10. Let W be a wff. If no free occurrence of the variable x in W(x) is in the scope of a quantifier that binds a variable in t, then we say the term t is... (A) out of bounds (C) bound to x (B) free to replace r in W(x) (D) a solitary term
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- 6. Let M = (Q,Sigma,s, q0, F) be a dfa and define cfg g= (v,sigma, R,S) as follows: 1. V=Q; 2. For each q in Q and a in sigma, define rule q->aq' where q' = s(q,a); 3. S = q0 Prove L(M) = L(G)Consider a system with input x(t) and output y(t) . The relationship between input and output is y(t) = x(t)x(t − 2) a. Is the system causal or non-causal?b. Determine the output of system when input is Aδ(t) , where A is any real orcomplex number?Simplify the following Boolean functions by means of K-map:a) f(A,B,C) =∑m(0,2,3,4,6)b) f(A,B,C,D) = ∑ m(0,2,3,7,8,10,12,13)c) F(X,Y,Z) = Y’Z+YZ+X’Y’Z’
- In each case below, show using the pumping lemma that the givenlanguage is not a CFL.e. L = {x ∈ {a, b, c}∗ | na(x) = max {nb(x), nc(x)}}For each of the following Boolean properties of theorems, state the set theory interpretation: E) x * x = xF) x + x = x G) x * 0 = 0For the transition probability matrix P = 1 2 3 4 5 1 0.2 0.1 0.15 0 0.55 2 0 1 0 0 0 3 0.35 0.2 0.2 0.1 0.15 4 0 0 0 1 0 5 0.25 0.2 0.15 0.25 0.15 (a) Rewrite P in the canonical form, clearly identifying R and Q. (b) For each state, i, calculate the mean number of times that the process is in a transient state j, given it started in i. (c) For each state i, find the mean number of transitions before the process hits an absorbing state, given that the process starts in a transient state i. (d) For each state i, find the probability of ending in each of the absorbing states.
- Create the K-maps and then simplify for the following functions (leave in sum-of-products form): F(w,x,y,z) = w′x′y′z + w′x′yz′+ w′xy′z + w′xyz + w′xyz′ + wxy′z + wxyz + wx′y′z F(w,x,y,z) = w′x′y′z′ + w′z + w′x′yz′ + w′xy′z′+ wx′y5. Let R = {1, 3, π, 4, 1, 9, 10}, S = {{1}, 3, 9, 10}, T = {1, 3, π}, and U = {{1, 3, π}, 1}. Which of thefollowing are true? For those that are not, why not? (d) 1 ⊆ U(e) {1} ⊆ T(f) {1} ⊆ S(g) {1} ∈ SShow that ( X→ Y ) ∨ ( Y →X) is a tautology.
- 1. Consider an impulse response h[n] such that h[n] = 0 for n < 0 and n > M, and h[n] =−h[M − n] for 0 ≤ n ≤ M where M is an odd integer.a) Express the Fourier transform of h[n] in the formH(ejω) = ejf(ω)A(ω) ,where f(ω) and A(ω) are real-valued functions of ω. Determine f(ω) and A(ω).b) Provide an example of such an impulse response h[n] for M = 7 and find the corresponding f(ω) and A(ω).) Show that ∀xP(x) ∧ ∃xQ(x) is logically equivalent to ∀x∃y(P(x) ∧ P(y)) The quantifiers have the same non empty domain. I know that to prove a proposition is logically equivalent to another one, I have to show that ∀xP(x) ∧ ∃xQ(x) ↔ ∀x∃y(P(x) ∧ P(y)) Which means I have to prove that (∀xP(x) ∧ ∃xQ(x)) → ∀x∃y(P(x) ∧ P(y)) ∧ ∀x∃y(P(x) ∧ P(y)) → (∀xP(x) ∧ ∃xQ(x)) I don't know the answer, so I saw the textbook answer. It says (1) Suppose that ∀xP(x) ∧ ∃xQ(x) is true. Then P(x) is true for all x and there is an element y for which Q(y) is true. I get this part. Because P(x) ∧ Q(x) is true for all x and there is a y for which Q(y) is true, ∀x∃y(P(x) ∧ P(y)) is true. Emm... I think ∀x∃y(P(x) ∧ P(y)) is true because ∀x only affects P(x) and ∃y only affects P(y) since their alphabets are different. So, it has the exact same meaning as ∀xP(x) ∧ ∃yQ(y). And since the domains are the same, ∀xP(x) ∧ ∃yQ(y) is actually equal to ∀xP(x) ∧ ∃xQ(x). But the textbook states that "P(x) ∧ Q(x) is…in the solution it says "Now, consider the model that satisfies all axioms in {A1, A2, ..., Ak} (since it's finite, it can be satisfied by a model), but also satisfies Fm because Fm implies Fm+1" but it is given that Fm doesnot imply fm+1