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

Chapter 4. Synthesis Algorithm Based on Predicates 252. (q p , q s ) F,let σ ∈ Σ u . There are two possible cases:• δ p (q p , σ)!, then δ s (q s , σ)! and (δ p (q p , σ), δ s (q s , σ)) P . Therefore σ ∈ Elig G (s)∩(Σ u ∪ {tick}) and also σ ∈ Elig K (s).• δ p (q p , σ) is undefined. Then σ /∈ Elig G (s) ∩ (Σ u ∪ {tick}).(Only if). Given K is controllable w.r.t G, suppose the contrary of the conclusion ,i.e.(∃(q p , q s ) P, σ ∈ Σ u )(δ p (q p , σ)! ∧ ((¬δ s (q s , σ)!) ∨ (δ p (q p , σ), δ s (q s , σ)) P ))It is obvious that now we have s ∈ K,sσ ∈ L(G), but sσ /∈ K, therefore K is notcontrollable w.r.t L(G). This contradicts the assumption.For the second condition, the proof is exactly the same.□Lemma 4.1.2 For the given G p and G s , P is nonblocking iff(∀(q p , q s ) P )((q p , q s ) is reachable) ⇒ ((q p , q s ) is coreachable)Proof: (If). We have to show that L m (G meet , P ) = L(G meet , P ).First we show L m (G meet , P ) ⊆ L(G meet , P ). given a string s ∈ L m (G meet , P ), there existss ′ ∈ L m (G meet , P ) such that (∃u ∈ Σ ∗ ) s ′ = su. Thus (δ p (q p,0 , s), δ s (q s,0 , s)) P andthen s ∈ L m (G meet , P ) because we always have s ∈ L(G meet ).Now we show L m (G meet , P ) ⊇ L(G meet , P ). Given a string s ∈ L(G meet , P ), we needto show there exists a string w, such that δ p (q p,0 , sw) ∈ Q p,m , δ s (q s,0 , sw) ∈ Q s,m . It isenough to show (∃w ∈ Σ ∗ )(δ p (δ p (q p,0 , s), w) ∈ Q p,m , δ s (δ s (q s,0 , s), w) ∈ Q s,m ) . This isobviously true because (δ p (q p,0 , s), δ s (q s,0 , s)) is reachable, and thus coreachable.(Only If). Suppose the contrary, i.e. (∃(q p , q s ) ∈ Q p × Q s )), (q p , q s ) is reachable but not