Solve this proof with no premises: (~Q->~P)->((~Q->P)->Q)

Well-formed-formula (wff) Prove (p a q) and (T→ (¬ (p → q))) are both wff using top-down parsing:

Prove using truth table

Simplify (~Q) --> (~P) using (~P)||(Q) via a truth table where II represents the "or" operator.

Pick the option that is correct for A, B, C, D (A) Element defintion or Hypothesis  (B) Exsistential generaliztion 1, 2 or Exsistential instantaion 1, 2 or Universal generaliztion 1, 2 or  Universal  instantaion 1, 2 (C) Q(b) ^ R(b) or  Q(b) ∨ R(b) or Q(x) ^ R(x)  or  Q(x) ∨ R(x) (D) ∀Q(x) or ∀¬Q(x) or ∃Q(x) or ∃¬Q(x)

Subject: algebra

In recurrence relation T(n) = T(n/3)+Θ(n2), the number of sub-problems are   a 2   b n/2   c 1   d 3

Let ap = 1, an = an-1 + 3 be the recurrence relation of algorithm A and T(1) = 1, T(n) = T(n/5)+n be the recurrence relation of algorithm B. 1. Find the solution for both relations. Use any method.

Prove ((A --> B) C) is a proposition.

Develop a closed form solution for the recurrence relation: P(n) = 5 if n = 0, P(n) = P(n-1)+3 if n > 0 Prove the closed form solution is valid for all n