State Based Control of Timed Discrete Event Systems using Binary ...

Chapter 4. Synthesis Algorithm Based on Predicates 29, P bad2 (i) for their values in cycle i.Now we want to show that:If(∃s ∈ K ′ )( (δ p (q p,0 , s), δ s (q s,0 , s)) P good2 (n) )then(∃s ′ ∈ K ′ )( (δ p (q p,0 , s ′ ), δ s (q s,0 , s ′ )) P good1 (n) )In order to show the above statement we get (ˆq p , ˆq s ) = (δ p (q p,0 , s), δ s (q s,0 , s)). If(ˆq p , ˆq p ) P good1 (n), then we get s ′ = s. Otherwise (i.e. (ˆq p , ˆq p ) P good1 (n) ) wehave (ˆq p , ˆq s ) P re (n) or (ˆq p , ˆq s ) P cr (n). Suppose we have (ˆq p , ˆq s ) P re (n). Theneither(ˆq p , ˆq s ) (P bad1 (n) ∪ P bad2 (n))or(∃w ≤ s)( (δ p (q p,0 , w), δ s (q s,0 , w)) (P bad1 (n) ∪ P bad2 (n)) )If the former is true, let (˜q p , ˜q s ) = (ˆq p , ˆq s ) and s 1 = s; otherwise let (˜q p , ˜q s ) =(δ p (q p,0 , w), δ s (q s,0 , w)) and s 1 = w ( w ∈ K ′ ).Similar reasoning can also be applied if (ˆq p , ˆq s ) P cr (n) Now we have(∃s 1 ∈ K ′ )( (˜q p , ˜q s ) = (δ p (q p,0 , s 1 ), δ s (q s,0 , s 1 )), (˜q p , ˜q s ) P bad1 (n) ∪ P bad2 (n) )If (˜q p , ˜q s ) P bad1 (n), we must have (˜q p , ˜q s ) P bad2 (n) thus(∃s 2 ∈ Σ ∗ )( (δ p (˜q p , s 2 ), δ s (˜q s , s 2 )) P bad1 (n) )In this case let (q ′ p, q ′ s) = (δ p (˜q p , s 2 ), δ s (˜q s , s 2 )) and s 3 = s 1 s 2 . Otherwise, i.e.(˜q p , ˜q s ) P bad1 (n), let (q ′ p, q ′ s) = (˜q p , ˜q s ) and s 3 = s 1 . Now we have(∃s 3 ∈ K ′ )((q ′ p, q ′ s) = (δ p (q p,0 , s 3 ), δ s (q s,0 , s 3 )), (q ′ p, q ′ s) P bad1 (n) )Based on the definition of P bad1 , there are 5 possible cases: