74	(3) DependencyPairsProof [EQUIVALENT, 7 ms]
3.83/1.74	(4) PiDP
3.83/1.74	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
3.83/1.74	(6) PiDP
3.83/1.74	(7) PiDPToQDPProof [SOUND, 0 ms]
3.83/1.74	(8) QDP
3.83/1.74	(9) QDPSizeChangeProof [EQUIVALENT, 0 ms]
3.83/1.74	(10) YES
3.83/1.74	
3.83/1.74	
3.83/1.74	----------------------------------------
3.83/1.74	
3.83/1.74	(0)
3.83/1.74	Obligation:
3.83/1.74	Clauses:
3.83/1.74	
3.83/1.74	dis(or(B1, B2)) :- ','(con(B1), dis(B2)).
3.83/1.74	dis(B) :- con(B).
3.83/1.74	con(and(B1, B2)) :- ','(dis(B1), con(B2)).
3.83/1.74	con(B) :- bool(B).
3.83/1.74	bool(0).
3.83/1.74	bool(1).
3.83/1.74	
3.83/1.74	
3.83/1.74	Query: dis(g)
3.83/1.74	----------------------------------------
3.83/1.74	
3.83/1.74	(1) PrologToPiTRSProof (SOUND)
3.83/1.74	We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
3.83/1.74	
3.83/1.74	dis_in_1: (b)
3.83/1.74	
3.83/1.74	con_in_1: (b)
3.83/1.74	
3.83/1.74	Transforming Prolog into the following Term Rewriting System:
3.83/1.74	
3.83/1.74	Pi-finite rewrite system:
3.83/1.74	The TRS R consists of the following rules:
3.83/1.74	
3.83/1.74	   dis_in_g(or(B1, B2)) -> U1_g(B1, B2, con_in_g(B1))
3.83/1.74	   con_in_g(and(B1, B2)) -> U4_g(B1, B2, dis_in_g(B1))
3.83/1.74	   dis_in_g(B) -> U3_g(B, con_in_g(B))
3.83/1.74	   con_in_g(B) -> U6_g(B, bool_in_g(B))
3.83/1.74	   bool_in_g(0) -> bool_out_g(0)
3.83/1.74	   bool_in_g(1) -> bool_out_g(1)
3.83/1.74	   U6_g(B, bool_out_g(B)) -> con_out_g(B)
3.83/1.74	   U3_g(B, con_out_g(B)) -> dis_out_g(B)
3.83/1.74	   U4_g(B1, B2, dis_out_g(B1)) -> U5_g(B1, B2, con_in_g(B2))
3.83/1.74	   U5_g(B1, B2, con_out_g(B2)) -> con_out_g(and(B1, B2))
3.83/1.74	   U1_g(B1, B2, con_out_g(B1)) -> U2_g(B1, B2, dis_in_g(B2))
3.83/1.74	   U2_g(B1, B2, dis_out_g(B2)) -> dis_out_g(or(B1, B2))
3.83/1.74	
3.83/1.74	The argument filtering Pi contains the following mapping:
3.83/1.74	dis_in_g(x1)  =  dis_in_g(x1)
3.83/1.74	
3.83/1.74	or(x1, x2)  =  or(x1, x2)
3.83/1.74	
3.83/1.74	U1_g(x1, x2, x3)  =  U1_g(x2, x3)
3.83/1.74	
3.83/1.74	con_in_g(x1)  =  con_in_g(x1)
3.83/1.74	
3.83/1.74	and(x1, x2)  =  and(x1, x2)
3.83/1.74	
3.83/1.74	U4_g(x1, x2, x3)  =  U4_g(x2, x3)
3.83/1.74	
3.83/1.74	U3_g(x1, x2)  =  U3_g(x2)
3.83/1.74	
3.83/1.74	U6_g(x1, x2)  =  U6_g(x2)
3.83/1.74	
3.83/1.74	bool_in_g(x1)  =  bool_in_g(x1)
3.83/1.74	
3.83/1.74	0  =  0
3.83/1.74	
3.83/1.74	bool_out_g(x1)  =  bool_out_g
3.83/1.74	
3.83/1.74	1  =  1
3.83/1.74	
3.83/1.74	con_out_g(x1)  =  con_out_g
3.83/1.74	
3.83/1.74	dis_out_g(x1)  =  dis_out_g
3.83/1.74	
3.83/1.74	U5_g(x1, x2, x3)  =  U5_g(x3)
3.83/1.74	
3.83/1.74	U2_g(x1, x2, x3)  =  U2_g(x3)
3.83/1.74	
3.83/1.74	
3.83/1.74	
3.83/1.74	
3.83/1.74	
3.83/1.74	Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
3.83/1.74	
3.83/1.74	
3.83/1.74	
3.83/1.74	----------------------------------------
3.83/1.74	
3.83/1.74	(2)
3.83/1.74	Obligation:
3.83/1.74	Pi-finite rewrite system:
3.83/1.74	The TRS R consists of the following rules:
3.83/1.74	
3.83/1.74	   dis_in_g(or(B1, B2)) -> U1_g(B1, B2, con_in_g(B1))
3.83/1.74	   con_in_g(and(B1, B2)) -> U4_g(B1, B2, dis_in_g(B1))
3.83/1.74	   dis_in_g(B) -> U3_g(B, con_in_g(B))
3.83/1.74	   con_in_g(B) -> U6_g(B, bool_in_g(B))
3.83/1.74	   bool_in_g(0) -> bool_out_g(0)
3.83/1.74	   bool_in_g(1) -> bool_out_g(1)
3.83/1.74	   U6_g(B, bool_out_g(B)) -> con_out_g(B)
3.83/1.74	   U3_g(B, con_out_g(B)) -> dis_out_g(B)
3.83/1.74	   U4_g(B1, B2, dis_out_g(B1)) -> U5_g(B1, B2, con_in_g(B2))
3.83/1.74	   U5_g(B1, B2, con_out_g(B2)) -> con_out_g(and(B1, B2))
3.83/1.74	   U1_g(B1, B2, con_out_g(B1)) -> U2_g(B1, B2, dis_in_g(B2))
3.83/1.74	   U2_g(B1, B2, dis_out_g(B2)) -> dis_out_g(or(B1, B2))
3.83/1.74	
3.83/1.74	The argument filtering Pi contains the following mapping:
3.83/1.74	dis_in_g(x1)  =  dis_in_g(x1)
3.83/1.74	
3.83/1.74	or(x1, x2)  =  or(x1, x2)
3.83/1.74	
3.83/1.74	U1_g(x1, x2, x3)  =  U1_g(x2, x3)
3.83/1.74	
3.83/1.74	con_in_g(x1)  =  con_in_g(x1)
3.83/1.74	
3.83/1.74	and(x1, x2)  =  and(x1, x2)
3.83/1.74	
3.83/1.74	U4_g(x1, x2, x3)  =  U4_g(x2, x3)
3.83/1.74	
3.83/1.74	U3_g(x1, x2)  =  U3_g(x2)
3.83/1.74	
3.83/1.74	U6_g(x1, x2)  =  U6_g(x2)
3.83/1.74	
3.83/1.74	bool_in_g(x1)  =  bool_in_g(x1)
3.83/1.74	
3.83/1.74	0  =  0
3.83/1.74	
3.83/1.74	bool_out_g(x1)  =  bool_out_g
3.83/1.74	
3.83/1.74	1  =  1
3.83/1.74	
3.83/1.74	con_out_g(x1)  =  con_out_g
3.83/1.74	
3.83/1.74	dis_out_g(x1)  =  dis_out_g
3.83/1.74	
3.83/1.74	U5_g(x1, x2, x3)  =  U5_g(x3)
3.83/1.74	
3.83/1.74	U2_g(x1, x2, x3)  =  U2_g(x3)
3.83/1.74	
3.83/1.74	
3.83/1.74	
3.83/1.74	----------------------------------------
3.83/1.74	
3.83/1.74	(3) DependencyPairsProof (EQUIVALENT)
3.83/1.74	Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
3.83/1.74	Pi DP problem:
3.83/1.74	The TRS P consists of the following rules:
3.83/1.74	
3.83/1.74	   DIS_IN_G(or(B1, B2)) -> U1_G(B1, B2, con_in_g(B1))
3.83/1.74	   DIS_IN_G(or(B1, B2)) -> CON_IN_G(B1)
3.83/1.74	   CON_IN_G(and(B1, B2)) -> U4_G(B1, B2, dis_in_g(B1))
3.83/1.74	   CON_IN_G(and(B1, B2)) -> DIS_IN_G(B1)
3.83/1.74	   DIS_IN_G(B) -> U3_G(B, con_in_g(B))
3.83/1.74	   DIS_IN_G(B) -> CON_IN_G(B)
3.83/1.74	   CON_IN_G(B) -> U6_G(B, bool_in_g(B))
3.83/1.74	   CON_IN_G(B) -> BOOL_IN_G(B)
3.83/1.74	   U4_G(B1, B2, dis_out_g(B1)) -> U5_G(B1, B2, con_in_g(B2))
3.83/1.74	   U4_G(B1, B2, dis_out_g(B1)) -> CON_IN_G(B2)
3.83/1.74	   U1_G(B1, B2, con_out_g(B1)) -> U2_G(B1, B2, dis_in_g(B2))
3.83/1.74	   U1_G(B1, B2, con_out_g(B1)) -> DIS_IN_G(B2)
3.83/1.74	
3.83/1.74	The TRS R consists of the following rules:
3.83/1.74	
3.83/1.74	   dis_in_g(or(B1, B2)) -> U1_g(B1, B2, con_in_g(B1))
3.83/1.74	   con_in_g(and(B1, B2)) -> U4_g(B1, B2, dis_in_g(B1))
3.83/1.74	   dis_in_g(B) -> U3_g(B, con_in_g(B))
3.83/1.74	   con_in_g(B) -> U6_g(B, bool_in_g(B))
3.83/1.74	   bool_in_g(0) -> bool_out_g(0)
3.83/1.74	   bool_in_g(1) -> bool_out_g(1)
3.83/1.74	   U6_g(B, bool_out_g(B)) -> con_out_g(B)
3.83/1.74	   U3_g(B, con_out_g(B)) -> dis_out_g(B)
3.83/1.74	   U4_g(B1, B2, dis_out_g(B1)) -> U5_g(B1, B2, con_in_g(B2))
3.83/1.74	   U5_g(B1, B2, con_out_g(B2)) -> con_out_g(and(B1, B2))
3.83/1.74	   U1_g(B1, B2, con_out_g(B1)) -> U2_g(B1, B2, dis_in_g(B2))
3.83/1.74	   U2_g(B1, B2, dis_out_g(B2)) -> dis_out_g(or(B1, B2))
3.83/1.74	
3.83/1.74	The argument filtering Pi contains the following mapping:
3.83/1.74	dis_in_g(x1)  =  dis_in_g(x1)
3.83/1.74	
3.83/1.74	or(x1, x2)  =  or(x1, x2)
3.83/1.74	
3.83/1.74	U1_g(x1, x2, x3)  =  U1_g(x2, x3)
3.83/1.74	
3.83/1.74	con_in_g(x1)  =  con_in_g(x1)
3.83/1.74	
3.83/1.74	and(x1, x2)  =  and(x1, x2)
3.83/1.74	
3.83/1.74	U4_g(x1, x2, x3)  =  U4_g(x2, x3)
3.83/1.74	
3.83/1.74	U3_g(x1, x2)  =  U3_g(x2)
3.83/1.74	
3.83/1.74	U6_g(x1, x2)  =  U6_g(x2)
3.83/1.74	
3.83/1.74	bool_in_g(x1)  =  bool_in_g(x1)
3.83/1.74	
3.83/1.74	0  =  0
3.83/1.74	
3.83/1.74	bool_out_g(x1)  =  bool_out_g
3.83/1.74	
3.83/1.74	1  =  1
3.83/1.74	
3.83/1.74	con_out_g(x1)  =  con_out_g
3.83/1.74	
3.83/1.74	dis_out_g(x1)  =  dis_out_g
3.83/1.74	
3.83/1.74	U5_g(x1, x2, x3)  =  U5_g(x3)
3.83/1.74	
3.83/1.74	U2_g(x1, x2, x3)  =  U2_g(x3)
3.83/1.74	
3.83/1.74	DIS_IN_G(x1)  =  DIS_IN_G(x1)
3.83/1.74	
3.83/1.74	U1_G(x1, x2, x3)  =  U1_G(x2, x3)
3.83/1.74	
3.83/1.74	CON_IN_G(x1)  =  CON_IN_G(x1)
3.83/1.74	
3.83/1.74	U4_G(x1, x2, x3)  =  U4_G(x2, x3)
3.83/1.74	
3.83/1.74	U3_G(x1, x2)  =  U3_G(x2)
3.83/1.74	
3.83/1.74	U6_G(x1, x2)  =  U6_G(x2)
3.83/1.74	
3.83/1.74	BOOL_IN_G(x1)  =  BOOL_IN_G(x1)
3.83/1.74	
3.83/1.74	U5_G(x1, x2, x3)  =  U5_G(x3)
3.83/1.74	
3.83/1.74	U2_G(x1, x2, x3)  =  U2_G(x3)
3.83/1.74	
3.83/1.74	
3.83/1.74	We have to consider all (P,R,Pi)-chains
3.83/1.74	----------------------------------------
3.83/1.74	
3.83/1.74	(4)
3.83/1.74	Obligation:
3.83/1.74	Pi DP problem:
3.83/1.74	The TRS P consists of the following rules:
3.83/1.74	
3.83/1.74	   DIS_IN_G(or(B1, B2)) -> U1_G(B1, B2, con_in_g(B1))
3.83/1.74	   DIS_IN_G(or(B1, B2)) -> CON_IN_G(B1)
3.83/1.74	   CON_IN_G(and(B1, B2)) -> U4_G(B1, B2, dis_in_g(B1))
3.83/1.74	   CON_IN_G(and(B1, B2)) -> DIS_IN_G(B1)
3.83/1.74	   DIS_IN_G(B) -> U3_G(B, con_in_g(B))
3.83/1.74	   DIS_IN_G(B) -> CON_IN_G(B)
3.83/1.74	   CON_IN_G(B) -> U6_G(B, bool_in_g(B))
3.83/1.74	   CON_IN_G(B) -> BOOL_IN_G(B)
3.83/1.74	   U4_G(B1, B2, dis_out_g(B1)) -> U5_G(B1, B2, con_in_g(B2))
3.83/1.74	   U4_G(B1, B2, dis_out_g(B1)) -> CON_IN_G(B2)
3.83/1.74	   U1_G(B1, B2, con_out_g(B1)) -> U2_G(B1, B2, dis_in_g(B2))
3.83/1.74	   U1_G(B1, B2, con_out_g(B1)) -> DIS_IN_G(B2)
3.83/1.74	
3.83/1.74	The TRS R consists of the following rules:
3.83/1.74	
3.83/1.74	   dis_in_g(or(B1, B2)) -> U1_g(B1, B2, con_in_g(B1))
3.83/1.74	   con_in_g(and(B1, B2)) -> U4_g(B1, B2, dis_in_g(B1))
3.83/1.74	   dis_in_g(B) -> U3_g(B, con_in_g(B))
3.83/1.74	   con_in_g(B) -> U6_g(B, bool_in_g(B))
3.83/1.74	   bool_in_g(0) -> bool_out_g(0)
3.83/1.74	   bool_in_g(1) -> bool_out_g(1)
3.83/1.74	   U6_g(B, bool_out_g(B)) -> con_out_g(B)
3.83/1.74	   U3_g(B, con_out_g(B)) -> dis_out_g(B)
3.83/1.74	   U4_g(B1, B2, dis_out_g(B1)) -> U5_g(B1, B2, con_in_g(B2))
3.83/1.74	   U5_g(B1, B2, con_out_g(B2)) -> con_out_g(and(B1, B2))
3.83/1.74	   U1_g(B1, B2, con_out_g(B1)) -> U2_g(B1, B2, dis_in_g(B2))
3.83/1.74	   U2_g(B1, B2, dis_out_g(B2)) -> dis_out_g(or(B1, B2))
3.83/1.74	
3.83/1.74	The argument filtering Pi contains the following mapping:
3.83/1.74	dis_in_g(x1)  =  dis_in_g(x1)
3.83/1.74	
3.83/1.74	or(x1, x2)  =  or(x1, x2)
3.83/1.74	
3.83/1.74	U1_g(x1, x2, x3)  =  U1_g(x2, x3)
3.83/1.74	
3.83/1.74	con_in_g(x1)  =  con_in_g(x1)
3.83/1.74	
3.83/1.74	and(x1, x2)  =  and(x1, x2)
3.83/1.74	
3.83/1.74	U4_g(x1, x2, x3)  =  U4_g(x2, x3)
3.83/1.74	
3.83/1.74	U3_g(x1, x2)  =  U3_g(x2)
3.83/1.74	
3.83/1.74	U6_g(x1, x2)  =  U6_g(x2)
3.83/1.74	
3.83/1.74	bool_in_g(x1)  =  bool_in_g(x1)
3.83/1.74	
3.83/1.74	0  =  0
3.83/1.74	
3.83/1.74	bool_out_g(x1)  =  bool_out_g
3.83/1.74	
3.83/1.74	1  =  1
3.83/1.74	
3.83/1.74	con_out_g(x1)  =  con_out_g
3.83/1.74	
3.83/1.74	dis_out_g(x1)  =  dis_out_g
3.83/1.74	
3.83/1.74	U5_g(x1, x2, x3)  =  U5_g(x3)
3.83/1.74	
3.83/1.74	U2_g(x1, x2, x3)  =  U2_g(x3)
3.83/1.74	
3.83/1.74	DIS_IN_G(x1)  =  DIS_IN_G(x1)
3.83/1.74	
3.83/1.74	U1_G(x1, x2, x3)  =  U1_G(x2, x3)
3.83/1.74	
3.83/1.74	CON_IN_G(x1)  =  CON_IN_G(x1)
3.83/1.74	
3.83/1.74	U4_G(x1, x2, x3)  =  U4_G(x2, x3)
3.83/1.74	
3.83/1.74	U3_G(x1, x2)  =  U3_G(x2)
3.83/1.74	
3.83/1.74	U6_G(x1, x2)  =  U6_G(x2)
3.83/1.74	
3.83/1.74	BOOL_IN_G(x1)  =  BOOL_IN_G(x1)
3.83/1.74	
3.83/1.74	U5_G(x1, x2, x3)  =  U5_G(x3)
3.83/1.74	
3.83/1.74	U2_G(x1, x2, x3)  =  U2_G(x3)
3.83/1.74	
3.83/1.74	
3.83/1.74	We have to consider all (P,R,Pi)-chains
3.83/1.74	----------------------------------------
3.83/1.74	
3.83/1.74	(5) DependencyGraphProof (EQUIVALENT)
3.83/1.74	The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 5 less nodes.
3.83/1.74	----------------------------------------
3.83/1.74	
3.83/1.74	(6)
3.83/1.74	Obligation:
3.83/1.74	Pi DP problem:
3.83/1.74	The TRS P consists of the following rules:
3.83/1.74	
3.83/1.74	   U1_G(B1, B2, con_out_g(B1)) -> DIS_IN_G(B2)
3.83/1.74	   DIS_IN_G(or(B1, B2)) -> U1_G(B1, B2, con_in_g(B1))
3.83/1.74	   DIS_IN_G(or(B1, B2)) -> CON_IN_G(B1)
3.83/1.74	   CON_IN_G(and(B1, B2)) -> U4_G(B1, B2, dis_in_g(B1))
3.83/1.74	   U4_G(B1, B2, dis_out_g(B1)) -> CON_IN_G(B2)
3.83/1.74	   CON_IN_G(and(B1, B2)) -> DIS_IN_G(B1)
3.83/1.74	   DIS_IN_G(B) -> CON_IN_G(B)
3.83/1.74	
3.83/1.74	The TRS R consists of the following rules:
3.83/1.74	
3.83/1.74	   dis_in_g(or(B1, B2)) -> U1_g(B1, B2, con_in_g(B1))
3.83/1.74	   con_in_g(and(B1, B2)) -> U4_g(B1, B2, dis_in_g(B1))
3.83/1.74	   dis_in_g(B) -> U3_g(B, con_in_g(B))
3.83/1.74	   con_in_g(B) -> U6_g(B, bool_in_g(B))
3.83/1.74	   bool_in_g(0) -> bool_out_g(0)
3.83/1.74	   bool_in_g(1) -> bool_out_g(1)
3.83/1.74	   U6_g(B, bool_out_g(B)) -> con_out_g(B)
3.83/1.74	   U3_g(B, con_out_g(B)) -> dis_out_g(B)
3.83/1.74	   U4_g(B1, B2, dis_out_g(B1)) -> U5_g(B1, B2, con_in_g(B2))
3.83/1.74	   U5_g(B1, B2, con_out_g(B2)) -> con_out_g(and(B1, B2))
3.83/1.74	   U1_g(B1, B2, con_out_g(B1)) -> U2_g(B1, B2, dis_in_g(B2))
3.83/1.74	   U2_g(B1, B2, dis_out_g(B2)) -> dis_out_g(or(B1, B2))
3.83/1.74	
3.83/1.74	The argument filtering Pi contains the following mapping:
3.83/1.74	dis_in_g(x1)  =  dis_in_g(x1)
3.83/1.74	
3.83/1.74	or(x1, x2)  =  or(x1, x2)
3.83/1.74	
3.83/1.74	U1_g(x1, x2, x3)  =  U1_g(x2, x3)
3.83/1.74	
3.83/1.74	con_in_g(x1)  =  con_in_g(x1)
3.83/1.74	
3.83/1.74	and(x1, x2)  =  and(x1, x2)
3.83/1.74	
3.83/1.74	U4_g(x1, x2, x3)  =  U4_g(x2, x3)
3.83/1.74	
3.83/1.74	U3_g(x1, x2)  =  U3_g(x2)
3.83/1.74	
3.83/1.74	U6_g(x1, x2)  =  U6_g(x2)
3.83/1.74	
3.83/1.74	bool_in_g(x1)  =  bool_in_g(x1)
3.83/1.74	
3.83/1.74	0  =  0
3.83/1.74	
3.83/1.74	bool_out_g(x1)  =  bool_out_g
3.83/1.74	
3.83/1.74	1  =  1
3.83/1.74	
3.83/1.74	con_out_g(x1)  =  con_out_g
3.83/1.74	
3.83/1.74	dis_out_g(x1)  =  dis_out_g
3.83/1.74	
3.83/1.74	U5_g(x1, x2, x3)  =  U5_g(x3)
3.83/1.74	
3.83/1.74	U2_g(x1, x2, x3)  =  U2_g(x3)
3.83/1.74	
3.83/1.74	DIS_IN_G(x1)  =  DIS_IN_G(x1)
3.83/1.74	
3.83/1.74	U1_G(x1, x2, x3)  =  U1_G(x2, x3)
3.83/1.74	
3.83/1.74	CON_IN_G(x1)  =  CON_IN_G(x1)
3.83/1.74	
3.83/1.74	U4_G(x1, x2, x3)  =  U4_G(x2, x3)
3.83/1.74	
3.83/1.74	
3.83/1.74	We have to consider all (P,R,Pi)-chains
3.83/1.74	----------------------------------------
3.83/1.74	
3.83/1.74	(7) PiDPToQDPProof (SOUND)
3.83/1.74	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
3.83/1.74	----------------------------------------
3.83/1.74	
3.83/1.74	(8)
3.83/1.74	Obligation:
3.83/1.74	Q DP problem:
3.83/1.74	The TRS P consists of the following rules:
3.83/1.74	
3.83/1.74	   U1_G(B2, con_out_g) -> DIS_IN_G(B2)
3.83/1.74	   DIS_IN_G(or(B1, B2)) -> U1_G(B2, con_in_g(B1))
3.83/1.74	   DIS_IN_G(or(B1, B2)) -> CON_IN_G(B1)
3.83/1.74	   CON_IN_G(and(B1, B2)) -> U4_G(B2, dis_in_g(B1))
3.83/1.74	   U4_G(B2, dis_out_g) -> CON_IN_G(B2)
3.83/1.74	   CON_IN_G(and(B1, B2)) -> DIS_IN_G(B1)
3.83/1.74	   DIS_IN_G(B) -> CON_IN_G(B)
3.83/1.74	
3.83/1.74	The TRS R consists of the following rules:
3.83/1.74	
3.83/1.74	   dis_in_g(or(B1, B2)) -> U1_g(B2, con_in_g(B1))
3.83/1.74	   con_in_g(and(B1, B2)) -> U4_g(B2, dis_in_g(B1))
3.83/1.74	   dis_in_g(B) -> U3_g(con_in_g(B))
3.83/1.74	   con_in_g(B) -> U6_g(bool_in_g(B))
3.83/1.74	   bool_in_g(0) -> bool_out_g
3.83/1.74	   bool_in_g(1) -> bool_out_g
3.83/1.74	   U6_g(bool_out_g) -> con_out_g
3.83/1.74	   U3_g(con_out_g) -> dis_out_g
3.83/1.74	   U4_g(B2, dis_out_g) -> U5_g(con_in_g(B2))
3.83/1.74	   U5_g(con_out_g) -> con_out_g
3.83/1.74	   U1_g(B2, con_out_g) -> U2_g(dis_in_g(B2))
3.83/1.74	   U2_g(dis_out_g) -> dis_out_g
3.83/1.74	
3.83/1.74	The set Q consists of the following terms:
3.83/1.74	
3.83/1.74	   dis_in_g(x0)
3.83/1.74	   con_in_g(x0)
3.83/1.74	   bool_in_g(x0)
3.83/1.74	   U6_g(x0)
3.83/1.74	   U3_g(x0)
3.83/1.74	   U4_g(x0, x1)
3.83/1.74	   U5_g(x0)
3.83/1.74	   U1_g(x0, x1)
3.83/1.74	   U2_g(x0)
3.83/1.74	
3.83/1.74	We have to consider all (P,Q,R)-chains.
3.83/1.74	----------------------------------------
3.83/1.74	
3.83/1.74	(9) QDPSizeChangeProof (EQUIVALENT)
3.83/1.74	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. 
3.83/1.74	
3.83/1.74	From the DPs we obtained the following set of size-change graphs:
3.83/1.74	*DIS_IN_G(or(B1, B2)) -> U1_G(B2, con_in_g(B1))
3.83/1.74	The graph contains the following edges 1 > 1
3.83/1.74	
3.83/1.74	
3.83/1.74	*U1_G(B2, con_out_g) -> DIS_IN_G(B2)
3.83/1.74	The graph contains the following edges 1 >= 1
3.83/1.74	
3.83/1.74	
3.83/1.74	*CON_IN_G(and(B1, B2)) -> DIS_IN_G(B1)
3.83/1.74	The graph contains the following edges 1 > 1
3.83/1.74	
3.83/1.74	
3.83/1.74	*CON_IN_G(and(B1, B2)) -> U4_G(B2, dis_in_g(B1))
3.83/1.74	The graph contains the following edges 1 > 1
3.83/1.74	
3.83/1.74	
3.83/1.74	*U4_G(B2, dis_out_g) -> CON_IN_G(B2)
3.83/1.74	The graph contains the following edges 1 >= 1
3.83/1.74	
3.83/1.74	
3.83/1.74	*DIS_IN_G(or(B1, B2)) -> CON_IN_G(B1)
3.83/1.74	The graph contains the following edges 1 > 1
3.83/1.74	
3.83/1.74	
3.83/1.74	*DIS_IN_G(B) -> CON_IN_G(B)
3.83/1.74	The graph contains the following edges 1 >= 1
3.83/1.74	
3.83/1.74	
3.83/1.74	----------------------------------------
3.83/1.74	
3.83/1.74	(10)
3.83/1.74	YES
4.02/1.80	EOF