1 00:00:02,480 --> 00:00:08,180 Formally we define a propositional net, or prop net, as a directed 2 00:00:08,180 --> 00:00:13,280 bipartite graph, consisting of propositions alternating 3 00:00:13,280 --> 00:00:18,320 with connectives. In this case there are six propositions. 4 00:00:19,440 --> 00:00:25,230 The round circles labeled a, b, p, q, r, and s. 5 00:00:26,340 --> 00:00:27,900 And there are four connectives. 6 00:00:29,210 --> 00:00:32,280 The grey and black nodes in the graph shown here. 7 00:00:35,870 --> 00:00:38,010 Basic prop nets, there are four types of 8 00:00:38,010 --> 00:00:40,500 connectives and they're all present in this case. 9 00:00:41,540 --> 00:00:47,480 There is an and gate on the upper left, an inverter in the upper right, 10 00:00:47,480 --> 00:00:52,410 an or gate on the lower right, and a transition on the lower left. 11 00:00:55,730 --> 00:00:59,990 Propositions are typically partitioned into three classes. 12 00:00:59,990 --> 00:01:04,330 Input propositions, those with, are those with no inputs. 13 00:01:04,330 --> 00:01:08,370 Base propositions are those with incoming arcs from transitions. 14 00:01:09,510 --> 00:01:11,670 And view propositions are those with 15 00:01:11,670 --> 00:01:14,820 incoming arcs from connectives other than transitions. 16 00:01:15,970 --> 00:01:20,860 In our example here, nodes a and b are input propositions. 17 00:01:20,860 --> 00:01:26,840 Note s is a base proposition, and notes p, q, and r are view propositions. 18 00:01:30,460 --> 00:01:33,110 An input marking, is a function from the 19 00:01:33,110 --> 00:01:36,809 input propositions of a propositional net to boolean values. 20 00:01:38,590 --> 00:01:40,130 a base marking is a function from the 21 00:01:40,130 --> 00:01:43,140 base propositions of a propositional net boolean values. 22 00:01:43,140 --> 00:01:45,840 And a view marking is a function from the 23 00:01:45,840 --> 00:01:49,710 view propositions of a propositional net to boolean values. 24 00:01:51,050 --> 00:01:52,670 And we use the word marking to refer to 25 00:01:52,670 --> 00:01:55,399 a combination of an input base and view marking. 26 00:01:59,090 --> 00:02:01,720 Now given a prop net, an input marking, and a 27 00:02:01,720 --> 00:02:06,800 base marking, determine a unique view marking for that prop net. 28 00:02:06,800 --> 00:02:08,150 This is based on the types of 29 00:02:08,150 --> 00:02:10,800 the connectives feeding into the view propositions. 30 00:02:11,910 --> 00:02:17,240 The output of an inverter is true, if and only if its input is false, for example. 31 00:02:17,240 --> 00:02:21,680 The output of an and gate is true if and only if all of its inputs are true. 32 00:02:21,680 --> 00:02:23,500 That's suggested by the second table. 33 00:02:24,660 --> 00:02:26,800 And the output of an or gate is true if 34 00:02:26,800 --> 00:02:29,370 and only if at least one of its inputs is true. 35 00:02:31,560 --> 00:02:35,120 Transitions behave just like and gates, except, 36 00:02:35,120 --> 00:02:37,220 that there's a one step time delay. 37 00:02:37,220 --> 00:02:40,530 The output occurs one step after the inputs. 38 00:02:45,140 --> 00:02:46,200 Here's an example. 39 00:02:46,200 --> 00:02:51,150 Suppose that we had an input marking that assigned the node 40 00:02:51,150 --> 00:02:54,099 A, the value one, and the node B, the value zero. 41 00:02:56,220 --> 00:02:59,560 That's the node in the upper left and the down node in the lower right. 42 00:02:59,560 --> 00:03:03,440 And suppose we had a base marking that assigned s, the value one. 43 00:03:03,440 --> 00:03:05,030 That's the one in the lower left. 44 00:03:08,070 --> 00:03:11,720 Then, the output of the and gate would be a one. 45 00:03:11,720 --> 00:03:14,170 The output of the inverter would be a zero. 46 00:03:14,170 --> 00:03:17,820 The output of the or gate would also be a zero. 47 00:03:19,200 --> 00:03:20,630 And at this point, we then have values for 48 00:03:20,630 --> 00:03:25,242 all of the view propositions in the prop net. 49 00:03:25,242 --> 00:03:28,140 Now let's move on to the next step. 50 00:03:28,140 --> 00:03:34,150 Supposing once again the input is, is, has A being one and B being zero. 51 00:03:34,150 --> 00:03:36,970 What's the value of our base proposition on this step? 52 00:03:36,970 --> 00:03:40,290 Since it's the output of a transition, its value on this step is 53 00:03:40,290 --> 00:03:44,310 the same as the value of that transitions input on the preceding step. 54 00:03:45,630 --> 00:03:49,360 In this case the transition input was zero on that preceding step as 55 00:03:49,360 --> 00:03:52,669 we just saw, and so the value is zero on this new step. 56 00:03:54,760 --> 00:03:57,030 As before, we can compute the view marking corresponding 57 00:03:57,030 --> 00:04:00,630 to this new input marking and this new base marking. 58 00:04:00,630 --> 00:04:04,950 In this case, since the second input to the and gate is now 0, the output is 0. 59 00:04:04,950 --> 00:04:11,090 The output of the inverter becomes 1, and the output of the or gate is 1 as well. 60 00:04:12,770 --> 00:04:15,000 Now, if we leave the inputs the same, for subsequent 61 00:04:15,000 --> 00:04:17,840 steps, the prop net will go on, alternating this way. 62 00:04:17,840 --> 00:04:19,890 If input A ever becomes false, it will stop 63 00:04:19,890 --> 00:04:24,440 alternating, however the alternation will begin again, as soon as it's set to true.