Problem 1. Use the Boolean algebra derivation (instead of truth table) to prove or answer the questions. pq-(p→q)^ (q→p) pvq-(p→q) →q (p+q)' -p'+q Is ((p = q) @ 1) = ((p =(q)) a tautology, why? Provide the proof or disproof. Is (pq) (p →q) a tautology, why? Provide the proof or disproof.

Use rules of inference to show that if Vx(P(x) V Q(x)) and Vx((-P(x) A Q(x)) → R(x)) are true, then Vx(-R(x)– P(x)) is also true, where the domains of all quantifiers are the same.

Please help me with this. I am having trouble understanding what

2.Use the method of proof by contradiction to show that [(pvq)^(p→r)^(q→r)]→r is a tautology.

For the function f(w,x,y,z) = E (m4, m7, m8, m9, m10, m12, mi13, m15), complete the parts of this question below. (a) Draw and fill in a correct, properly-labelled Karnaugh map for f(w,x,y,z). (b) List ALL sum-of-prod. prime implicants, and identify which ones are essential. (c) Write the complete least-cost sum-of-products expression for f(w,x,y,z). (d) Now consider m2, m5, and m11 as don't-care cases for f(w,x,y,z) in part (a). Provide the least-cost sum-of-products expression using this information.

Mathematical logic Is (((P→Q→P)→P) a tautology? Please be as clear as possible. Explain your answer. Thank you very much.

Please explain, as simplistically (but thoroughly) as possible, how to develop a k-map from a truth table to be used to simplify an equation.Examples: xz + (xy + ~z) ~xyz + yz + x~y

1. Prove or Disprove (by the use of truth tables)  p→(¬q↔r)≡¬p∨[(q→r)∧(¬r→q)]

Show that the compound statement is a tautology.((p ∨ q) ∧ ￢p) → q

a) Show that: (Vx)(A → (B = C)) → ((\x)(A → B) = (Vx)(A → → C)) b) Show that: (\x)((A v B) → C) → (Vx) (A → C) c) [Show that: (\x)(A → (B ^ C)) → (\x)(A → B) d) Prove (AV (Vx) BC = (\x)B(AV C), as long as x not free in A. (Hint: Translate first to Standard notation)