我创建了一个通用SMV程序,并使用check_property
和msat_check_ltlspec_bmc
检查了一对LTL属性。这两个命令的一个属性为true
。相反,另一个属性使用第一个命令和一个单状态给出 14 状态的反例
与后一个命令相反的示例。
问:为什么第二个反例仅包含一个状态,应如何解释?
MODULE Seq_19(T_41, T_41_PRESENT)
VAR
_next_t : { Init_46_idle, LO_51_idle, BO_73_idle, State_120_idle, LOSense_118, BOSense_115, TransitionSegment_125 };
FlagLO : boolean;
FlagBO : boolean;
tt : -256..255;
DEFINE
LOOut_68 :=
case
(_next_t = LOSense_118) : TRUE;
TRUE : FALSE;
esac;
BOOut_84 :=
case
(_next_t = BOSense_115) : TRUE;
TRUE : FALSE;
esac;
LOOut_68_PRESENT :=
case
(_next_t = LOSense_118) : TRUE;
TRUE : FALSE;
esac;
BOOut_84_PRESENT :=
case
(_next_t = BOSense_115) : TRUE;
TRUE : FALSE;
esac;
guard_LOSense_118 := (tt > 5);
guard_BOSense_115 := (tt > 10);
guard_TransitionSegment_125 := TRUE;
ASSIGN
init(_next_t) := { Init_46_idle, LOSense_118 };
init(FlagLO) := FALSE;
init(FlagBO) := FALSE;
init(tt) := 0;
TRANS _next_t in { Init_46_idle }
-> next(_next_t) in { Init_46_idle, LOSense_118 };
TRANS _next_t in { LO_51_idle, LOSense_118 }
-> next(_next_t) in { LO_51_idle, BOSense_115 };
TRANS _next_t in { BO_73_idle, BOSense_115 }
-> next(_next_t) in { BO_73_idle, TransitionSegment_125 };
TRANS _next_t in { State_120_idle, TransitionSegment_125 }
-> next(_next_t) in { State_120_idle };
TRANS (_next_t = Init_46_idle)
-> next(tt) = (tt + 1) &
next(FlagLO) = FlagLO &
next(FlagBO) = FlagBO;
TRANS (_next_t = LO_51_idle)
-> next(tt) = (tt + 1) &
next(FlagLO) = FlagLO &
next(FlagBO) = FlagBO;
TRANS (_next_t = BO_73_idle)
-> next(tt) = (tt + 1) &
next(FlagLO) = FlagLO &
next(FlagBO) = FlagBO;
TRANS (_next_t = State_120_idle)
-> next(FlagLO) = FlagLO &
next(FlagBO) = FlagBO &
next(tt) = tt;
TRANS (_next_t = LOSense_118)
-> next(FlagLO) = TRUE &
next(tt) = (tt + 1) &
next(FlagBO) = FlagBO;
TRANS (_next_t = BOSense_115)
-> next(FlagBO) = TRUE &
next(tt) = (tt + 1) &
next(FlagLO) = FlagLO;
TRANS (_next_t = TransitionSegment_125)
-> next(FlagLO) = FlagLO &
next(FlagBO) = FlagBO &
next(tt) = tt;
INVAR ((_next_t = LOSense_118) -> guard_LOSense_118)
INVAR ((_next_t = BOSense_115) -> guard_BOSense_115)
INVAR ((_next_t = TransitionSegment_125) -> guard_TransitionSegment_125)
INVAR ((_next_t = Init_46_idle) -> !(guard_LOSense_118))
INVAR ((_next_t = LO_51_idle) -> !(guard_BOSense_115))
INVAR ((_next_t = BO_73_idle) -> !(guard_TransitionSegment_125))
INVAR ((_next_t = State_120_idle) -> TRUE)
MODULE main
VAR
T_41 : -256..255;
T_41_PRESENT : boolean;
module : Seq_19(T_41,T_41_PRESENT);
通过标准LTL模型检查,我得到以下输出:
nuXmv > read_model -i Seq19-1210063772855777412.smv
nuXmv > go
nuXmv > check_property -l -p "G (!(((FlagLO = TRUE) & (tt < 5)))) IN module"
-- specification G !(FlagLO = TRUE & tt < 5) IN module is true
nuXmv > check_property -l -p "G (!(((FlagLO = FALSE) & (tt < 5)))) IN module"
-- specification G !(FlagLO = FALSE & tt < 5) IN module is false
-- as demonstrated by the following execution sequence
Trace Description: LTL Counterexample
Trace Type: Counterexample
-> State: 1.1 <-
T_41 = -256
T_41_PRESENT = FALSE
module._next_t = Init_46_idle
module.FlagLO = FALSE
module.FlagBO = FALSE
module.tt = 0
module.guard_TransitionSegment_125 = TRUE
module.guard_BOSense_115 = FALSE
module.guard_LOSense_118 = FALSE
module.BOOut_84_PRESENT = FALSE
module.LOOut_68_PRESENT = FALSE
module.BOOut_84 = FALSE
module.LOOut_68 = FALSE
-> State: 1.2 <-
module.tt = 1
-> State: 1.3 <-
module.tt = 2
-> State: 1.4 <-
module.tt = 3
-> State: 1.5 <-
module.tt = 4
-> State: 1.6 <-
module.tt = 5
-> State: 1.7 <-
module._next_t = LOSense_118
module.tt = 6
module.guard_LOSense_118 = TRUE
module.LOOut_68_PRESENT = TRUE
module.LOOut_68 = TRUE
-> State: 1.8 <-
module._next_t = LO_51_idle
module.FlagLO = TRUE
module.tt = 7
module.LOOut_68_PRESENT = FALSE
module.LOOut_68 = FALSE
-> State: 1.9 <-
module.tt = 8
-> State: 1.10 <-
module.tt = 9
-> State: 1.11 <-
module.tt = 10
-> State: 1.12 <-
module._next_t = BOSense_115
module.tt = 11
module.guard_BOSense_115 = TRUE
module.BOOut_84_PRESENT = TRUE
module.BOOut_84 = TRUE
-> State: 1.13 <-
module._next_t = TransitionSegment_125
module.FlagBO = TRUE
module.tt = 12
module.BOOut_84_PRESENT = FALSE
module.BOOut_84 = FALSE
-- Loop starts here
-> State: 1.14 <-
module._next_t = State_120_idle
-> State: 1.15 <-
相反,通过基于msat_
的LTL模型检查,我得到以下输出:
nuXmv > reset
nuXmv > read_model -i Seq19-1210063772855777412.smv
nuXmv > go_msa
nuXmv > msat_check_ltlspec_bmc -p "G (!(((FlagLO = TRUE) & (tt < 5)))) IN module"
-- no counterexample found with bound 0
-- no counterexample found with bound 1
-- no counterexample found with bound 2
-- no counterexample found with bound 3
-- no counterexample found with bound 4
-- no counterexample found with bound 5
-- no counterexample found with bound 6
-- no counterexample found with bound 7
-- no counterexample found with bound 8
-- no counterexample found with bound 9
-- no counterexample found with bound 10
nuXmv > msat_check_ltlspec_bmc -p "G (!(((FlagLO = FALSE) & (tt < 5)))) IN module"
-- specification G !(FlagLO = FALSE & tt < 5) IN module is false
-- as demonstrated by the following execution sequence
Trace Description: MSAT BMC counterexample
Trace Type: Counterexample
-> State: 1.1 <-
T_41 = -256
T_41_PRESENT = FALSE
module._next_t = Init_46_idle
module.FlagLO = FALSE
module.FlagBO = FALSE
module.tt = 0
module.guard_TransitionSegment_125 = TRUE
module.guard_BOSense_115 = FALSE
module.guard_LOSense_118 = FALSE
module.BOOut_84_PRESENT = FALSE
module.LOOut_68_PRESENT = FALSE
module.BOOut_84 = FALSE
module.LOOut_68 = FALSE
答案 0 :(得分:1)
在第一种情况下,将使用经典的基于Tableau的模型检查来检查LTL属性。通过这种方法,引擎可以对模型中表示的有限状态机进行全局查看,因此它可以提供一个反例跟踪,该跟踪代表违反了给定属性的(无限)执行跟踪。
在第二种情况下,将使用有界模型检查来检查LTL属性,这意味着引擎将通过考虑越来越长的执行跟踪来推进搜索,并且缺少模型中表示的有限状态机的全局视图。结果,此引擎返回的反示例始终包含最小长度的某些(有限)执行跟踪。
在给定的代码示例中,在执行跟踪的第一状态下,属性G (!(((FlagLO = FALSE) & (tt < 5)))) IN module
已被违反:
-> State: 1.1 <-
T_41 = -256
T_41_PRESENT = FALSE
module._next_t = Init_46_idle
module.FlagLO = FALSE
module.FlagBO = FALSE
module.tt = 0
module.guard_TransitionSegment_125 = TRUE
module.guard_BOSense_115 = FALSE
module.guard_LOSense_118 = FALSE
module.BOOut_84_PRESENT = FALSE
module.LOOut_68_PRESENT = FALSE
module.BOOut_84 = FALSE
module.LOOut_68 = FALSE
因此,返回的解决方案是正确的。